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Stresses in Beams (Basic Topics)
Longitudinal Strains in Beams
Problem 5.4-1 Determine the maximum normal strain max produced in a steel wire of diameter d 1/16 in. when it is bent around a cylindrical drum of radius R 24 in. (see figure).
R d
Solution 5.4-1
Steel wire
R 24 in. d 1 in. 16 Substitute numerical values: max 1/16 in. 2(24 in.) + 1/16 in. 1300 * 10
6
From Eq. (5-4): y max r d/2 R + d/2 d 2R + d
;
Problem 5.4-2 A copper wire having diameter d 3 mm is bent into a circle and held with the ends just touching (see figure). If the maximum permissible strain in the copper is max 0.0024, what is the shortest length L of wire that can be used?
d = diameter L = length
Solution 5.4-2
Copper wire d 3 mm max
L 2pr r L 2p
0.0024
From Eq. (5-4): max Lmin y r pd max d/2 L/2p pd L 3.93 m ;
p(3 mm) 0.0024
389
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Problem 5.4-3 A 4.5 in. outside diameter polyethylene pipe designed to carry chemical wastes is placed in a trench and bent around a quarter-circular 90 bend (see figure). The bent section of the pipe is 46 ft long. Determine the maximum compressive strain max in the pipe.
90
Solution 5.4-3
Polyethylene pipe
Angle equals 90 or p/2 radians, L L d L length of 90 bend 46 ft 4.5 in. 2pr 4 pr 2 max 552 in. r r r L p/2 pd 4L radius of curvature 2L p max y r d/2 2L/p 6400 * 10
6
p 4.5 in. a b 4 552 in.
;
Problem 5.4-4 A cantilever beam AB is loaded by a couple M0 at its free end (see figure.) The length of the beam is L 1.5 m and the longitudinal normal strain at the top surface is 0.001. The distance from the top surface of the beam to the neutral surface is 75 mm. Calculate the radius of curvature r, the curvature k, and the vertical deflection d at the end of the beam.
A B L
d M0
Solution 5.4-4 NUMERICAL DATA max L 2.0 m
c 82.5 mm
0.0012
Deflection: constant curvature for pure bending so gives a circular arc; assume flat deflection curve (small defl.) so BC L sin(u) L r u L asina b r L r
4
RADIUS OF CURVATURE c r r 68.8 m max CURVATURE k 1 r k 1.455 * 10
; u
5
0.029 radians cos(u) r (1
0.029 r 29.1 mm 6.875 * 104 mm ;
m
1
;
1 d
4.232 * 10 cos(u)) d
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Problem 5.4-5 A thin strip of steel of length L 20 in. and thickness t 0.2 in. is bent by couples M0 (see figure). The deflection at the midpoint of the strip (measured from a line joining its end points) is found to be 0.20 in. Determine the longitudinal normal strain at the top surface of the strip.
M0
M0
t L 2
d L 2
Solution 5.4-5 NUMERICAL DATA
L d 28 inches 0.20 inches t 0.25 inches
solving for r:
r 1
d cosa r 1 L b 2r 0.20 cos a 14 b r
LONGITUDINAL NORMAL STRAIN AT TOP SURFACE t 2 t r 2r L 2 r L 2r
insert numerical data:
numerical solution for radius of curvature r gives r 489.719 inches strain at top (compressive): t 2.552 * 10 2r 255 110
6
d
r (1
cos(u))
sin(u)
sin(u)
4
assume angle is small so that u L 2r d r a1 cosa L bb 2r
2
;
Problem 5.4-6 A bar of rectangular cross section is loaded and supported as shown in the figure. The distance between supports is L 1.5 m and the height of the bar is h 120 mm. The deflection at the midpoint is measured as 3.0 mm. What is the maximum normal strain at the top and bottom of the bar?
h P
d P
a
L 2
L 2
a
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Solution 5.4-6 NUMERICAL DATA
L d 1.5 m 3.0 mm h 120 mm
d
r a1
cosa cosa
L bb 2r d 0
r a1
L bb 2r
NORMAL STRAIN AT TOP OF BAR: h 2 r h 2r
numerical solution for radius of curvature r gives r 93.749 m radius of strain at top (compressive): h 640 * 10 2r
6
tensile strain, r curvature u
;
SMALL DEFLECTION SO SMALL ANGLE
sin(u)
L 2 r
u
L 2r
Normal Stresses in Beams
Problem 5.5-1 A thin strip of hard copper (E 16,000 ksi) having length L 90 in. and thickness t 3/32 in. is bent into a circle and held with the ends just touching (see figure).
(a) Calculate the maximum bending stress smax in the strip. (b) By what percent does the stress increase or decrease if the thickness of the strip is increased by 1/32 in.?
3 t = in. 32
Solution 5.5-1 (a) MAXIMUM BENDING STREES
E 16000 ksi t 2 EP Q r L 90 inches t 3 inches 32
smaxnew
69.813 ksi 33.3 ;
smax smaxnew (100) smax
s r smax
r
L 2p smax ; Et 2r
14.324 inches 52.4 ksi
33% increase (linear) in max.stress due to increase in t; same as % increase in thickness t 3 4 32 32 (100) 33.3 3 32
(b) % CHANGE IN STRESS tnew 4 32 smaxnew Etnew 2r
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Problem 5.5-2 A steel wire (E 200 GPa) of diameter d is bent around a pulley of radius R0 500 mm (see figure).
1.25 mm
(a) What is the maximum stress smax in the wire? (b) By what percent does the stress increase or decrease if the radius of the pulley is increased by 25%?
R0 d
Solution 5.5-2 (a) MAX. NORMAL STRESS IN WIRE
E 200 GPa E s d 2 r d 1.25 mm E smax d 2 R0 500 mm
(b) % CHANGE IN MAX. STRESS DUE TO INCREASE IN PULLEY RADIUS BY 25% E snew d 2 d 2
snew
199.8 MPa
d R0 + 2 ;
1.25 R0 + smax snew (100) smax
smax
250 MPa
20%
;
Problem 5.5-3 A thin, high-strength steel rule (E 30 106 psi) having thickness t 0.175 in. and length L 48 in. is bent by couples M0 into a circular are subtending a central angle a 40 (see figure).
(a) What is the maximum bending stress smax in the rule? (b) By what percent does the stress increase or decrease if the central angle is increased by 10%?
M0
L = length t M0 a
Solution 5.5-3 (a) MAX. BENDING STRESS
a L r 40 a L a E p b 180 t a 0.698 radians
(b) % CHANGE IN STRESS DUE TO 10% INCREASE IN ANGLE a snew 0.175 in. E 30 (106) psi E t (1.1a) 2L snew 10% 41997 psi ;
48 inches r
68.755 inches Et 2r Eta 2L
smax snew (100) smax linear increase (%) smax
smax smax
t 2 smax r 38.2 ksi ;
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Problem 5.5-4 A simply supported wood beam AB with span length L 4 m carries a uniform load of intensity q 5.8 kN/m (see figure).
(a) Calculate the maximum bending stress smax due to the load q if the beam has a rectangular cross section with width b 140 mm and height h 240 mm. (b) Repeat (a) but use the trapezoidal distuibuted load shown in the figure part (b).
A
q B h
L (a) q 2 A q B
b
L (b)
Solution 5.5-4 (a) MAX. BENDING STRESS DUE TO UNIFORM LOAD q
Mmax
3
(b)
MAX. BENDING STRESS DUE TO TRAPEZOIDAL LOAD
q
qL 8
2
S
I h 2
RA
c
1q 1 q1 a bL+ a b Ld 22 3 22
uniform load (q/2) & triang. load (q/2) RA find x qL2 8 1 a bh2 b 6 RA 1 qL 3 location of zero shear q x 2 1184 L22 2(3) 4L2 0 1 184b 6 1 xq a bx 2 L2 0
S
bh 12 h 2
S
12 bh 6
smax
Mmax S 3 L2 q 4 bh2 kN 5.8 m 240 mm qL2 8 L
smax
3x2 + 6Lx x x L xmax a 6L
smax q h
4m
b
140 mm
1+
0.52753 L
Mmax Mmax smax
11.6 kN # m 8.63 MPa ;
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Mmax Mmax Mmax
RAxmax
q xmax2 22
2
2 1 xmax q xmax a b 2 L2 3
smax smax
Mmax S 6.493 * 103 N m2 smax 6.49 MPa ;
9.40376 * 10 8.727 kN # m
qL2
Problem 5.5-5 Each girder of the lift bridge (see figure) is 180 ft long and simply supported at the ends. The design load for each girder is a uniform load of intensity 1.6 k/ft. The girders are fabricated by welding tree steel plates so as to form an I-shaped cross section (see figure) having section modulus S 3600 in.3. What is the maximum bending stress smax in a girder due to the uniform load?
Solution 5.5-5
Bridge girder
L S 180 ft 3600 in. qL2 8 Mmax S qL2 8S 8(3600 in.3) 21.6 ksi ;
3
q
1.6 k/ft
Mmax smax smax
(1.6 k/ft)(180 ft)2(12 in./ft)
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Problem 5.5-6 A freight-car axle AB is loaded approximately as shown in the figure, with the forces P representing the car loads (transmitted to the axle through the axle boxes) and the forces R representing the rail loads (transmitted to the axle through the wheels). The diameter of the axle is d 80 mm, the distance between centers of the rails is L, and the distance between the forces P and is R is b 200 mm. Calculate the maximum bending stress smax in the axle if P 47 kN.
P A d R b L R
P B d
b
Solution 5.5-6 NUMERICAL DATA
d P I 82 mm 50 pd 64
4
MAX. BENDING STRESS 220 mm smax m4 smax Md 2I 203 MPa ;
b
kN I Pb 2.219 * 10 Mmax
6
Mmax
11 kN # m
Problem 5.5-7 A seesaw weighing 3 lb/ft of length is occupied by two children, each weighing 90 lb (see figure). The center of gravity of each child is 8 ft from the fulcrum. The board is 19 ft long, 8 in. wide, and 1.5 in. thick. What is the maximum bending stress in the board?
Solution 5.5-7
Seesaw
b q 8 in. 3 lb/ft Pd + h P qL 2
2
1.5 in. 90 lb d 8.0 ft L 9.5 ft
Mmax
720 lb-ft + 135.4 lb-ft 855.4 lb-ft 10,264 lb-in.
S smax
bh 6
2
3.0 in.3. M S 10,264 lb-in. 3.0 in.3 3420 psi ;
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Problem 5.5-8 During construction of a highway bridge, the main girders are cantilevered outward from one pier toward the next (see figure). Each girder has a cantilever length of 48 m and an I-shaped cross section with dimensions shown in the figure. The load on each girder (during construction) is assumed to be 9.5 kN/m, which includes the weight of the girder. Determine the maximum bending stress in a girder due to this load.
52 mm
2600 mm 28 mm
620 mm
Solution 5.5-8 NUMERICAL DATA tw tf 52 mm
h L I I 2600 mm 48 m 1 (b ) h3 12 f q bf
28 mm 620 mm kN 9.5 m
Mmax smax smax
L qL a b 2 Mmax h 2I 101 MPa
Mmax
1.094 * 104 kN m
;
1 (b 12 f
tw) [ h
2 (tf)]3
1.41 * 1011 mm4
Problem 5.5-9 The horizontal beam ABC of an oil-well pump has the cross section shown in the figure. If the vertical pumping force acting at end C is 9 k and if the distance from the line of action of that force to point B is 16 ft, what is the maximum bending stress in the beam due to the pumping force?
Horizontal beam transfers loads as part of oil well pump C B A 0.875 in.
0.625 in.
22 in.
8.0 in.
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Solution 5.5-9 NUMERICAL DATA
FC Mmax I 9k BC 16 ft Mmax 1 (8 12
3
MAX. BENDING STRESS AT B 144 k-ft 0.625) I 1.995 * 10 in.
3 4
F C (BC)
smax smax
Mmax (12) a I 9.53 ksi
22 b 2 ;
1 (8) (22)3 12 * [22
2 (0.875)]
Problem 5.5-10 A railroad tie (or sleeper) is subjected to two rail loads, each of magnitude P 175 kN, acting as shown in the figure. The reaction q of the ballast is assumed to be uniformly distributed over the length of the tie, which has cross-sectional dimensions b 300 mm and h 250 mm. Calculate the maximum bending stress smax in the tie due to the loads P, assuming the distance L 1500 mm and the overhang length a 500 mm.
P a L
P a b h q
Solution 5.5-10 Railroad tie (or sleeper) P 175 kN h b 300 mm DATA
L q 1500 mm 2P L + 2a S a bh2 6 500 mm 3.125 * 10
250 mm
Substitute numerical values: M1 17,500 N # m 21,875 N # m M2 21,875 N # m Mmax
3
m3
MAXIMUM BENDING STRESS BENDING-MOMENT DIAGRAM smax Mmax 5 21,875 N # m 3.125 * 10
3
m3
7.0 MPa
;
(Tension on top; compression on bottom) qa2 2 Pa2 L + 2a PL 2 PL 2
M1 M2
2 qL a + ab 22
2 L P a + ab L + 2a 2
P (2a 4
L)
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Problem 5.5-11 A fiberglass pipe is lifted by a sling, as shown in the figure. The outer diameter of the pipe is 6.0 in., its thickness is 0.25 in., and its weight density is 0.053 lb/in.3 The length of the pipe is L 36 ft and the distance between lifting points is s 11 ft. Determine the maximum bending stress in the pipe due to its own weight.
s L
Solution 5.5-11
Pipe lifted by a sling
L s g a
36 ft 11 ft
432 in. 132 in.
d2
6.0 in.
t
0.25 in.
I q
0.053 lb/in.3 (L s)/2
d1 d2 2t p2 A (d d2) 1 42
5.5 in. 4.5160 in.2
p4 (d 64 2 gA
d14)
18.699 in.4 0.23935 lb/in.
(0.053 lb/in.3)(4.5160 in.2)
150 in.
MAXIMUM BENDING STRESS smax smax Mmax c I c d2 2 3.0 in. 432 psi ;
BENDING-MOMENT DIAGRAM
(2,692.7 lb-in.)(3.0 in.) 18.699 in.4
(Tension on top) M1 M2 Mmax qa2 2 qL L a 42 2,692.7 lb-in. sb 2,171.4 lb-in.
2,692.7 lb-in.
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Problem 5.5-12 A small dam of height h 2.0 m is constructed of vertical wood beams AB of thickness t 120 mm, as shown in the figure. Consider the beams to be simply supported at the top and bottom. Determine the maximum bending stress smax in the beams, assuming that the weight density of water is g 9.81 kN.m3
h t
A
B
Solution 5.5-12
Vertical wood beam
MAXIMUM BENDING MOMENT
RA M h t g 2.0 m 120 mm 9.81kN/ m3(water) dM dx
q0 L 6 RAx q0 x 3 6L q0 Lx q0 x 3 6 6L q0L q0x 2 0x 6 2L q0 L L a b 6 13
L 13 9 13
Let b = width of beam perpendicular to the plane of the figure Let q0 = maximum intensity of distributed load q0 gbh S bt2 6
Substitute x Mmax
L/13 into the equation for M: q0 L3 a b 6L 313 q0 L2 q0 h 2 913
For the vertical wood beam: L Maximum bending stress smax Mmax S 2q0 h2 3 13 bt2 ;
h; Mmax 2gh3
313 t2
SUBSTITUTE NUMERICAL VALUES: smax 2.10 MPa
NOTE: For b 1.0 m, we obtain q0 19,620 N/m, S 0.0024 m3, Mmax 5,034.5 N # m, and smax Mmax/S 2.10 MPa
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Problem 5.5-13 Determine the maximum tensile stress st (due to pure bending about a horizontal axis through C by positive bending moments M) for beams having cross sections as follows (see figure).
(a) A semicircle of diameter d (b) An isosceles trapezoid with bases b1 and altitude h (c) A circular sector with p/3 and r
C xc
y x b1 C b2 (b) xc h y a O (c) C a r x xc x
b and b2 d/2
4b/3,
d (a)
Solution 5.5-13 MAX. TENSILE STRESS DUE TO POSITIVE BENDING MOMENT
IS ON BOTTOM OF BEAM CROSS-SECTION
Ix ybar d For a A ; c
r4 (a + sin (a) cos(a)) 4 2r sin (a) a b 3 a 1 p/3, r p b 12 d/2: A A a d2p ba b 2 3 c ybar
(a) SEMICIRCLE From Appendix D, Case 10: Ic c st (9p2 4r 3p Mc Ic 64)r4 72p 2d 3p 768M (9p2 64)d3 30.93 M d3 64)d4 (9p2 1152p
d2 a 2a
0.2618 d2
d b 2
(b) ISOSCELES TRAPEZOID From Appendix D, Case 8: IC h3(b2 + 4b1b2 + b2) 1 2 36(b1 + b2) 73bh3 756 c st h(2b1 + b2) 3(b1 + b2) Mc Ic 360M 73bh2 10h 21 ;
3 a d4 b 2 4 Ix cd 4
p sin a b 3 p 3 a
c
0.276 d
Ix Ix IC IC IC p/3, r d/2
p p p + sina b cosa b b 3 3 3
0.02313 d 4 A y2 bar (4p 313) 768
3
p d 13 2 d 2 a 12 b c a bd d 2p Mc IC M d3
3.234 * 10
d4 st 85.24 ;
(c) CIRCULAR SECTOR WITH a From Appendix D, Case 13: A r 2 (a)
max. tensile stress st
Problem 5.5-14 Determine the maximum bending stress smax (due to pure bending by a moment M) for a beam having a cross section in the form of a circular core (see figure). The circle has diameter d and the angle b 60. (Hint: Use the formulas given in Appendix D, Cases 9 and 15.)
C
b b
d
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Solution 5.5-14
Circular core From Appendix D, Cases 9 and 15:
Iy r pr 4 d 2
4
r aa 2 p 2 r sin b b b
4
ab r2 +
2ab r4
3
b
pd 4 64 d4 (4b 128
d4 p a 32 2
b+
1 sin 4b b 4
sin4b)
a
MAXIMUM BENDING STRESS r cos b smax smax For b smax Mc Iy d (4b 60
3
b Iy
radians pd 64
4
a dp a 32 2 dp a 32 2 d4 p a 32 2
4 4
radians a b b b
sin b cos b + 2 sin b cos3 b b (sin b cos b)(1 1 a sin 2b b( 2 2 cos b)b cos 2b)b
2
c
r sin b
d sin b 2 ;
64M sin b (8p 13 + 9)d sin 4b) p/3 rad: 576M
3
pd 64
4
pd 4 64
10.96
M d3
;
Problem 5.5-15 A simple beam AB of span length L 24 ft is subjected to two wheel loads acting at distance d = 5 ft apart (see figure). Each wheel transmits a load P = 3.0 k, and the carriage may occupy any position on the beam. Determine the maximum bending stress smax due to the wheel loads if the beam is an I-beam having section modulus S 16.2 in.3
P
d
P C
A
B
L
Solution 5.5-15
Wheel loads on a beam
Substitute x into the equation for M: Mmax L d P S 24 ft 5 ft 3k 16.2 in.
3
288 in. 60 in.
P aL 2L
d2 b 2
MAXIMUM BENDING STRESS smax Mmax S P aL 2LS 3k 2(288 in.)(16.2 in.3) 21.4 ksi ; d2 b 2 ;
MAXIMUM BENDING MOMENT RA M dM dx P L L RA x P (2L L x+ P (L L dx 4x) 0 x d) 2x2) x L 2 d 4 P (2L L d 2x)
Substitute numerical values: smax (288 in. 30 in.)2
P (2L x L d
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Problem 5.5-16 Determine the maximum tensile stress st and maximum compressive stress sc due to the load P acting on the simple beam AB (see figure). Data are as follows: P 6.2 kN, L 3.2 m, d 1.25 m, b 80 mm, t 25 mm, h 120 mm, and h1 90 mm.
t P A d B h h1
L
b
Solution 5.5-16 NUMERICAL DATA
P d t h1 6.2 kN 1.25 m 25 mm 90 mm L b h 3.2 m 80 mm 120 mm
MAX. MOMENT & NORMAL STRESSES Mmax Pd (L L d) Mmax 4.7 kN # m c1) ;
MAX. COMPRESSIVE STRESS AT TOP (c sc Mmax c1 I sc 61.0 MPa
Beam cross section properties: centroid and moment of inertia Af b (h Aw c1 c2 I h h1) Aw th1 (h 2 h1) d h1 + Af c h 2 Af + Aw c1 c2
MAX. TENSILE STRESS AT BOTTOM (c st c1 76 mm Mmax c2 I st 35.4 MPa
c2) ;
44 mm dist. to C from bottom
1 1 t h3 + b (h h1)3 1 12 12 (h h 1) 2 + Af cc2 d + Aw ac1 2 5879395.2 mm4
h1 2 b 2
I
250 lb
Problem 5.5-17 A cantilever beam AB, loaded by a uniform load and a concentrated load (see figure), is constructed of a channel section. Find the maximum tensile stress st and maximum compressive stress sc if the cross section has the dimensions indicated and the moment of inertia about the z axis (the neutal axis) is I 3.36 in.4 (Note: The uniform load represents the weight of the beam.)
22.5 lb/ft B 5.0 ft y z C 3.0 ft
A
0.617 in. 2.269 in.
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Solution 5.5-17 NUMERICAL DATA
I c2 3.36 in.
4
MAXIMUM STRESSES c1 0.617 in. st sc MAmax c1 I MAmax c2 I st sc 4341 psi 15964 psi ; ;
2.269 in. 22.5 (8)2 + 250 (5) ft-lb 2 1970 ft-lb 23640 in.-lb
MAmax MAmax
MAmax (12)
Problem 5.5-18 A cantilever beam AB of isosceles trapezoidal cross section has length L 0.8 m, dimensions b1 80 mm, b2 90 mm, and height h 110 mm (see figure). The beam is made of brass weighing 85 kN/m3.
(a) Determine the maximum tensile stress st and maximum compressive stress sc due to the beams own weight. (b) If the width b1 is doubled, what happens to the stresses? (c) If the height h is doubled, what happens to the stresses?
q
b1 C h
L b2
Solution 5.5-18 NUMERICAL DATA
L b1 h 0.8 m 80 mm 110 mm g b2 85 kN m
3
MAX. TENSILE STRESS AT SUPPORT (TOP) st Mmax (h I ybar) st 1.514 MPa ;
90 mm
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM) sc Mmax ybar I sc 1.456 MPa ;
(a) MAX. STRESSES DUE TO BEAMS OWN WEIGHT Mmax A q q L2 2 q gA A 1 (b + b 2) h 21
(b) DOUBLE b1& RECOMPUTE STRESSES b1 A q 160 mm 1 (b + b2) h 21 gA q A 1.375 * 104 mm2 N m
9.35 * 103 mm2 7.9475 * 102 N m
Mmax ybar I I h3
254.32 N # m h (2b1 b2) 3 (b1 b2) 4 b1 b2 b2)
4 6
1.169 * 103
ybar b2 2 2
53.922 mm
1 b2 1
Mmax Mmax ybar
qL2 2 374 N # m h (2 b1 + b2) 3 (b1 + b2) ybar 60.133 mm
36 (b1
9.417 * 10 mm
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I I
h3
1b2 + 4 b1 b2 + b22 1 2 36 (b1 + b2)
Mmax ybar I h3
qL2 2
Mmax
508.64 N # m ybar 107.843 mm
1.35 * 107 mm4
h (2b1 + b2) 3 (b1 + b2) 36 1b1 + b22
MAX. TENSILE STRESS AT SUPPORT (TOP) st Mmax (h I ybar) st 1.381 MPa ; I
(b12 + 4b1 b2 + b22)
7.534 * 107 mm4
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM) sc Mmax ybar 2 MAX. TENSILE STRESS AT SUPPORT (TOP) sc 1.666 MPa ; st Mmax (h I ybar) st 0.757 MPa ;
(c) DOUBLE h & RECOMPUTE STRESSES b1 A q 80 mm h 1 (b + b2) h 21 gA q 220 mm A 1.87 * 104 mm2 N m
200 lb/ft A B 12 ft C
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM) sc Mmax ybar I sc 0.728 MPa ;
1.589 * 103
Problem 5.5-19 A beam ABC with an overhang from B to C supports a uniform load of 200 lb/ft throughout its length (see figure). The beam is a channel section with dimensions as shown in the figure. The moment of inertia about the z axis (the neutral axis) equals 8.13 in.4 Calculate the maximum tensile stress st and maximum compressive stress sc due to the uniform load.
6 ft
y z C
0.787 in. 2.613 in.
Solution 5.5-19 NUMERICAL DATA
q c1 200 lb I 8.13 in.4 ft 0.787 in. c2 2.613 in.
LOCATON OF ZERO SHEAR IN SPAN AB & MAX. (+) MOMENT IN SPAN AB xmax MmaxAB MmaxAB RB 2700 lb RA 900 lb RA q xmax RA xmax 2025 ft-lb (6)2 2 q 4.5 ft xmax2 2
COMPUTE SUPPORT REACTIONS q a MA a Fv 0 0 RB RA (18)2 2 12 RB
max. ( ) moment at B MB q MB 3600 ft-lb
q (18)
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MAX. STRESSES IN SPAN AB sC st st MmaxAB (12) c1 I MmaxAB (12) c2 I 7810 psi ; sc 2352 psi
MAX. STRESSES IN SPAN BC sc sc MB (12) c2 I 13885 psi MB (12) c1 I
; st
max. compressive stress 4182 psi
max. tensile stress
st
A
Problem 5.5-20 A frame ABC travels horizontally with an acceleration a0 (see figure). Obtain a formula for the maximum stress smax in the vertical arm AB, which had length L, thickness t , and mass density r.
t L B a0 = acceleration C
Solution 5.5-20
Accelerating frame
TYPICAL UNITS FOR USE
IN THE PRECEDING EQUATION
L length of vertical arm t thickness of vertical arm r mass density a0 acceleration Let b width of arm perpendicular to the plane of the figure Let q inertia force per unit distance along vertical arm VERTICAL ARM
SI units: r L a0 t smax m/s2
kg/m3
N # s2/m4
meters (m) meters (m) N/m2 (pascals) slug/ft3 ft/s
2 2
q S
rbta0 bt2 6
Mmax smax
qL2 2 Mmax S
rbta0L2 2 3rL2a0 t ;
USCS units: r L smax ft a0
lb-s2/ft4 ft
t
lb/ft (Divide by 144 to obtain psi)
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Problem 5.5-21 A beam of T-section is supported and loaded as shown in the figure. The cross section has width b 2 1/2 in., height h 3 in., and thickness t 3/8 in. Determine the maximum tensile and compressive stresses in the beam.
P = 700 lb L1 = 3 ft q = 100 lb/ft
3
t = in. 8 t = in. 8
3
h= 3 in.
L2 = 8 ft
L3 = 5 ft
b = 2 in. 2
1
Solution 5.5-21
NUMERICAL DATA L1 P t 3 ft 700 lb 3 in. 8 h L2 q 8 ft 100 3 in. lb ft b 2.5 in. L3 5 ft Moment diagram (843.75 ft-lb at load P, at right support)
8.438E+02
a Fv
0
Rlf
P + qL3
Rrt
R lf
281 lb
1250 ft-lb
Find centroid of cross section (c2 from bottom, c1 from top) t) Af t b Aw t (h Af c2 c1 h t h t + Aw at + b 2 2 Af + Aw c2 c1 Aw a h 2 3 2 in. t b + Af ah equals h t b 2
1.250E+03
c2
1 in. MP Mrt 843.75 ft-lb 1250 ft-lb
MAX. STRESSES IN BEAM at load P sc MP (12) c1 sc 12494 psi I (max. compressive stress) MP (12) c2 I st 5842 psi ;
check c1 c1 2
Af + Aw c1 + c2
MOMENT OF INERTIA I 1 t (h 12 + Aw cc1
SUPPORT
t)3 + (h
1 b t 3 + Af ac2 12 t) 2 d
2
t2 b 2
st
I
2 in.4
at right support Mrt (12) c2 sc I Mrt (12) c1 st st I (max. tensile stress) sc 8654 psi 18509 psi ;
FIND SUPPORT REACTIONS-SUM MOMENTS ABOUT LEFT PL1 + qL3 aL2 + L2 L3 b 2
a Mlf Rrt
0
Rrt
919 lb
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Problem 5.5-22 A cantilever beam AB with a rectangular cross section has a longitudinal hole drilled throughout its length (see figure). The beam supports a load P 600 N. The cross section is 25 mm wide and 50 mm high, and the hole has a diameter of 10 mm. Find the bending stresses at the top of the beam, at the top of the hole, and at the bottom of the beam.
10 mm A 50 mm B 12.5 mm
P = 600 N L = 0.4 m 25 mm
37.5 mm
Solution 5.5-22
Rectangular beam with a hole
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS (THE z AXIS) All dimensions in millimeters. Rectangle: Iz Ic + Ad2 1 (25)(50)3 + (25)(50)(25 12 240 N # m 260,420 + 878 Hole: Iz 1250 mm2 Ic + Ad2 p (10)4 + (78.54)(37.5 64 490.87 + 13,972 14,460 mm4 14,460 246,800 mm4 24.162)2 24.162)2
MAXIMUM BENDING MOMENT M PL (600 N)(0.4 m)
261,300 mm4
PROPERTIES OF THE CROSS SECTION A1 A2 area of rectangle (25 mm)(50 mm) area of hole p (10 mm)2 4 A A1 Aiyi y A2 78.54 mm
2
Cross-section: I 261,300
area of cross section 1171.5 mm B as reference axis: A2(37.5 mm) 28,305 mm3
STRESS AT THE TOP OF THE BEAM (240 N # m)(25.838 mm) Mc1 s1 I 246,800 mm4 25.1 MPa (tension) ;
Using line B
A1(25 mm)
a Aiyi A y
28,305 mm3 1171.5 mm2
24.162 mm
Distances to the centroid C: c2 c1 24.162 mm c2 25.838 mm 50 mm
STRESS AT THE TOP OF THE HOLE My s2 y c1 7.5 mm 18.338 mm I (240 N # m)(18.338 mm) 17.8 MPa s2 246,800 mm4 (tension) STRESS AT THE BOTTOM OF THE BEAM s3 (240 N # m)(24.162 mm) Mc2 I 246,800 mm4 23.5 MPa ; (compression)
;
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Problem 5.5-23 A small dam of height h 6 ft is constructed of vertical wood beams AB, as shown in the figure. The wood beams, which have thickness t 2.5 in., are simply supported by horizontal steel beams at A and B. Construct a graph showing the maximum bending stress smax in the wood beams versus the depth d of the water above the lower support at B. Plot the stress smax (psi) as the ordinate and the depth d (ft) as the abscissa. (Note: The weight density g of water equals 62.4 lb/ft3.)
Steel beam A Wood beam t h d B Steel beam t Wood beam
Side view
Top view
Solution 5.5-23
Vertical wood beam in a dam h 6 ft
t g 2.5 in. 62.4 lb/ft3
MAXIMUM BENDING STRESS 12 bt Section modulus: S 6 smax q0 smax Mmax S g bd gd3 t2 a1 6 q0d2 c a1 bt2 6 d 2d d + bd L 3LA 3L
Let b width of beam (perpendicular to the figure) Let q0 intensity of load at depth d q0 gbd ANALYSIS OF BEAM L RA RB x0 Mc h 6 ft q0d2 6L q0d a3 6 d d A 3L
d 2d d + b L 3LA 3L
;
SUBSTITUTE NUMERICAL VALUES: d depth of water (ft) (Max. d h 6 ft) L h 6 ft g 62.4 lb/ ft3 t 2.5 in. smax psi smax (62.4)d3 (2.5)2 a1 d dd + b 6 9A 18 9d + d12d )
smax(psi) 0 9 59 171 347 573 830
d b L
0.1849d3(54 q0d2 a1 6 d b L
d(ft) 0 1 2 3 4 5 6
;
RA(L
d)
Mmax
q 0 d2 a1 6
d 2d d + b L 3LA 3L
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Problem 5.5-24 Consider the nonprismatic cantilever beam of circular cross section shown. The beam has an internal cylindrical hole in segment 1; the bar is solid (radius r) in segment 2. The beam is loaded by a downward triangular load with maximum intensity q0 as shown. Find expressions for maximum tensile and compressive flexural stresses at joint 1.
y
q0
Linea r q(x )
P = q0L/2 x
M1 1 R1 2L 3 Segment 1 0.5 EI 2 3 L 3 Segment 2 EI
Solution 5.5-24 STATICS
a Fv R1 g M1 M1 M1 0 1 6 q0 L R1 1 2L q0 a b 2 3 q0 L 2
MAX. STRESSES AT JOINT 1 MAX. COMPRESSION AT TOP (RADIUS r) 23 q L2 (r) M1 r 54 0 sc sc 0.5 EI EI 2 sc 23 q0 L2 r 27 EI ; 23 27 0.852 same magnitude as
0 1 2L 1 2L c q0 a b a b 2 3 33 23 q0 L2 54 23 54 q0 L Ld 2 0.426
Max. tensile stress at bottom compressive stress at top
Problem 5.5-25 A steel post (E 30 106 psi) having thickness t 1/8 in. and height L 72 in. supports a stop sign (see figure: s 12.5 in.). The height of the post L is measured from the base to the centroid of the sign. The stop sign is subjected to wind pressure p 20 lb/ft2 normal to its surface. Assume that the post is fixed at its base.
(a) What is the resultant load on the sign? [See Appendix D, Case 25, for properties of an octagon, n 8]. (b) What is the maximum bending stress smax in the post?
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411
s
L
Section AA
y 5/8 in.
Circular cutout, d = 0.375 in. Post, t = 0.125 in. c1
z
1.5 in.
C
c2
Stop sign 0.5 in. 1.0 in. 1.0 in. 0.5 in. Wind load
Numerical properties of post A = 0.578 in.2, c1 = 0.769 in., c2 = 0.731 in., Iy = 0.44867 in.4, Iz = 0.16101 in.4
A
A
Elevation view of post
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Solution 5.5-25 (a) RESULTANT LOAD F ON SIGN
p b A or A F pA 20 psf s 12.5 in. b A n 8 360 p a b n 180 b ns2 cota b 4 2 5.239 ft2 F 104.8 lb ;
(b) MAX. BENDING STRESS IN POST L c1 Mmax 754.442 in.2 sc 72 in. 0.769 in. FL Mmax c1 Iz Mmax c2 Iz IZ c2 Mmax 12 sc 0.16101 in.4 0.731 in. 628.701 ft-lb 36.0 ksi ;
0.785 rad
(max. bending stress at base of post) st st 34.2 ksi
Design of Beams
Problem 5.6-1 The cross section of a narrow-gage railway bridge is shown in part (a) of the figure. The bridge is constructed with longitudinal steel girders that support the wood cross ties. The girders are restrained against lateral buckling by diagonal bracing, as indicated by the dashed lines. The spacing of the girders is s1 50 in. and the spacing of the rails is s2 30 in. The load transmitted by each rail to a single tie is P 1500 lb. The cross section of a tie, shown in part (b) of the figure, has width b 5.0 in. and depth d. Determine the minimum value of d based upon an allowable bending stress of 1125 psi in the wood tie. (Disregard the weight of the tie itself.)
P Wood tie b Steel girder s1 (a) s2 P Steel rail d
(b)
Solution 5.6-1
Railway cross tie
Mmax S bd 2 6 P(s1 2 s2) 15,000 lb-in. 5d 2 6 d inches 5d 2 b 6 ;
1 (50 in.)(d 2) 6 s allow S d2 15,000 16.0 in.
Mmax s1 d 50 in. b 5.0 in. P s2 30 in. sallow 1125 psi Solving,
(1125)a dmin
4.0 in. 3P(s1 s2) bsallow
depth of tie
1500 lb
NOTE: Symbolic solution: d 2
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Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has the shape and dimensions shown in the figure. A vertical load p 40 N acts at the free end D. Determine the minimum permissible diameter dmin of the bracket if the allowable bending stress in the material is 30 MPa and b 37 mm. (Note: Disregard the weight of the bracket itself.)
6b A B
2b D P 2b C
Solution 5.6-2
(3Pb) a a dmin b 2 b
1
sa
pdmin4 64
dmin3
96Pb psa
dmin dmin
96Pb 3 a b psa 11.47 mm
dmin ;
96 (40) (37) 3 c d p (30)
1
P
2750 lb
Problem 5.6-3 A cantilever beam of length L 7.5 ft supports a uniform load of intensity q 225 lb/ft and a concentrated load P 2750 lb (see figure). Calculate the required section modulus S if sallow 17,000 psi. Then select a suitable wide-flange beam (W shape) from Table E-1(a), Appendix E, and recalculate S taking into account the weight of beam. Select a new beam size if necessary.
q
225 lb/ft
L = 7.5 ft
Solution 5.6-3 sa 17000 psi
q 225 lb ft PL + L qL2 2
P
2750 lb
Mmax2 smax 2.695 * 104 lb-ft
2.774 * 104 lb-ft Mmax2 (12) Sact smax 13699 psi
7.5 ft Mmax1
Mmax1
below allowable -OK Repeat for W14 * 26 which is lighter than W8 * 28 w
Find Sreqd without beam weight Sreqd Mmax1 (12) sa Sreqd 19.026 in.
3 3
26
lb ft PL +
Sact
35.3 in.3
try W 8 * 28 (S Check - add weight per ft for beam W 28 lb ft PL + Sact 24.3 in.3
24.3 in. )
Mmax3 Mmax3 smax
(q + w) L2 2
2.768 * 104 lb-ft M max3 (12) Sact smax 9411 psi
Mmax2
(q + w) L2 2
well below allowable - OK use W 14 * 26 ;
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Problem 5.6-4
A simple beam of length L 5 m carries a uniform load kN of intensity q 5.8 and a concentrated load 22.5 kN (see figure). m Assuming sallow 110 MPa, calculate the required section modulus S. Then select an 200 mm wide-flange beam (W shape) from Table E-1(b) Appendix E, and recalculate S taking into account the weight of beam. Select a new 200 mm beam if necessary.
P = 22.5 kN 1.5 m q = 5.8 kN/m
L=5m
Solution 5.6-4 NUMERICAL DATA
L P a 5m q 22.5 kN L b a 5.8 b kN m 1.5 m
RECOMPUTE MAX. MOMENT WITH BEAM MASS INCLUDED &
THEN CHECK ALLOWABLE STRESS
w w
a41.7
kg M b a9.81 2 b m s N m Sact 398 * 103 mm3
3.5 m qL Pb + 2 L qL Pa + 2 L
409.077 aq +
sallow statics
110 MPa RA RB RA RB 21.25 kN RA 30.25 kN 51.5 kN RA xm
W bL 1000 2 + xm
Pd L RA q+W
22.273 kN
qL + P
51.5 kN
RA + RB
3.587 m greater than a so max. moment at load pt RA a (q + W ) a2 2
LOCATE POINT OF ZERO SHEAR xm RA q Mmax xm 3.664 m Mmax smax smax Mmax S act
39.924 kN # m
greater than dist. a to load P so zero shear is at load point Mmax RA a q a2 2 Mmax 38.85 kN # m
100.311 MPa
OK, less than 110 MPa
FIND REQUIRED SECTION MODULUS Sreqd Mmax sallow Sreqd 353.182 * 103 mm3 ; (Sact 398 * 103 mm3)
q A P q B
select W 200 * 41.7
Problem 5.6-5 A simple beam AB is loaded as shown in the figure. Calculate the required section modulus S if sallow 17,000 psi, L 28 ft, P 2200 lb, and q 425 lb/ft. Then select a suitable I-beam (S shape) from Table E-2(a), Appendix E, and recalculate S taking into account the weight of the beam. Select a new beam size if necessary.
L 4
L 4
L 4
L 4
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Solution 5.6-5 NUMERICAL DATA
sa P 17000 psi 2200 lb L q 28 ft 425 lb ft
RECOMPUTE REACTIONS AND MAX. MOMENT THEN CHECK lb MAX. STRESS w 25.4 ft RA Mmax P L L +q +w 2 4 2 RA L 2 RA 4.431 * 103 lb w L 1L a b 2 22
FIND REACTIONS (EQUAL DUE TO SYMMETRY) THEN MAX.
MOMENT AT CENTER OF BEAM
RA Mmax Mmax
P L +q 2 4 RA L 2
RA
4.075 * 103 lb Mmax smax smax
qL L 1L a+ b 44 24
2.83 * 104 ft-lb Mmax (12) Sact 13,806 psi less than allowable so OK
qL L 1L a+ b 44 24
2.581 * 104 ft-lb Mmax (12) sa 24.6 in. , w
3
Compute Sreqd & then select S shape Sreqd Sreqd ; 25.4 lb/ft) 18.221 in.3
select S 10 * 25.4 (Sact
Problem 5.6-6 A pontoon bridge (see figure) is constructed of two longitudinal wood beams, known an balks, that span between adjacent pontoons and support the transverse floor beams, which are called chesses. For purposes of design, assume that a uniform floor load of 8.0 kPa acts over the chesses. (This load includes an allowance for the weights of the chesses and balks.) Also, assume that the chesses are 2.0 m long and that the balks are simply supported with a span of 3.0 m. The allowable bending stress in the wood is 16 MPa. If the balks have a square cross section, what is their minimum required width bmin?
Chess Pontoon
Balk
Solution 5.6-6
Pontoon bridge
FLOOR LOAD: W 8.0 kPa 16 MPa length of balks 3.0 m
ALLOWABLE STRESS: sallow Lc length of chesses 2.0 m Lb
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LOADING DIAGRAM FOR ONE BALK
Section modulus S Mmax S qL2 b 8
b3 6 9,000 N # m
6
(8.0 kN/m)(3.0 m)2 8 9,000 N # m 16 MPa
6
Mmax sallow b3 6
562.5 * 10
m3
6
W
total load wLbLc W 2Lb wLc 2
562.5 * 10
m3 and b3 150 mm
3375 * 10 ;
m3
Solving, bmin
0.150 m
q
(8.0 kPa)(2.0 m) 2 8.0 kN/m
Problem 5.6-7 A floor system in a small building consists of wood planks supported by 2 in. (nominal width) joists spaced at distance s, measured from center to center (see figure). The span length L of each joist is 10.5 ft, the spacing s of the joists is 16 in., and the allowable bending stress in the wood is 1350 psi. The uniform floor load is 120 lb/ft2, which includes an allowance for the weight of the floor system itself. Calculate the required section modulus S for the joists, and then select a suitable joist size (surfaced lumber) from Appendix F, assuming that each joist may be represented as a simple beam carrying a uniform load.
Planks
s L Joists s s
Solution 5.6-7
Floor joists
Mmax qL2 8 1 (13.333 lb/in.)(126 in.)2 8 Mmax sallow 26,460 lb/in. 1350 psi 26,460 lb-in. ;
Required S
19.6 in.3 ;
From Appendix F: Select 2 * 10 in. joists sallow L w s q 1350 psi 126 in. 120 lb/ft2 16 in. 0.8333 lb/in.2 10.5 ft floor load ws
spacing of joists 13.333 lb/in.
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Problem 5.6-8 The wood joists supporting a plank floor (see figure) are 40 mm * 180 mm in cross section (actual dimensions) and have a span length L 4.0 m. The floor load is 3.6 kPa, which includes the weight of the joists and the floor. Calculate the maximum permissible spacing s of the joists if the allowable bending stress is 15 MPa. (Assume that each joist may be represented as a simple beam carrying a uniform load.)
Solution 5.6-8
Spacing of floor joists
L w s
4.0 m floor load 3.6 kPa sallow 15 MPa spacing of joists
q S
ws bh 6
2
SPACING OF JOISTS
smax
4 bh2sallow 3wL2
;
Substitute numerical values: qL2 8 wsL2 8 wsL 8sallow
2
Mmax S
smax bh 6
2
4(40 mm)(180 mm)2(15 MPa) 3(3.6 kPa)(4.0 m)2 0.450 m 450 mm ;
Mmax sallow
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Problem 5.6-9 A beam ABC with an overhang from B to C is constructed of a C 10 30 channel section (see figure). The beam supports its own weight (30 lb/ft) plus a triangular load of maximum intensity q0 acting on the overhang. The allowable stresses in tension and compression are 20 ksi and 11 ksi, respectively. Determine the allowable triangular load intensity q0,allow if the distance L equals 3.5 ft.
q0 A L C L
B
3.033 in.
C 10.0 in.
2.384 in. 0.649 in.
Solution 5.6-9 NUMERICAL DATA
w L c1 30 lb ft sat 20 ksi sac 11 ksi
check tension on top st MB c1 I22 3 L
2
MB csat a
s at
I22 c1 1 wL2 d 2 governs 1 wL2 d 2
3.5 ft 2.384 in. c2 I22 0.649 in. 3.93 in.
4
q0allow q0allow
I22 b c1 ; I22 b c2
from Table E-3(a)
628 lb/ft 3 L
2
MAX. MOMENT IS AT B (TENSION TOP, COMPRESSION BOTTOM) MB MB wL L 1 2 + q0L a Lb 2 2 3
check compression on bottom q0allow q0allow csac a lb ft
1 1 wL2 + q0L2 2 3
1314
Problem 5.6-10 A so-called trapeze bar in a hospital room provides a means for patients to exercise while in bed (see figure). The bar is 2.1 m long and has a cross section in the shape of a regular octagon. The design load is 1.2 kN applied at the midpoint of the bar, and the allowable bending stress is 200 MPa. Determine the minimum height h of the bar. (Assume that the ends of the bar are simply supported and that the weight of the bar is negligible.)
C
h
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Solution 5.6-10
Trapeze bar (regular octagon)
P
1.2 kN L
2.1 m
sallow
200 MPa
b
0.41421h Ic h/2
Ic
1.85948(0.41421h)4
0.054738h4
Determine minimum height h. MAXIMUM BENDING MOMENT Mmax PL 4 (1.2 kN)(2.1 m) 4 630 N # m
SECTION MODULUS S 0.054738h4 h/2 0.109476h3
MINIMUM HEIGHT h s M S S M s 630 N # m 200 MPa
6
PROPERTIES OF THE CROSS SECTION Use Appendix D, Case 25, with n b b b tan 2 b cot 2 360 n 8 360 8 length of one side 45
0.109476h3 h3
3.15 * 10
6
m3
28.7735 * 10 30.6 mm
m3 h ;
0.030643 m 12 12
b (from triangle) h h b
hmin
ALTERNATIVE SOLUTION (n M b PL 4 (12 a a b 1)h 45 tan h b 2 a h3
8)
1
cot
b 2
1
(12 + 1)b 412 12 2(4 12 5 b h4 ;
For b
b 45: h h b
45 tan 2 45 cot 2
Ic 0.41421 2.41421 S
11 + 812 4 bb 12 5 bh3
6
412 6
3PL 5)sallow
Substitute numerical values: MOMENT OF INERTIA Ic Ic b b nb acot b a3 cot2 192 2 2
4
h3 1b 1] 1.85948b4
28.7735 * 10
m3 hmin
30.643 mm
;
8b4 (2.41421)[3(2.41421)2 192
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Problem 5.6-11 A two-axle carriage that is part of an overhead traveling crane in a testing laboratory moves slowly across a simple beam AB (see figure). The load transmitted to the beam from the front axle is 2200 lb and from the rear axle is 3800 lb. The weight of the beam itself may be disregarded.
3800 lb
5 ft
2200 lb
A 18 ft
B
(a) Determine the minimum required section modulus S for the beam if the allowable bending stress is 17.0 ksi, the length of the beam is 18 ft, and the wheelbase of the carriage is 5 ft. (b) Select the most economical I-beam (S shape) from Table E-2(a), Appendix E.
Solution 5.6-11 NUMERICAL DATA
L P2 sa 18 ft 3800 lb 17 ksi P1 d 2200 lb 5 ft
xm RA RA Mmax Mmax Sreqd
P1d (P1 + P2) L 2 (P1 + P2) P2 a L L xm b + P1 c xm L b L
xm
8.083 ft
(xm + d) d L L (xm L d) dd
(a) FIND REACTION RA THEN AN EXPRESSION FOR MOMENT UNDER LARGER LOAD P2; LET X DIST. FROM A TO LOAD P2 RA M2 M2 M2 P2 a RA x x cP2 a xP2 L L L P2 x2 x b + P1 c L (x + d ) dd L L L x b + P1 c L (x + d ) d L
2694 lb xm cP2 a Mmax sa L P1 c
21780 ft-lb Sreqd 15.37 in.3 ;
(b) SELECT MOST ECONOMICAL S SHAPE FROM TABLE E-2(A) select S8 * 23 ; Sact 16.2 in.3
xP1L P1x2 xP1d L Take derivative of MA & set to zero to find max. bending moment at x x m d xP2L a dx P2L P2L P2x2 xP1L L P1x2 xP1d b
2P2x + P1L L 2P1x
2P1x P1d
P1d 0
2P2x + P1L
Problem 5.6-12 A cantilever beam AB of circular cross section and length L 450 mm supports a load P 400 N acting at the free end (see figure). The beam is made of steel with an allowable bending stress of 60 MPa. Determine the required diameter dmin of the beam, considering the effect of the beams own weight.
A B d P L
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Solution 5.6-12 Cantilever beam L 450 mm DATA P 400 N sallow 60 MPa g weight density of steel
77.0 kN/m
3
MINIMUM DIAMETER Mmax sallow S PL + pgd 2L2 8 sallow a pd 3 b 32
WEIGHT OF BEAM PER UNIT LENGTH q ga pd 2 b 4
Rearrange the equation: sallow d 3 4gL2 d 2 32 PL p 0
(Cubic equation with diameter d as unknown.) Substitute numerical values (d (60 * 10 N/m )d
6 2 3
MAXIMUM BENDING MOMENT Mmax q L2 PL + 2 S pgd3L2 PL + 8 pd 3 32
meters):
4(77,000 N/m3)(0.45 m)2d2 0 0
SECTION MODULUS
32 (400 N)(0.45 m) p 60,000d 3 62.37d 2
1.833465
Solve the equation numerically: d 0.031614 m dmin 31.61 mm ;
Problem 5.6-13 A compound beam ABCD (see figure) is supported at points A, B, and D and has a splice at point C. The distance a 6.25 ft, and the beam is a S 18 70 wide-flange shape with an allowable bending stress of 12,800 psi.
q A
B
C Splice a 4a
D
(a) If the splice is a moment release, find the allowable 4a uniform load qallow that may be placed on top of the beam, taking into account the weight of the beam itself. [See figure part (a).] (b) Repeat assuming now that the splice is a shear release, as in figure part (b).
(a) (b) Moment Shear release release
Solution 5.6-13 NUMERICAL DATA lb S 103 in.3 w 70 ft a 6.25 ft sa 12800 psi
(a) MOMENT RELEASE AT C-GIVES MAX. MOMENT AT B (SEE MOMENT DIAGRAM) 2.5 q a2 sa Mmax S and Mmax 70 Mmax saS [1qallow + w2 a2 (2.5)]
MZ 9.453E01 @ 1.375E+00
MZ 2.000E+00 @ 2.000E+00
MZ 2.500E+00 @ 4.000E+00
w a
lb S 103 in.3 ft 6.25 ft sa 12800 psi
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qallow qallow
sa S 12 in./ft 2.5 a2 lb 1055 ft
MZ 8.00E+00
w ; for moment release
(b) SHEAR RELEASE AT C-GIVES MAX. MOMENT AT C (SEE MOMENT DIAGRAM) 8 q a2 sa S 12 in./ft 8a2 qallow w 282
lb ft
;
for shear release
qallow
Problem 5.6-14 A small balcony constructed of wood is supported by three identical cantilever beams (see figure). Each beam has length L1 2.1 m, width b, and height h 4b/3. The dimensions of the balcon floor are L1 * L2, with L2 2.5 m. The design load is 5.5 kPa acting over the entire floor area. (This load accounts for all loads except the weights of the cantilever beams, which have a weight density g 5.5 kN/m3.) The allowable bending stress in the cantilevers is 15 MPa. Assuming that the middle cantilever supports 50% of the load and each outer cantilever supports 25% of the load, determine the required dimensions b and h. Solution 5.6-14 Compound beam
4b h= 3 L2 L1 b
MAXIMUM BENDING MOMENT (q q0)L2 1 1 (6875 N/m Mmax 2 2 15,159 + 16,170b2 (N # m) L1 2.1 m L2 2.5 m Floor dimensions: L1 * L2 Design load w 5.5 kPa g 5.5 kN/m3 (weight density of wood beam) sallow 15 MPa MIDDLE BEAM SUPPORTS 50% OF THE LOAD. q wa L2 b 2 4gb2 3 (5.5 kPA) a 2.5 m b 2 6875 N/m S bh2 6 8b3 27 sallow S
7333b2)(2.1 m)2
Mmax
15,159 + 16,170b2
(15 * 106 N/m2)a
8b3 b 27
Rearrange the equation: (120 * 106)b3 436,590b2 409,300 0
WEIGHT OF BEAM q0 gbh 4 (5.5 kN/m2) b2 3 (b meters)
SOLVE NUMERICALLY FOR DIMENSION b 4b b 0.1517 m h 0.2023 m 3 REQUIRED DIMENSIONS b 152 mm h 202 mm ;
7333b2 (N/m)
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Problem 5.6-15 A beam having a cross section in the form of an unsymmetric wide-flange shape (see figure) is subjected to a negative bending moment acting about the z axis. Determine the width b of the top flange in order that the stresses at the top and bottom of the beam will be in the ratio 4:3, respectively.
y b 1.5 in. 1.25 in. z C 12 in. 1.5 in. 16 in.
Solution 5.6-15
Unsymmetric wide-flange beam
AREAS OF THE CROSS SECTION (in.2) A1 A3 A 1.5b A2 (12)(1.25) 24 in.2 39 + 1.5b (in.2) 15 in.2 (16)(1.5)
A1 + A2 + A3
FIRST MOMENT OF THE CROSS-SECTIONAL AREA ABOUT THE LOWER EDGE B-B QBB Stresses at top and bottom are in the ratio 4:3. Find b (inches) h 15 in. height of beam LOCATE CENTROID stop c1 4 sbottom c2 3 4 60 h 8.57143 in. c1 7 7 3 45 c2 h 6.42857 in. 7 7 gyi Ai (14.25)(1.5b) + (7.5)(15) + (0.75)(24) 130.5 + 21.375b (in.3) DISTANCE c2 FROM LINE B-B TO THE CENTROID C c2 QBB A 130.5 + 21.375b 39 + 1.5b 45 in. 7
SOLVE FOR b (39 + 1.5b)(45) 82.125b 841.5 (130.5 + 21.375b)(7) b 10.25 in. ;
Problem 5.6-16 A beam having a cross section in the form of a channel (see figure) is subjected to a bending moment acting about the z axis. Calculate the thickness t of the channel in order that the bending stresses at the top and bottom of the beam will be in the ratio 7:3, respectively.
z
y t t t 55 mm
C 152 mm
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Solution 5.6-16 NUMERICAL DATA
h 152 mm b 55 mm take 1st moments to find distances c1 & c2 1st moments about base t (h 2 b c2 2t) (t) + 2.55t a 2t) 55 b 2 b 2t) (t) + 2bt a b 2 2bt + t (h t (152 2 2t)
ratio of top to bottom stresses 1 11385 2 t (152 2 186 t + t2 131 + t 55 b 2
c1/c2
7/3
c2 c1
J
2t) (t) + 2.55 t a 186 t + t22 3025 t2 b d 2 t)
111385
2.55 t + t (152 76 t + t2
7/3
K
c3 c
a11385 7a 76 t
186 t t2
c2
2.55t + t (152 t (152 2 1 11385 2
c1 c1
55
55 2t) (t) + 2.55t a b 2 2.55t + t(152 2t)
2
t2 t
109 t + 1298 109
186 t + t 131 + t
11092 2
3025 b d 0 4 (1298)
0
t
13.61 mm
;
Problem 5.6-17 Determine the ratios of the weights of three beams that have the same length, are made of the same material, are subjected to the same maximum bending moment, and have the same maximum bending stress if their cross sections are (1) a rectangle with height equal to twice the width, (2) a square, and (3) a circle (see figures).
h = 2b
a
b
a
d
Solution 5.6-17 Ratio of weights of three beams Beam 1: Rectangle (h 2b) Beam 2: Square (a side dimension) Beam 3: Circle (d diameter) L, g, Mmax, and smax are the same in all three beams. M S section modulus S s Since M and s are the same, the section moduli must be the same.
(1) RECTANGLE: S A1 bh2 6 2b2 2b3 3 2a b 3S 1/3 ab 2 2.6207S2/3
(2) SQUARE: S A2
a3 6 a2 pd 3 32 pd 2 4
a
(6S)1/3 (6S)2/3 a 3.3019 S2/3
(3) CIRCLE: S
d
32S 1/3 b p 3.6905 S2/3
A3
p 32S 2/3 a b 4p
3S 2/3 b 2
Weights are proportional to the cross-sectional areas (since L and g are the same in all 3 cases). W1 : W2 : W3 A1 : A2 : A3 W1 : W2 : W3 A1 : A2 : A3 2.6207 : 3.3019 : 3.6905 1 : 1.260 : 1.408 ;
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Design of Beams
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Problem 5.6-18 A horizontal shelf AD of length L 915 mm, width b 305 mm, and thickness t 22 mm is supported by brackets at B and C [see part (a) of the figure]. The brackets are adjustable and may be placed in any desired positions between the ends of the shelf. A uniform load of intensity q, which includes the weight of the shelf itself, acts on the shelf [see part (b) of the figure]. Determine the maximum permissible value of the load q if the allowable bending stress in the shelf is sallow 7.5 MPa and the position of the supports is adjusted for maximum load-carrying capacity.
t A B L (a) q A B L (b) C D C D b
Solution 5.6-18 NUMERICAL DATA
L 915 mm 7.5 MPa b 305 mm t 22 mm sallow
Substitute x into the equation for either M1 or |M2|: Mmax Mmax qL2 (3 8 sallow S 212) sallow a bt 2 b 6 Eq. (1) Eq. (2)
MOMENT DIAGRAM
Equate Mmax from Eqs. (1) and (2) and solve for q: qmax 4bt2sallow 3L2(3 212) ;
Substitute numerical values: For maximum load-carrying capacity, place the supports so that M1 |M2|. Let x M1 length of overhang qL (L 8 qL (L 8 4x) 4x) qx2 2 L (12 2 1) |M2| qx2 2 qmax 10.28 kN/m
Solve for x: x
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Problem 5.6-19 A steel plate (called a cover plate) having cross-sectional dimensions 6.0 in. * 0.5 in. is welded along the full length of the bottom flange of a W 12 * 50 wide-flange beam (see figure, which shows the beam cross section). What is the percent increase in the smaller section modulus (as compared to the wide-flange beam alone)?
W 12
50
6.0
0.5 in. cover plate
Solution 5.6-19 NUMERICAL PROPERTIES FOR W 12 * 50 (FROM TABEL E-1(a))
A c1 I 14.6 in. c2 391 in.4
2
FIND I ABOUT HORIZ. CENTROIDAL AXIS Ih I + A ac1 d2 1 b+ (6) (0.5)3 2 12 0.5 2 b 2
d d 2 S
12.2 in.
c1
+ (6) (0.5) a c2 64.2 in.3 Ih 491.411in.4
FIND CENTROID OF BEAM WITH COVER PLATE (TAKE 1ST MOMENTS ABOUT TOP TO FIND c1 7 c2) A c1 c2 d 0.5 + (6) (0.5) ad + b 2 2 A + (6) (0.5) (d + 0.5) c1 c2
c1
7.182 in.
5.518 in.
FIND SMALLER SECTION MODULUS Ih Stop Stop 68.419 in.3 c1 % increase in smaller section modulus S Stop (100) 6.57% ; S
Problem 5.6-20 A steel beam ABC is simply supported at A and B and has an overhang BC of length L 150 mm (see figure). The beam supports a uniform load of intensity q 4.0 kN/m over its entire span AB and 1.5q over BC. The cross section of the beam is rectangular with width b and height 2b. The allowable bending stress in the steel is 77.0 kN/m3. sallow 60 MPa, and its weight density is
1.5 q q A B 2L L b C 2b
(a) Disregarding the weight of the beam, calculate the required width b of the rectangular cross section. (b) Taking into account the weight of the beam, calculate the required width b.
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Solution 5.6-20 NUMERICAL DATA
L sa 150 mm 60 MPa q g kN 4 m kN 77 3 m
(b) NOW MODIFY-INCLUDE BEAM WEIGHT w gA w g 12b22 L2 2 2 s a a b3 b 3
Mmax and
(1.5q + w) Mmax
(a) IGNORE BEAM SELF WEIGHT-FIND bmin Mmax1 and 1.5 q Mmax2 L2 2 at B saS S 23 b 3 2 a sa b b3 3
Equate Mmax1 to Mmax2 & solve for bmin 1gL22 b2 32 qL 4 0
Insert numerical values, then solve for b bmin 11.92 mm ;
Equate Mmax1 to Mmax2 & solve for bmin bmin bmin 9 qL2 3 a b 8 sa 11.91 mm ;
1
Problem 5.6-21 A retaining wall 5 ft high is constructed of horizontal wood planks 3 in. thick (actual dimension) that are supported by vertical wood piles of 12 in. diameter (actual dimension), as shown in the figure. The lateral earth pressure is p1 100 lb/ft2 at the top of the wall and p2 400 lb/ft2 at the bottom. Assuming that the allowable stress in the wood is 1200 psi, calculate the maximum permissible spacing s of the piles. (Hint: Observe that the spacing of the piles may be governed by the load-carrying capacity of either the planks or the piles. Consider the piles to act as cantilever beams subjected to a trapezoidal distribution of load, and consider the planks to act as simple beams between the piles. To be on the safe side, assume that the pressure on the bottom plank is uniform and equal to the maximum pressure.)
12 in. diam.
3 in. p1 = 100 lb/ft2 12 in. diam.
5 ft
s
3 in.
Top view p2 = 400 lb/ft2 Side view
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Solution 5.6-21
Retaining wall
(1) PLANK AT THE BOTTOM OF THE DAM t thickness of plank 3 in. b width of plank (perpendicular to the plane of the figure) p2 maximum soil pressure 400 lb/ft2 2.778 lb/in.2 s spacing of piles q p2b sallow 1200 psi S section modulus Mmax Mmax qs 8
2
q1 q2 d
p1s p2s diameter of pile
12 in.
Divide the trapezoidal load into two triangles (see dashed line). Mmax S 1 2h (q1) (h)a b 2 3 Mmax 1 h (q2)(h)a b 2 3 or sh2 (2p1 6 p2 )
p2bs 8
2
S p2bs 2 8
bt 6
2
Solve for s: s 4sallow t 2 3p 2 72.0 in.
A
s allow S
or
bt 2 sallow a b 6
pd 3 32
sallow S pd 3 b 32
sh 2 (2p1 + p2) 6 Solve for s: s
sallow a
3psallow d 3 16h2 (2p1 + p2) smax
81.4 in. 72.0 in. ;
(2) VERTICAL PILE h 5 ft 60 in. p1 soil pressure at the top 100 lb/ft 2 0.6944 lb/in.2
PLANK GOVERNS
Problem 5.6-22 A beam of square cross section (a length of each side) is bent in the plane of a diagonal (see figure). By removing a small amount of material at the top and bottom corners, as shown by the shaded triangles in the figure, we can increase the section modulus and obtain a stronger beam, even though the area of the cross section is reduced.
(a) Determine the ratio b defining the areas that should be removed in order to obtain the strongest cross section in bending. (b) By what percent is the section modulus increased when the areas are removed?
z
y
a C a
ba
ba
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Solution 5.6-22 removed
Beam of square cross section with corners
RATIO OF SECTION MODULI S S0 (1 + 3b)(1 b)2 Eq. (1)
GRAPH OF EQ. (1)
a length of each side ba amount removed Beam is bent about the z axis. ENTIRE CROSS SECTION (AREA 0) I0 a4 12 c0 S0 I0 c0 a 12
a3 12 12
(a) VALUE OF b dS ab db S0 1 9
FOR A MAXIMUM VALUE OF
S/S0
SQUARE mnpq (AREA 1) I1 (1 b)4a4 12
0
Take the derivative and solve this equation for b . b ;
PARALLELOGRAM mm, n, n (AREA 2) 1 I2 (base)(height)3 3 I2 (1 b)a 3 1 (ba 12)c d 3 12 ba4 (1 6 b)
3
(b) MAXIMUM VALUE OF S/S0 Substitute b 1/9 into Eq. (1). (S/S0)max 1.0535 The section modulus is increased by 5.35% when ; the triangular areas are removed.
REDUCED CROSS SECTION (AREA qmm, n, p, pq) a4 I I1 + 2I2 (1 + 3b)(1 b)3 12 c (1 12 b) a S I c 12 a3 (1 + 3b)(1 12 b)2
Problem 5.6-23 The cross section of a rectangular beam having width b and height h is shown in part (a) of the figure. For reasons unknown to the beam designer, it is planned to add structural projections of width b/9 and height d to the top and bottom of the beam [see part (b) of the figure]. For what values of d is the bending-moment capacity of the beam increased? For what values is it decreased?
b (a)
b 9 d
h
h
b 9 (b)
d
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CHAPTER 5
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Solution 5.6-23
Beam with projections
Graph of S2 d versus S1 h d h 0 0.25 0.50 0.75 1.00 h 2 S1 I1 c1 bh2 6 S2 S1 1.000 0.8426 0.8889 1.0500 1.2963
(1) ORIGINAL BEAM I1 bh3 12 c1
(2) BEAM WITH PROJECTIONS I2 1 8b 3 1b a bh + a b (h + 2d)3 12 9 12 9 b [8h3 + (h + 2d)3] 108 h 1 +d (h + 2d) 2 2 I2 c2 b[8h3 + (h + 2d)3] 54(h + 2d)
c2 S2
RATIO OF SECTION MODULI S2 S1 b [8h + (h + 2d) ] 9(h + 2d)(bh2)
3 3
8 + a1 +
2d 3 b h
Moment capacity is increased when d 7 0.6861 ; h Moment capacity is decreased when d 6 0.6861 ; h NOTES: S2 S1 or d h 1 when a1 + 2d 3 b h 9a1 +
3 14 2
2d 9 a1 + b h
2d b+8 h
0
EQUAL SECTION MODULI Set d h S2 S1 d 1 and solve numerically for . h and d h 0
0.6861 and 0
S2 d is minimum when S1 h a S2 b S1 min 0.8399
1
0.2937
0.6861
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SECTION 5.7
Nonprismatic Beams
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Nonprismatic Beams
Problem 5.7-1 A tapered cantilever beam AB of length L has square cross sections and supports a concentrated load P at the free end [see figure part (a)]. The width and height of the beam vary linearly from hA at the free end to hB at the fixed end. Determine the distance x from the free end A to the cross section of maximum bending stress if hB 3hA.
(a) What is the magnitude smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress B at the support? (b) Repeat (a) if load P is now applied as a uniform load of intensity q P/L over the entire beam, A is restrained by a roller support and B is a sliding support [see figure, part (b)].
q = P/L B hA A hB x P L (a) x L (b) Sliding support A B
Solution 5.7-1 (a) FIND MAX. BENDING STRESS FOR TAPERED
CANTILEVER
sB
s(L) 4PL
sB
2PL 9hA 3
h(x) s(x)
hA a1 + M(x) S(x)
2x b L s(x)
S(x)
h(x)3 6 6(P)(x) 2x bd L
3
chA a 1 +
smax sB
9hA 3 2PL 9hA 3
smax sB
2
;
s(x) d s(x) dx
6PxL3 hA3 (L 0 + 2x)3 then solve for xmax 0 36Px so L3 hA3 (L x L 4 L 6P L3 4 L3 hA3 aL + 2 b 4 2x)
4
(b) REPEAT (A) BUT NOW FOR DISTRIBUTED UNIFORM LOAD OF P/L OVER ENTIRE BEAM a Fv M(x) d M(x) 0 c c RA x Px RA P P x xa b d d L 2
6PxL3 d c3 d dx hA (L + 2x)3 c6P L3 hA3 (L L + 4x hA3 (L + 2x)4 L sa b 4 4PL 9hA 3 2x)
3
0
1 2P x 2L Px s(x) 1 2P x 2L 2x 3 bd L
0
s(x)
M(x) S(x)
c hA a1 + 6 L2
smax
smax
s(x)
3xP (
2L + x)
hA3 (L + 2x)3
smax
;
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d s(x) dx d c dx c
0
then solve for xmax 2L x) L hA3 (L L2 hA3(L + 2x)
3 2
xmax smax d 0 smax sB
0.20871 L s (0.20871 L) PL 0.394 3 ; hA s(L) So PL 9hA3 smax sB a 0.39385 PL 9hA3 PL hA3 b
3xP ( 3P ( 3xP
2x)3
2L + x) L2
hA3 (L + 2x)3 2L + x) L hA3 (L
2 4
sB d 0
+ 18xP ( Simplifying
5xL + x 2 0 so L2 152 xmax 4 5 L 2
+ 2x)
smax sB
3.54
;
Problem 5.7-2 A tall signboard is supported by two vertical beams consisting of thin-walled, tapered circular tubes [see figure]. For purposes of this analysis, each beam may be represented as a cantilever AB of length L 8.0 m subjected to a lateral load P 2.4 kN at the free end. The tubes have constant thickness t 10.0 mm and average diameters dA 90 mm and dB 270 mm at ends A and B, respectively. Because the thickness is small compared to the diameters, the moment of inertia at any cross section may be obtained from the formula I pd3t/8 (see Case 22, Appendix D), and therefore, the section modulus may be obtained from the formula S pd2t/4.
(a) At what distance x from the free end does the maximum bending stress occur? What is the magnitude smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at the support? (b) Repeat (a) if concentrated load P is applied upward at A and downward uniform load q(x) 2P/L is applied over the entire beam as shown. What is the ratio of the maximum stress to the stress at the location of maximum moment?
2P q(x) = L B A x L = 8.0 m t = 10.0 mm x L = 8.0 m (b) dA = 90 mm dB = 270 mm P B d t
P = 2.4 kN Wind load A
(a)
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Nonprismatic Beams
433
Solution 5.7-2 (a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER
d(x) P L dA dB s(x) d A a1 + 2.4 kN 8m t 90 mm 270 mm M(x) S(x) s(x) 4P pt x smax 2x bd L K
2
PL smax sB smax sB 2ptdA2 a 9 8 4P L b 9 pt dA2 ;
2x b L
S(x)
pd(x) t 4
2
10 mm
Evaluate using numerical data (2400) (8) 2p (0.010) (0.090)2 37.7 MPa ;
s(x) d s(x) dx
4P xL2 c2 d pt dA (L + 2x)2 0
J
c dA a 1 +
smax
(b) REPEAT (A) BUT NOW ADD DISTRIBUTED LOAD M(x) M(x) aPx Pxa M(x) S(x) 2 Px xb L2 L+x b L Px a L+x b L
then solve for xmax 0
d 4P xL2 c c2 dd dx pt dA (L + 2x)2 L2 P c4 2 pt dA (L + 2x)2 16 or so xL2 P d pt dA2 (L + 2x)3 L + 2x
s(x)
s(x)
pt 2x 2 cdA a1 + bd 4 L x) L ptdA2 (L 2x)2
0 d 0
s(x)
4Px (
L
c 4PL2 xmax
ptdA2 (L + 2x)3 4m ;
tension on top, compression on bottom of beam d s(x) dx d c dx 0 then solve for xmax L x) L ptdA2 (L L ptdA2 (L + 2x)2 2x)2 d 0
L 2
smax
L sa b 2 4P pt PL 2ptdA2 s(L) L2 L 2 dA2 L2 aL + 2 b 2
4Px ( 4P ( 4Px
c
L + x) L ptdA2 (L L
smax
+ 2x)2 L ptdA2 (L 4PL2 2x)3 d 0 d 0
smax
16Px (
x) c
Stress at support sB sB 4P L 9 pt dA2
OR simplifying so xmax xmax L 4 2m
L + 4x ptdA2 (L + 2x)3
;
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smax
L sa b 4
stress at support sB L b 4 L p t dA2 aL 2 L2 bK 4 sB sB 0 s(L) 4PL ( ptdA2 (L + 2L2) so no ratio of smax/sB is possible L + L) L
smax
3 p t dA2 evaluate using numerical data L P 2.4 kN dA t 10 mm dB 270 mm smax smax 8m 90 mm
smax
J
L 4P a L 4 PL
MAX. MOMENT AT L/2 SO COMPARE Stress at location of max. moment L sa b 2 L sa b 2 4P L aL 2 L b 2 L ptdA2 aL 2 L2 b 2
(2400) (8) 3p (0.010) (0.090)2 25.2 MPa ;
1 L P 4 ptdA2 PL 3ptdA2 1 L aP b 4 ptdA2 4 3
smax/s(L/2)
;
Problem 5.7-3 A tapered cantilever beam AB having rectangular cross sections is subjected to a concentrated load P 50 lb and a couple M0 800 lb-in. acting at the free end [see figure part (a)]. The width b of the beam is constant and equal to 1.0 in., but the height varies linearly from hA 2.0 in. at the loaded end to hB 3.0 in. at the support.
(a) At what distance x from the free end does the maximum bending stress smax occur? What is the magnitude smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at the support? (b) Repeat (a) if, in addition to P and M0, a triangular distributed load with peak intensity q0 3P/L acts upward over the entire beam as shown. What is the ratio of the maximum stress to the stress at the location of maximum moment?
P = 50 lb A M0 = 800 lb-in. hA = 2.0 in. x b = 1.0 in. L = 20 in. (a) b = 1.0 in. B hB = 3.0 in.
P = 50 lb
3P q0 = L B
A M0 = 800 lb-in.
x L = 20 in. (b)
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Nonprismatic Beams
435
Solution 5.7-3 (a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER FIG. (A) x b h(x) hA a1 + 2L
numerical data P hA M0 I(x) 50 lb L 20 in. 2 in. hB 3 in. b 4 PL M0 800 in.-lb 5 I(x) bh(x)3 S(x) h(x) 12 2 bh(x)2 6
2 x b chA a1 + bd 2L
s(x)
Px + M0 b c hA a1 + 6
2 x bd 2L
s(x) 1 in. d s(x) dx
24 1Px + M02 0
L2 bhA2 (2L + x)2
then solve for xmax 0
d L2 c24 1Px + M02 d dx bhA2 (2L + x)2 24P L2 bhA2 (2L + x)2 24M02 L2
S(x)
S(x) M(x) s(x)
6 Px + M0 M(x) S(x)
2000
0 bhA2 (2L x)3 2PL + Px + 2M0 d OR simplifying c 24L2 bhA2 (2L + x)3 2 1PL M02 so x P xmax 8 in. ; agrees with plot at left Evaluate max. stress & stress at B using numerical data smax s(8) smax sB ; 1250 psi 1200 psi ; sB s(20) smax 1.042 sB
2 124Px
0
1500 M(x) (in.-lb) 1000
500
0
10 x (in.)
20
(b) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER, FIG. (B) h(x) M0 hA a1 + 4 PL 5 P 3 L bh(x)3 12 x b 2L M0 800 in.-lb
1260
1240 (x) (psi) 1220
q0 I(x)
0 10 x (in.) 20
S(x)
1200
I(x) h(x) 2
S(x)
bh(x)2 6
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S(x)
b chA a1 + 6 Px + M0 + M(x) S(x)
1500
2 x bd 2L
d s(x) dx d s(x) dx
0 then solve for xmax c124PL 12x2 q02 2 (24PxL + 24M0L L bhA2 (2L + x)3 d 0
M(x) s(x)
1 2
a
x x q0 b x L 3
L bhA2 (2L + x)2 4x3q0) * Simplifying
12PL2 + 6PxL + 6x2 q0L
M(x) 1000 (in.-lb)
+ x3 q0 + 12M0 L Solve for xmax xmax 4.642 in. ;
0
Max. stress & stress at B
500 0 10 x (in.) 20
smax smax sB
s (xmax) 1235 psi s (20) sB ; 867 psi
1400
FIND MAX. MOMENT AND STRESS AT LOCATION OF MAX.
1200 (x) (psi) 1000
MOMENT
d M(x) dx xm
800
0
10 x (in.)
20
A q0
0 s(xm)
d aPx + M0 dx xm sm ; 16.33 in. 1017 psi
q0x3 b 6L
0
P (2L)
sm smax sm
Px + M0 s(x) b chA a 1 + 6 s(x) 41 * 6PxL L bhA2 (2L
q0 x3 6L
2 x bd 2L
1.215
6 M0 L + x3 q02 + x)2
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SECTION 5.7
Nonprismatic Beams
437
Problem 5.7-4 The spokes in a large flywheel are modeled as beams fixed at one end and loaded by a force P and a couple M0 at the other (see figure). The cross sections of the spokes are elliptical with major and minor axes (height and width, respectively) having the lengths shown in the figure part (a). The cross-sectional dimensions vary linearly from end A to end B. Considering only the effects of bending due to the loads P and M0, determine the following quantities.
(a) (b) (c) (d) (e) The largest bending stress sA at end A The largest bending stress sB at end B The distance x to the cross section of maximum bending stress The magnitude smax of the maximum bending stress Repeat (d) if uniform load q(x) 10P/3L is added to loadings P and M0, as shown in the figure part (b).
P = 12 kN M0 = 10 kNm A x L = 1.25 m M0 A x L = 1.25 m hA = 90 mm hB = 120 mm (b) bA = 60 mm bB = 80 mm (a) B B P 10P q(x) = 3L
Solution 5.7-4 (a-d) FIND MAX. BENDING STRESS FOR TAPERED
CANTILEVER
30
numerical data L bB P h(x) I(x) 1.25 m bA 80 mm hB 12 kn hA a1 M0 x b 3L 60 mm hA 120 mm 10 kN # m b(x) S(x) bA a1 + I(x) h(x) 2 x b 3L
10 0 0.5 x (m) 1
90 mm
M(x) 20 (kNm)
p b(x) h(x)3 64 p b(x) h(x)2 32 p bA hA2 a1 + 32
240 230 (x) 220 (MPa) 210 200 0 0.5 x (m) 1
S(x)
S(x)
x3 b 3L
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M(x) s(x)
Px + M0
s(x)
M(x) S(x)
I(x)
p b(x) h(x)3 64 p b(x) h(x)2 32 p bA hA2 a1 + 32 P x + M0 M(x) S(x)
15
S(x)
Px + M0 p bA hA2 a1 + 32 864 a 0 x3 b 3L ba L3 (3L + x)3 b S(x)
I(x) h(x) 2
s(x) d s (x) dx
Px + M0 p bA hA2
S(x) M(x) s(x)
x3 b 3L 10 P x2 3L2
then solve for xmax 0
Px + M0 d L3 c864 d 2 dx p bAhA (3L + x)3 864 P L3 p bAhA2 (3L + x)3 Px + M0 L3 2592 2 p bAhA (3L + x)4 d
0
10 M(x) (kNm) 5
OR simplfying c 864L3 3PL + 2Px + 3M0 p bAhA2 (3L + x)4 0
0
0
3(PL M0) so xmax 2P xmax 0.625 m ; agrees with plot above Evaluate using numerical data smax s(xmax) smax sA sB 231 MPa 210 MPa 221 MPa ; ; ; sA s(0) sB s(L) smax 1.045 sB
(x) (MPa)
0.5 x (m)
1
300
200
100
0
0
0.5 x (m)
1
(e) FIND MAX. BENDING STRESS INCLUDING
UNIFORM LOAD
P x + M0 s(x) 90 mm s(x)
10 P x2 3L2 x3 b 3L
L bB P h(x) b(x)
1.25 m 80 mm 12 kN
bA hB M0
60 mm 120 mm 10 kN # m
hA
p bA hA2 a1 + 32 288 1 3PxL
x b hA a1 + 3L bA a1 + x b 3L d s(x) dx
3 M0 L L2
+ 5 P x22 0
p bA hA2 (3 L + x)3
then solve for xmax
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SECTION 5.7
Nonprismatic Beams
439
d c dx *
288 1
3 Px L L
2
3 M0 L + 5 P x22 d 0 L2 p bA hA2 (3 L d 1440Px22 x)3
OR 9PL2 36PxL + 5Px2 9M0L 0.105m 0
pbAhA2 (3L + x)3 c(864 PL
Solving for x max: xmax
d s (x) dx
2880 P x)
solution agrees with plot above, evaluate using numerical data smax s(xmax) sA s(0) sB s(L) smax 214 MPa ; sA 210 MPa ; sB 0 MPa ;
3 1864 P x L L2 pbAhA2 (3L
864 M0 L + x)4 0
* OR simplifying c9PL2
36PxL + 5Px2 cpbAhA2 (3L + x)4 d
(288 L )
2
9M0L d
0
Problem 5.7-5
Refer to the tapered cantilever beam of solid circular cross section shown in Fig. 5-24 of Example 5-9.
(a) Considering only the bending stresses due to the load P, determine the range of values of the ratio dB/dA for which the maximum normal stress occurs at the support. (b) What is the maximum stress for this range of values?
Solution 5.7-5
Tapered cantilever beam
FROM EQ. (5-32), EXAMPLE 5-9 s1 32Px pcdA + (dB x3 dA)a b d L [32Px][p] [3]cdA + (dB Eq. (1) After simplification: N FIND THE VALUE OF x THAT MAKES s1 A MAXIMUM u v ds1 dx va du b dx v
2
x 21 dA)a b d c (dB L L
dA) d x dA) d L
32pPcdA + (dB p 2 cdA + (dB ds1 dx N D
x2 dA)a b d cdA L x6 dA) d L 32PcdA 2(dB
2(dB
Let s1 N
ua
dv b dx
D N D
pcdA + (dB
x3 dA)a b d [32P] L
x dA) d L
p cdA + (dB
x4 dA)a b d L
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ds1 dx x L
0
dA dA
2(dB
x dA)a b L 1 2a dB dA 1b
Maximum bending stress occurs at the support when 0 1 Eq. (2) dB 1.5 dA ;
2(dB
dA)
(b) MAXIMUM STRESS (AT SUPPORT B) Substitute x/L 1 into Eq. (1): smax 32PL pdB3 ;
(a) GRAPH OF x/L VERSUS dB/dA (EQ. 2)
Fully Stressed Beams
Problems 5.7-6 to 5.7-8 pertain to fully stressed beams of rectangular cross section. Consider only the bending stresses obtained from the flexure formula and disregard the weights of the beams.
q
Problem 5.7-6 A cantilever beam AB having rectangular cross sections with constant width b and varying height hx is subjected to a uniform load of intensity q (see figure). How should the height hx vary as a function of x (measured from the free end of the beam) in order to have a fully stressed beam? (Express hx in terms of the height hB at the fixed end of the beam.)
B A hx x L hB
hx b b
hB
Solution 5.7-6 Fully stressed beam with constant width and varying height hx height at distance x AT THE FIXED END (x L): hB height at end B 3q b width (constant) hB L A bsallow qx 2 bhx2 hB x hx x S AT DISTANCE x: M hx Therefore, 2 6 hB L L 3qx 2 M sallow S bhx2
hx x 3q A bsallow
;
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SECTION 5.7
Fully Stressed Beams
441
Problem 5.7-7 A simple beam ABC having rectangular cross sections with constant height h and varying width bx supports a concentrated load P acting at the midpoint (see figure). How should the width bx vary as a function of x in order to have a fully stressed beam? (Express bx in terms of the width bB at the midpoint of the beam.)
P A B h C
x L 2 L 2
h bx bB
h
Solution 5.7-7 Fully stressed beam with constant height and varying width h height of beam (constant) AT MIDPOINT B (x L/2) L bx width at distance x from end Aa0 x b 3PL bB 2 2sallowh2 bB width at midpoint B (x L/2) bx 2bB x 2x ; Therefore, and bx Px 1 bb L L 2 S bx h AT DISTANCE x M L 2 6 NOTE: The equation is valid for 0 x and the 2 M 3Px 3Px sallow bx 2 2 beam is symmetrical about the midpoint. S bx h sallow h
Problem 5.7-8 A cantilever beam AB having rectangular cross sections with varying width bx and varying height hx is subjected to a uniform load of intensity q (see figure). If the width varies linearly with x according to the equation bx bB x/L, how should the height hx vary as a function of x in order to have a fully stressed beam? (Express hx in terms of the height hB at the fixed end of the beam.) A
x
q
B hx hB
L
hx bx bB
hB
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Solution 5.7-8
hx hB bx bB bx
Fully stressed beam with varying width and varying height
hx 3qLx
height at distance x height at end B width at distance x width at end B x bB a b L
A bB sallow A bB sallow
3qL
2
AT THE FIXED END (x hB x AL
L) x AL
AT DISTANCE x M sallow qx 2 2 M S S bx h2 x 6 3qLx bB h2 x bB x (hx)2 6L Therefore, hx hB
hx
hB
;
Shear Stresses in Rectangular Beams
Problem 5.8-1 Eq. (5-39):
The shear stresses t in a rectangular beam are given by V h2 a 2I 4 y2 b 1
t
in which V is the shear force, I is the moment of inertia of the cross-sectional area, h is the height of the beam, and y1 is the distance from the neutral axis to the point where the shear stress is being determined (Fig. 5-30). By integrating over the cross-sectional area, show that the resultant of the shear stresses is equal to the shear force V.
Solution 5.8-1
Resultant of the shear stresses
V R R shear force acting on the cross section resultant of shear stresses t
h/2 h/2
L h/2 12V
tbdy1
h/2
2 a h2 4 V
L 0
V h2 a 2I 4
y2 bbdy1 1
bh L 0 3 12V 2h a b h3 24 I t bh3 12 V h2 a 2I 4 y2 b 1 R
(b) 3
y2 bdy1 1
V Q.E.D.
;
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Shear Stresses in Rectangular Beams
443
Problem 5.8-2 Calculate the maximum shear stress tmax and the maximum bending stress smax in a wood beam (see figure) carrying a uniform load of 22.5 kN/m (which includes the weight of the beam) if the length is 1.95 m and the cross section is rectangular with width 150 mm and height 300 mm, and the beam is (a) simply supported as in the figure part (a) and (b) has a sliding support at right as in the figure part (b).
22.5 kN/m 300 mm
1.95 (a)
150 m mm
22.5 kN/m
1.95 m (b)
Solution 5.8-2 kN q 22 b m
h 300 mm L
150 mm 1.95 m
smax
M S
smax
4.65 MPa
;
(b) MAXIMUM SHEAR STRESS V tmax qL 3V 2A qL2 2 M S smax 18.59 MPa ; tmax 1430 kPa ;
(a) MAXIMUM SHEAR STRESS V tmax qL 2 A 3V 2A qL2 8 bh tmax 715 kPa ;
MAXIMUM BENDING STRESS M smax
MAXIMUM BENDING STRESS M S bh2 6
Problem 5.8-3 Two wood beams, each of rectangular cross section (3.0 in. 4.0 in., actual dimensions) are glued together to form a 4.0 in. solid beam of dimensions 6.0 in. 4.0 in. (see figure). The beam is simply supported with a span of 8 ft. What is the maximum moment Mmax that may be 6.0 in. applied at the left support if the allowable shear stress in the glued joint is 200 psi? (Include the effects of the beams own weight, assuming that the wood weighs 35 lb/ft3.)
M
8 ft
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Solution 5.8-3
L h g q q 8 ft 6 in. 35 lb ft3 1b ft b 4 in. 200 psi A b#h t allow V tmax M Mmax Mmax qL M + L 2 3V 2A qL 3M a+ b 2A L 2 qL2 2 qL2 2 ;
g A weight of beam per unit distance 5.833
2 AL t 3 max 2 AL t 3 allow 25.4 k-ft
Maximum load Mmax
Problem 5.8-4 A cantilever beam of length L 2 m supports a load P 8.0 kN (see figure). The beam is made of wood with cross-sectional dimensions 120 mm * 200 mm. Calculate the shear stresses due to the load P at points located 25 mm, 75 mm, and 100 mm from the top surface of the beam. from these results, plot a graph showing the distribution of shear stresses from top to bottom of the beam.
P = 8.0 kN 200 mm L=2m 120 mm
Solution 5.8-4
Shear stresses in a cantilever beam
Distance from the top surface (mm) 0 25 50
2
y1 (mm) 100 75 50 25 0
t (MPa) 0 0.219 0.375 0.469 0.500
t (kPa) 0 219 375 469 500
Eq. (5-39): t V I h t t P bh3 12
Vh a 2I 4
y2 b 1
75 100 (N.A.)
8.0 kN
8,000 N
6 4
GRAPH OF SHEAR STRESS t
80 * 10 mm (y1 c
200 mm 8,000
mm) y2 d 1 (t N/mm2 mm; t MPa) MPa)
(200)2 4 2(80 * 106) 50 * 10
6
(10,000
y2) (y1 1
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SECTION 5.8
Shear Stresses in Rectangular Beams
445
Problem 5.8-5 A steel beam of length L 16 in. and crosssectional dimensions b 0.6 in. and h 2 in. (see figure) supports a uniform load of intensity q 240 lb/in., which includes the weight of the beam. Calculate the shear stresses in the beam (at the cross section of maximum shear force) at points located 1/4 in., 1/2 in., 3/4 in., and 1 in. from the top surface of the beam. From these calculations, plot a graph showing the distribution of shear stresses from top to bottom of the beam.
q = 240 lb/in. h = 2 in.
L = 16 in.
b = 0.6 in.
Solution 5.8-5
Shear stresses in a simple beam
Distance from the top surface (in.) 0 0.25 0.50 0.75 1.00 (N.A.) 0.4 in.4 GRAPH OF SHEAR STRESS t y1 (in.) 1.00 0.75 0.50 0.25 0 t (psi) 0 1050 1800 2250 2400
Eq. (5-39): t V qL 2
V h2 a 2I 4
y2 b 1 bh3 12
1920 lb I
UNITS: POUNDS AND INCHES t (t 1920 (2)2 c 2(0.4) 4 psi; y1 in.) y2 1 (2400)(1 y2) d 1
Problem 5.8-6 A beam of rectangular cross section (width b and height h) supports a uniformly distributed load along its entire length L. The allowable stresses in bending and shear are sallow and tallow, respectively.
(a) If the beam is simply supported, what is the span length L0 below which the shear stress governs the allowable load and above which the bending stress governs? (b) If the beam is supported as a cantilever, what is the length L0 below which the shear stress governs the allowable load and above which the bending stress governs?
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Solution 5.8-6 Beam of rectangular cross section b width h height L length
Uniform load q intensity of load sallow and tallow ALLOWABLE STRESSES (a) SIMPLE BEAM BENDING Mmax smax qallow SHEAR Vmax tmax qallow qL 2 3V 2A A bh qL2 8 Mmax S S bh2 6 3qL
3
(b) CANTILEVER BEAM BENDING Mmax smax qallow SHEAR qL2 2 Mmax S 3L2 S bh2 6 3qL2 bh2 (3)
sallowbh2
4bh2 4sallow bh2 3L
2
(1)
Vmax tmax qallow
qL A 3V 2A
bh 3qL 2bh (4)
2tallow bh 3L h sallow a b 2 tallow
3qL 4bh (2)
Equate (3) and (4) and solve for L0: L0 ;
4tallowbh 3L sallow b tallow
Equate (1) and (2) and solve for L0: L0 ha ;
NOTE: If the actual length is less than L 0, the shear stress governs the design. If the length is greater than L0, the bending stress governs.
Problem 5.8-7 A laminated wood beam on simple supports is built up by gluing together four 2 in. 4 in. boards (actual dimensions) to form a solid beam 4 in. 8 in. in cross section, as shown in the figure. The allowable shear stress in the glued joints is 65 psi, and the allowable bending stress in the wood is 1800 psi. If the beam is 9 ft long, what is the allowable load P acting at the one-third point along the beam as shown? (Include the effects of the beams own weight, assuming that the wood weighs 35 lb/ft3.)
3 ft
P
2 in. 2 in. 2 in. 2 in.
L
9 ft 4 in.
Solution 5.8-7 L 9 ft
h t allow 8 in. 65 psi
b A
4 in. bh 1800 psi s allow
q
7.778
1b ft
ALLOWABLE LOAD BASED UPON SHEAR STRESS IN THE
GLUED JOINTS; MAX. SHEAR STRESS AT NEUTRAL AXIS
WEIGHT OF BEAM PER UNIT DISTANCE g 35 lb ft3 q gA
t
VQ Ib
tmax
3V 2A
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SECTION 5.8
Shear Stresses in Rectangular Beams
447
V tmax P Pmax Pmax
P
qL 2 + 3 2 qL 3 2 aP + b 2A 3 2 3 qL b 4 3 qL 4
ALLOWABLE LOAD BASED UPON BENDING STRESS M S P qL 2 3 ft + 3 ft 3 2 q (3 ft)2 2
3V 2A
a A tmax A t allow
b h2 6 qL q 2 3 ft + 3 ft (3 ft)2 M 3 2 2 S S q sallow S3 3 qL a (3ft)b (3 ft) 2 22 2 P 3.165 k 2.03 k ;
smax Pmax Pmax P allow
2.03 k (governs)
Problem 5.8-8 A laminated plastic beam of square cross section is built up by gluing together three strips, each 10 mm 30 mm in cross section (see figure). The beam has a total weight of 3.6 N and is simply supported with span length L 360 mm. Considering the weight of the beam (q) calculate the maximum permissible CCW moment M that may be placed at the right support.
(a) If the allowable shear stress in the glued joints is 0.3 MPa. (b) If the allowable bending stress in the plastic is 8 MPa.
M q 10 mm 10 mm 30 mm 10 mm 30 mm L
Solution 5.8-8 (a) FIND M BASED ON ALLOWABLE SHEAR STRESS IN GLUED
JOINT
b W q q
30 mm 3.6 N W L 10 N m
h L
30 mm 360 mm
ta
0.3 MPa
Q
bh h 33 4 3bh L cta a L cta a
Q
b h2 9
Q Ib
b h2 9 b2 h3 12
Q Ib beam distributed weight M M ta a bh 12
23
Ib b Q 3bh b 4
qL d 2 qL d 2 ;
MAX. SHEAR ST LEFT SUPPORT Vm ta qL M + 2 L Vm Q Ib I and Vm bh 12
3
Ib b Q
Mmax
72.2 N # M
Ib
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(b) FIND M BASED ON ALLOWABLE BENDING STRESS AT h/2 FROM NA AT LOCATION (xm) OF MAX. BENDING MOMENT, Mm M(x) qL M a + bx 2 L qx2 2 d M(x) dx 0
Mm
a
qL M L M + ba + b 2 L 2 qL qa L M2 + b 2 qL 2
use to find location of zero shear where max. moment occurs M d qL ca + bx dx 2 L M 1 qL + 2 L xm L M + 2 qL qL M a + b xm 2 L
2
simplifying Mm
2 2 1 1qL + 2 M2 8q L2
qx2 d 2 qx 0
also
Mm
sa S
Mm
sa a
bh2 b 6
Equating both Mm expressions & solving for M where sa 8 MPa qxm 2
MAX. MOMENT Mm Mm M Mmax
A
sa a
bh2 b a8 qL2 b 6 2 ;
qL2
9.01 N # m
Problem 5.8-9 A wood beam AB on simple supports with span length equal to 10 ft is subjected to a uniform load of intensity 125 lb/ft acting along the entire length of the beam, a concentrated load of magnitude 7500 lb acting at a point 3 ft from the right-hand support, and a moment at A of 18,500 ft-lb (see figure). The allowable stresses in bending and shear, respectively, are 2250 psi and 160 psi.
7500 lb 18,500 ft-lb 125 lb/ft A 10 ft
3 ft
B
(a) From the table in Appendix F, select the lightest beam that will support the loads (disregard the weight of the beam). (b) Taking into account the weight of the beam (weight density 5 35 lb/ft3), verify that the selected beam is satisfactory, or if it is not, select a new beam.
Solution 5.8-9 1b (a) q 125 P ft d L 10 ft sAllow 2250 psi qL d +P RA 2 L
RA RB
75001b M 3 ft t allow M L
18500 ft-b
RB Vmax
7.725 * 103 1b RB RB d Vmax qd2 2 7.725 * 103 1b
160 psi
Mmax Mmax tmax Areq
2.261 * 104 1b-ft 3V 2A Areq 3Vmax 2tallow
1.025 * 103 1b qL L d M +P + 2 L L
72.422 in.2
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SECTION 5.8
Shear Stresses in Rectangular Beams
449
smax
M S
Sreq
Mmax sallow
Sreq
120.6 in.3
Vmax Areq
RB
Areq
3Vmax 2 tallow
From Appendix F: Select 8 * 12 in. beam (nominal dimensions) ; A 86.25 in.2 S 165.3 in.3
73.405 in.2 < A 1b ft
8 * 12 beam is still satisfactory for shear. qtotal q + qbeam q total RB d qd2 2 145.964
(b) REPEAT (A) CONSIDERING THE WEIGHT OF THE BEAM g 35 1b ft3 qbeam gA Mmax Mmax Sreq qbeam L 2
q beam RB RB
1b 20.964 ft 7.725 * 103 1b + 7.83 * 103 1b
2.293 * 104 1b-ft Mmax sallow Sreq 122.3 in.3 < S
8 * 12 beam is still satisfactory for moment. Use 8 * 12 in. beam ;
Problem 5.8-10 A simply supported wood beam of rectangular cross section and span length 1.2 m carries a concentrated load P at midspan in addition to its own weight (see figure). The cross section has width 140 mm and height 240 mm. The weight density of the wood is 5.4 kN/m3. Calculate the maximum permissible value of the load P if (a) the allowable bending stress is 8.5 MPa, and (b) the allowable shear stress is 0.8 MPa.
P 240 mm 140 mm
0.6 m
0.6 m
Solution 5.8-10
Simply supported wood beam
P 240 mm 140 mm
(a) ALLOWABLE P BASED UPON BENDING STRESS sallow Mmax + 8.5 MPa s qL2 PL + 4 8 Mmax S P(1.2 m) 4
0.6 m
0.6 m
b A S g L
140 mm bh bh 6
2
h
240 mm
33,600 mm2 1344 * 103 mm3 (P Mmax gbh 181.44 N/m
(181.44 N/m)(1.2 m)2 8 N # m) (1344 * 103 mm3)(8.5 MPa)
0.3 P + 32.66 N # m newtons; M Ssallow 11,424 N # m Equate values of Mmax and solve for P: 0.3P + 32.66 or P 11,424 ; P 37,970 N 38.0 kN
5.4 kN/m3 1.2 m q
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(b) ALLOWABLE LOAD P BASED UPON SHEAR STRESS tallow V 0.8 MPa t 3V 2A
Equate values of V and solve for P: P + 108.86 2 or P 17,920 ; P 35,622 N
qL P + 2 2
(181.44 N/m)(1.2 m) P + 2 2
35.6 kN
P + 108.86 (N) 2 2At 2 V (33,600 mm2)(0.8 MPa) 3 3
NOTE: The shear stress governs and Pallow 35.6 kN
17,920 N
Problem 5.8-11 A square wood platform, 8 ft * 8 ft in area, rests on masonry walls (see figure). The deck of the platform is constructed of 2 in. nominal thickness tongue-and-groove planks (actual thickness 1.5 in.; see Appendix F) supported on two 8-ft long beams. The beams have 4 in. * 6 in. nominal dimensions (actual dimensions 3.5 in. * 5.5 in.). The planks are designed to support a uniformly distributed load w (lb/ft2) acting over the entire top surface of the platform. The allowable bending stress for the planks is 2400 psi and the allowable shear stress is 100 psi. When analyzing the planks, disregard their weights and assume that their reactions are uniformly distributed over the top surfaces of the supporting beams.
(a) Determine the allowable platform load w1 (lb/ft2) based upon the bending stress in the planks. (b) Determine the allowable platform load w2 (lb/ft2) based upon the shear stress in the planks. (c) Which of the preceding values becomes the allowable load wallow on the platform? (Hints: Use care in constructing the loading diagram for the planks, noting especially that the reactions are distributed loads instead of concentrated loads. Also, note that the maximum shear forces occur at the inside faces of the supporting beams.)
8 ft
8 ft
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SECTION 5.8
Shear Stresses in Rectangular Beams
451
Solution 5.8-11
Wood platform with a plank deck
Load on one plank: q c w(lb/ft2) 144 in. / ft
2 2
d(b in.) 96 in. b 2 lb; w
wb (lb/in.) 144 a wb b(48) 144 wb 3 in.)
Reaction R
qa
(R
lb/ft2; b
Mmax occurs at midspan. Mmax Ra 3.5 in. 89 in. + b 2 2 q(48 in.)2 3 89 wb 12 in.)
wb (46.25) 3 Platform: 8 ft * 8 ft t w sallow tallow thickness of planks 1.5 in. uniform load on the deck (lb/ft ) 2400 psi 100 psi
2 2
wb (1152) 144 lb/ft2; b
(M Mallow
lb-in.; w
Allowable bending moment: s allow S (2400 psi)(0.375 b) 900 b (lb-in.) Equate Mmax and Mallow and solve for w: 89 wb 12 900 b w1 121 lb/ft2 ;
Find wallow (lb/ft ) (a) ALLOWABLE LOAD BASED UPON BENDING STRESS IN THE
PLANKS
(b) ALLOWABLE
PLANKS
LOAD BASED UPON SHEAR STRESS IN THE
Let b
width of one plank (in.) A S 1.5b (in.2) b (1.5 in.)2 6 0.375b (in.3)
See the free-body diagram in part (a). Vmax occurs at the inside face of the support. Vmax qa 89 in. b 2 wb b 144 44.5q 89 wb 288 lb/ft2; b 2Atallow 3 100 b (lb) in.)
(44.5)a (V 3V 2A
Free-body diagram of one plank supported on the beams:
lb; w
Allowable shear force: t Vallow
2(1.5 b)(100 psi) 3 89wb 288
Equate Vmax and Vallow and solve for w: 100b w2 324 lb/ft2 ;
(c) ALLOWABLE LOAD Bending stress governs. wallow 121 lb/ft2 ;
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Problem 5.8-12 A wood beam ABC with simple supports at A and B and an overhang BC has height h 300 mm (see figure). The length of the main span of the beam is L 3.6 m and the length of the overhang is L/3 1.2 m. The beam supports a concentrated load 3P 18 kN at the midpoint of the main span and a moment PL/2 10.8 kN . m at the free end of the overhang. The wood has weight density g 5.5 kN/m3.
(a) Determine the required width b of the beam based upon an allowable bending stress of 8.2 MPa. (b) Determine the required width based upon an allowable shear stress of 0.7 MPa.
L 2 A
3P
PL M = 2 B L 3 C h= 300 mm b
L
Solution 5.8-12 Numerical data:
L A g 3.6 m bh 5.5 kN m3 P h 300 mm 6 kN qbeam M gA PL 2
s b
Mmax S 3PL sallow h2
6 Mmax bh2 b 87.8 mm ;
(b) REQUIRED WIDTH b BASED UPON SHEAR STRESS tallow 0.7 MPa Vmax t 2P + 3 Vmax 2A 8 q L 9 beam 3 Vmax 2 bh 3P 4 + gL bh 3 89.074 mm
Reactions, max. shear and moment equations RA RB Vmax MD MB 3P 2 M 4 + qbeam L L 9 P 4 q L 9 beam 8 q L 9 beam
3P M 8 + + qbeam L 2 L 9 RB L RA 2 PL 2 2P + 8 q L 9 beam PL 2
2P +
8 3 a2 P + qbeam Lb 2 bh 9 b 3P h a t allow 89.1 mm 4 gLb 3 b
L2 qbeam 2
17 q L2 18 beam b
Shear stress governs ; (governs)
(a) REQUIRED WIDTH b BASED UPON BENDING STRESS sallow 8.2 MPa Mmax MB PL 2
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SECTION 5.9
Shear Stresses in Circular Beams
453
Shear Stresses in Circular Beams
Problem 5.9-1 A wood pole of solid circular cross section (d diameter) is subjected to a horizontal force P 450 lb (see figure). The length of the pole is L 6 ft, and the allowable stresses in the wood are 1900 psi in bending and 120 psi in shear. Determine the minimum required diameter of the pole based upon (a) the allowable bending stress, and (b) the allowable shear stress.
q0 = 20 lb/in.
d L d
Solution 5.9-1 1b L q 20 in
s allow t allow Vmax Mmax 1900 psi 120 psi qL 2 qL 2 L 23
6 ft
dmin dmin 720 1b t 2.88 * 103 1b-ft dmin
A p sallow
3
32 Mmax
5.701 in.
(b) BASED UPON SHEAR STRESS Vmax 4V 3A
Mmax
(a) BASED UPON BENDING STRESS s M S 32 M pd
3
A 3p tallow
3pd2 16 Vmax
16V
dmin dmin
3.192 in. 5.70 in. ;
Bending stress governs
Problem 5.9-2 A simple log bridge in a remote area consists of two parallel logs with planks across them (see figure). The logs are Douglas fir with average diameter 300 mm. A truck moves slowly across the bridge, which spans 2.5 m. Assume that the weight of the truck is equally distributed between the two logs. Because the wheelbase of the truck is greater than 2.5 m, only one set of wheels is on the bridge at a time. Thus, the wheel load on one log is equivalent to a concentrated load W acting at any position along the span. In addition, the weight of one log and the planks it supports is equivalent to a uniform load of 850 N/m acting on the log. Determine the maximum permissible wheel load W based upon (a) an allowable bending stress of 7.0 MPa, and (b) an allowable shear stress of 0.75 MPa.
x
W
850 N/m 300 mm
2.5 m
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Solution 5.9-2
Log bridge
Diameter d 300 mm sallow 7.0 MPa tallow 0.75 MPa Find allowable load W (a) BASED UPON BENDING STRESS Maximum moment occurs when wheel is at midspan (x L/2). Mmax qL WL + 4 8
2
(b) BASED UPON SHEAR STRESS Maximum shear force occurs when wheel is adjacent to support (x 0). Vmax W+ qL 2 W+ 1 (850 N/m)(2.5 m) 2 newtons)
W + 1062.5 N (W A tmax Vmax pd 4
2
0.070686 m2
1 W (2.5 m) + (850 N/m)(2.5 m)2 4 8 0.625W + 664.1 (N # m) (W S pd3 32 2.651 * 10 Ssallow
3
4Vmax 3A 3Atallow 4 39,760 N 3 (0.070686 m2)(0.75 MPa) 4 39,760 N 38.7 kN ;
newtons)
m3
3
Mmax
(2.651 * 10 18,560
m3)(7.0 MPa)
W + 1062.5 N W 38,700 N
18,560 N # m 0.625W + 664.1 W 28,600 N 28.6 kN ;
Problem 5.9-3 A sign for an automobile service station is supported by two aluminum poles of hollow circular cross section, as shown in the figure. The poles are being designed to resist a wind pressure of 75 lb/ft2 against the full area of the sign. The dimensions of the poles and sign are h1 20 ft, h2 5 ft, and b 10 ft. To prevent buckling of the walls of the poles, the thickness t is specified as one-tenth the outside diameter d.
(a) Determine the minimum required diameter of the poles based upon an allowable bending stress of 7500 psi in the aluminum. (b) Determine the minimum required diameter based upon an allowable shear stress of 2000 psi.
b
h2
Wind load
d t= 10
d h1
Probs. 5.9.3 and 5.9.4
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SECTION 5.9
Shear Stresses in Circular Beams
455
Solution 5.9-3
Wind load on a sign
b b p sallow tallow d t
width of sign 10 ft 75 lb/ft2 7500 psi 2000 psi diameter d 10 W W wind force on one pole 1875 lb b ph2 a b 2 h2 b 2
(b) REQUIRED DIAMETER BASED UPON SHEAR STRESS Vmax t r1 W 4V a 3A d 2 r22 1875 lb + r2r1 + r12 r2 + r1 t d 2
2 2
b 2d 5
r2
d 2
d 10
r2 + r2r1 + r2 2 1 r2 2 + r1 2 d 2d 2d 2 d2 a b + a ba b + a b 2 2 5 5 d2 2d 2 a b +a b 2 5 A t d2 p2 (d2 4 d2) 1 p2 cd 4 a 4d 2 bd 5 V d2
(a) REQUIRED DIAMETER BASED UPON BENDING STRESS Mmax I I Wah1 + 506,250 lb-in. d d1 d 2t 4 d 5
p (d 4 64 2 p4 cd 64
61 41 9pd2 100
d24) d2 4d 4 a bd 5
pd 4 369 a b 64 625
369pd 4 (in.4) 40,000 c s d3 d 2 Mc I (d inches) M(d/2)
4
4V 61 100 a ba b 3 41 9pd2 7.0160 Vmax tallow (7.0160)(1875 lb) 2000 psi
7.0160
17.253 M
3
6.5775 in.2
369pd /40,000 d (17.253)(506,250 lb-in.) 17.253 Mmax sallow 7500 psi
3
d
2.56 in.
;
(Bending stress governs.)
1164.6 in.
d
10.52 in.
;
Problem 5.9-4 Solve the preceding problem for a sign and poles having the following dimensions: h1 6.0 m, h2 1.5 m, b 3.0 m, and t d/10. The design wind pressure is 3.6 kPa, and the allowable stresses in the aluminum are 50 MPa in bending and 14 MPa in shear.
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Stresses in Beams (Basic Topics)
Solution 5.9-4
Wind load on a sign
b b p
width of sign 3.0 m 3.6 kPa 50 MPa 16 MPa W wind force on one pole 8.1 kN b ph2 a b 2 h2 Wah1 + b 2 I d1 p4 (d 64 2 d a 2t 4d 4 bd 5 369pd 4 (m 4) 40,000
(b) REQUIRED DIAMETER BASED UPON SHEAR STRESS Vmax t r1 W
2
8.1 kN r2 d 2
sallow tallow d t d 10
4V r2 + r1 r2 + r1 2 a b 3A r2 2 + r1 2 d 2 t d 2 d 10 2d 5
diameter W
r2 2 + r1r2 + r1 2 r2 2 + r1 2 d 2d 2d 2 d2 a b + a ba b + a b 2 5 5 5 d2 2d 2 a b +a b 2 5 A p (d 2 42 p2 cd 4 t d2 d1 2 ) a 4d 2 bd 5 9pd2 100 7.0160 V
(a) REQUIRED DIAMETER BASED UPON BENDING STRESS Mmax s d2 I Mc I d 54.675 kN # m d4) 1 4 d 5
61 41
p4 cd 64
100 4V 61 a ba b 3 41 9pd2 7.0160 Vmax tallow 0.004059 m2 0.06371 m
pd 4 369 a b 64 625 c s d3 d 2 Mc I (d
d2 (7.0160)(8.1 kN) 14 MPa ;
meters) d M(d/2) 17.253 M
63.7 mm
369pd 4/40,000 d3 (17.253)(54.675 kN # m) 17.253Mmax sallow 50 MPa 266 m ;
Bending stress governs
0.018866 m3 d 0.266 m
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SECTION 5.10
Shear Stresses in Beams with Flanges
457
Shear Stresses in Beams with Flanges
Problem 5.10-1 through 5.10-6 A wide-flange beam (see figure) having the cross section described below is subjected to a shear force V. Using the dimensions of the cross section, calculate the moment of inertia and then determine the following quantites:
(a) The maximum shear stress tmax in the web. (b) The minimum shear stress tmin in the web. (c) The average shear stress taver (obtained by dividing the shear force by the area of the web) and the ratio tmax/taver. (d) The shear force Vweb carried in the web and the ratio Vweb /V. NOTE: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles.
z t b Probs 5.10.1through 5.-10.6 y
O h1
h
Problem 5.10-1 Dimensions of cross section: b h 12 in., h1 10.5 in., and V 30 k.
6 in., t
0.5 in.,
Solution 5.10-1
Wide-flange beam
b t h h1 V 6.0 in. 0.5 in. 12.0 in. 10.5 in. 30 k taver tmax taver (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) tmin Vb 2 (h 8It V th1 1.014 h12) 4555 psi ;
(c) AVERAGE SHEAR STREAR IN THE WEB (Eq. 5-50) 5714 psi ; ;
MOMENT OF INERTIA (Eq.5-47) I 1 (bh3 12 bh3 + th3) 1 1 333.4 in.4
(d) SHEAR FORCE IN THE WEB (Eq. 5-49) Vweb Vweb V th1 (2tmax + tmin) 3 0.942 ; 28.25 k ;
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax V (bh2 8It bh2 + th2) 1 1 5795 psi ;
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Problem 5.10-2 Dimensions of cross section: b h 420 mm, h1 380 mm, and V 125 kN.
180 mm, t
12 mm,
Solution 5.10-2
Wide-flange beam b 180 mm
t h h1 V 12 mm 420 mm 380 mm 125 kN
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) Vb 2 (h h2) 21.86 MPa ; 1 8It (c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) tmin taver tmax taver V th1 1.037 27.41 MPa ; ;
MOMENT OF INERTIA (Eq. 5-47) I 1 (bh3 12 bh3 + th3) 1 1 343.1 * 106 mm4
(d) SHEAR FORCE IN THE WEB (Eq. 5-49) Vweb ; Vweb V th1 (2tmax + tmin) 3 0.957 ; 119.7 kN ;
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax V (bh2 8It bh2 + th2) 1 1 28.43 MPa
Problem 5.10-3 Wide-flange shape, W 8 * 28 (see Table E-1(a), Appendix E); V 10 k.
Solution 5.10-3
Wide-flange beam
W 8 * 28 b t h h1 V 6.535 in. 0.285 in. 8.06 in. 7.13 in. 10 k tmax taver (b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b) tmin Vb 2 (h 8It h2 ) 1 4202 psi ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) taver V th1 0.988 4921 psi ; ;
MOMENT OF INERTIA (Eq. 5-47) I 1 (bh3 12 bh3 + th3) 1 1 96.36 in.4
(d) SHEAR FORCE IN THE WEB (EQ. 5-49) Vweb Vweb V th1 (2tmax + tmin) 3 0.943 ; 9.432 k ;
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax V (bh2 8It bh2 + th2) 1 1 4861 psi ;
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Shear Stresses in Beams with Flanges
459
Problem 5.10-4 Dimensions of cross section: b h 600 mm, h1 570 mm, and V 200 kN .
220 mm, t
12 mm,
Solution 5.10-4
Wide-flange beam b 220 mm
t h h1 V 12 mm 600 mm 570 mm 200 kN
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-50) taver tmax taver V th1 1.104 29.24 MPa ;
(d) SHEAR FORCE IN THE WEB (Eq. 5-49) MOMENT OF INERTIA (Eq. 5-47) I 1 (bh3 12 bh3 + th3) 1 1 750.0 * 106 mm4 Vweb Vweb V th1 (2tmax + tmin) 3 0.981 ; 196.1 kN ;
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax V (bh2 8It bh2 + th2) 1 1 32.28 MPa ;
(b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-48b) tmin Vb 2 (h 8It h2) 1 21.45 MPa ;
Problem 5.10-5 Wide-flange shape, W 18 * 71 (see Table E-1(a), Appendix E); V 21 k.
Solution 5.10-5
Wide-flange beam W 18 * 71
b t h h1 V 7.635 in. 0.495 in. 18.47 in. 16.85 in. 21 k
(b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-48b) tmin Vb 2 (h 8It h2) 1 1993 psi ;
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-50) taver tmax taver V th1 1.046 2518 psi ; ;
MOMENT OF INERTIA (Eq. 5-47) I 1 (bh3 12 bh3 + th3) 1 1 1162 in.4
(d) SHEAR FORCE IN THE WEB (EQ. 5-49) Vweb Vweb V th1 (2tmax + tmin) 3 0.961 ; 20.19 k ;
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax V (bh2 8It bh2 + th2) 1 1 2634 psi ;
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Problem 5.10-6 Dimensions of cross section: b h 350 mm, h1 330 mm, and V 60 kN
120 mm, t
7 mm,
Solution 5.10-6
Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48) b t h h1 V 120 mm 7 mm 350 mm 330 mm 60 kN taver tmax taver tmin Vb 2 (h 8It V th1 1.093 h2) 1 19.35 MPa ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50) 25.97 MPa ; ;
MOMENT OF INERTIA (Eq. 5-47) I 1 (bh3 12 bh3 + th3) 1 1 90.34 * 106 mm4
(d) SHEAR FORCE IN THE WEB (Eq. 5-49) Vweb Vweb V th1 (2tmax + tmin) 3 0.977 ; 58.63 kN ;
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a) tmax ; V (bh2 8It bh2 + th2) 1 1 28.40 MPa
Problem 5.10-7 A cantilever beam AB of length L 6.5 ft supports a trapezoidal distributed load of peak intensity q, and minimum intensity q/2, that includes the weight of the beam (see figure). The beam is a steel W 12 14 wide-flange shape (see Table E-1(a), Appendix E). Calculate the maximum permissible load q based upon (a) an allowable bending stress sallow 18 ksi and (b) an allowable shear stress tallow 7.5 ksi. (Note: Obtain the moment of inertia and section modulus of the beam from Table E-1(a))
q 2
q
A L = 6.5 ft
B
W 12
14
Solution 5.10-7
b t tf S h h1 h1 L h 2 tf 3.97 in. 0.2 in. 0.225 in. 14.9 in.3 11.9 in. Mmax Mmax I 88.6 # in.4 Vmax a q + qb L 2 2 3 qL 4
Vmax
1q 2 1 q 2L L+ L 22 22 3 5 qL2 12
11.45 in. 6.5 ft s allow 18 ksi t allow 7.5 ksi
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SECTION 5.10
Shear Stresses in Beams with Flanges
461
(a) MAXIMUM LOAD BASED UPON BENDING STRESS M S 52 qL 12 S 12S sallow 5L2
s q
q
q q
3 qL 1 bh2 bh2 + th22 1 1 32It tallow32It 3 L1 bh2 3210 lb ft q 1270 lb/ft ; bh2 + th22 1 1
1270 lb/ft Vmax 1 bh2 8It bh2 + th22 1 1
(b) MAXIMUM LOAD UPON SHEAR STRESS tmax
Shear stress governs
Problem 5.10-8 A bridge girder AB on a simple span of length L 14 m supports a distributed load of maximum intensity q at midspan and minimum intensity q/2 at supports A and B that includes the weight of the girder (see figure). The girder is constructed of three plates welded to form the cross section shown. Determine the maximum permissible load q based upon (a) an allowable bending stress sallow 110 MPa and (b) an allowable shear stress tallow 50 MPa.
q q 2 q 2
450 mm 32 mm
A L = 14 m
B
16 mm 1800 mm
32 mm 450 mm
Solution 5.10-8 L 14 m
h b I I S RA 1864 mm h1 450 mm 1 1 bh3 12 3.194 * 10 2I h RB S tf 1800 mm 32 mm tw 16 mm
(a) MAXIMUM LOAD BASED UPON BENDING STRESS sallow Mmax 110 MPa 3 L qL 8 2 qLL 224 qLL 246
bh3 + tw h32 1 1
10
mm
4
3.427 * 107 mm3 s 3 qL 8
qL qL + 22 42
5 qL2 48 5 qL2 Mmax 48 S S sallow S 52 L 48 184.7 kN m ;
qmax
qmax
;
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(b) MAXIMUM LOAD BASED UPON SHEAR STRESS tallow Vmax tmax 50 MPa RA 3 qL 8 bh2 + th22 1 1 qmax qmax
3 qL 1 bh2 64It 3 L 1bh2
bh2 + th22 1 1 bh12 ; 184.7 kN/m ; tw h122
64 tallow Itw
Vmax 1 bh2 8It
247 kN/m
Bending stress governs: qmax
Problem 5.10-9 A simple beam with an overhang supports a uniform load of intensity q 1200 lb/ft and a concentrated load P 3000 lb (see figure). The uniform load includes an allowance for the wight of the beam. The allowable stresses in bending and shear are 18 ksi and 11 ksi, respectively. Select from Table E-2 (a), Appendix E, the lightest I-beam (S shape) A that will support the given loads.
(Hint: Select a beam based upon the bending stress and then calculate the maximum shear stress. If the beam is overstressed in shear, select a heavier beam and repeat.)
8 ft
P = 3000 lb
P = 3000 lb
q = 1200 lb/ft
B
12 ft 4 ft
C
Solution 5.10-9 sallow 18 ksi
q 1200 lb ft
Beam with an overhand t allow 11 ksi L 12 ft
P 3000 lb
Find moment at D (at Load P between A and B)
MD MD Mmax R A 8 ft q (8 ft)2 2 2.16 * 104 lb-ft
Sum moments about A & Solve for RB
4 1 qa Lb + P(8 ft + 16 ft) 3 2 12 ft 1.88 * 104 lb
2
1.28 * 104 lb-ft | MB| Mmax sallow 64.7 in.4 4.17 in. 0.425 in. h 2 tf h1 Vmax 1 bh2 8 It t h Mmax
Required section modulus:
S S 14.4 in.3
RB RB
Lightest beam is S 8 * 23 (from Table E-2(a))
I b tf h1 S 16.2 in.3 0.441 in. 8 in. 7.15 in. RB
Sum forces in vertical direction
RA RA Vmax Vmax MB MB q (16 ft) + 2P 6.4 * 10 lb RB (P + q4 ft) (4 ft)2 q 2
4 3
1.1 * 104 lb at B P (4 ft)
Check max. shear stress
tmax tmax bh2 + th22 1 1
2.16 * 10 lb-ft
3674 6 11,000 psi so ok for shear ;
Select S 8 * 23 beam
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463
Problem 5.10-10 A hollow steel box beam has the rectangular cross section shown in the figure. Determine the maximum allowable shear force V that may act on the beam if the allowable shear stress in 36 Mpa.
10 mm
20 mm
450 10 mm mm 20 mm
200 mm
Solution 5.10-10 tallow 36 MPa
Find Vallow t Vallow I VQ It tallowIt Q
Rectangular box beam
Q (200)a 450 450 ba b 2 4 (180)a 410 410 ba b 2 4
1.280 * 106 mm3 Vallow tallow It Q (36 MPa)(484.9 * 106 mm4)(20 mm) 1 (180)(410)3 12 1.280 * 106 mm3 273 kN ;
1 (200)(450)3 12 484.9*106mm4 2(10 mm)
t
20 mm
Problem 5.10-11 A hollow aluminum box beam has the square cross section shown in the figure. Calculate the maximum and minimum shear stresses tmax and tmin in the webs of the beam due to a shear force V 28 k.
1.0 in.
1.0 in.
12 in.
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Stresses in Beams (Basic Topics)
Solution 5.10-11
Square box beam
Q a b2 b ba b 2 4 b3) 1 a b2 b1 1 ba b 2 4
V t1 b b1 t VQ It t 2t1 2.0 in .
28 k 1.0 in . 12 in. 10 in.
28,000 lb tmax
13 (b 8 VQ It
91.0 in.3 1424 psi
(28,000 lb)(91.0 in.3) (894.67 in.4)(2.0 in.) ;
1.42 ksi
MINIMUM SHEAR STRESS IN THE WEB (AT LEVEL A.A) Q t1 Q tmin Ay b 2 (bt1)a b1 b 2 t1 b 2 b2 (b 8 a bt1 b(b 2 t1)
MOMENT OF INERTIA I 14 (b 12 b4 ) 1 894.67 in.
4
Q
b2) 1 (10 in.)2] 66.0 in.3 1033 psi
MAXIMUM SHEAR STRESS IN THE WEB (AT NEUTRAL AXIS) Q A2 y1 A1y1 b1 a A2y2 A1 b2 1 2 b 4 y2 1 b1 ab 22 b1 4 b ba b 2 b2 2
(12 in.) [(12 in.)2 8 VQ It
(28,000 lb)(66.0 in.3) (894.67 in4)(2.0 in.) ;
b1 b 2
1.03 ksi
1b ab 22
Problem 5.10-12 The T-beam shown in the figure has cross-sectional dimensions as follows: b 220 mm, t 15 mm, h 300 mm, and h1 275 mm . The beam is subjected to a shear force V 60 kN. Determine the maximum shear stress tmax in the web of the beam.
z c
y t h1 C b Probs 5.10.12 and 5.-10.13 h
Solution 5.10-12 h 300 mm h1
b tf tf 210 mm h h1 20 mm t V
280 mm 16 mm 68 kN
LOCATION OF NEUTRAL AXIS b1 h c c c1 c2 h 12 a h 2 h1 b + t h1 a h h1 b 2
b1 h 87.419 mm c h c1 c c2
h12 + t h1
87.419 mm 212.581 mm
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SECTION 5.10
Shear Stresses in Beams with Flanges
465
MOMENT OF INERTIA ABOUT THE z-AXIS Iweb Iweb Iflange 13 1 t c + t1 c1 32 3 5.287 * 107 mm4 1 b t 3 + b tf ac1 12 f tf 2 b
2
Iflange I
2.531 * 107 mm4 Iweb + Iflange I 7.818 * 107 mm4
tf 23
FIRST MOMENT OF AREA ABOVE THE z AXIS c2 Q tc2 2 VQ tmax tmax 19.7 MPa ; It
Problem 5.10-13 Calculate the maximum shear stress tmax in the web of the T-beam shown in the figure if b 10 in., t 0.5 in., h 7 in., h1 6.2 in., and the shear force V 5300 lb.
Solution 5.10-13 T-beam h 7 in. h1 6.2 in.
b tf V 10 in. h h1 5300 lb b1h 1.377 in. c h c1 c 1.377 in. c2 5.623 in. h 12 a h 2 h1 b + t h1 a h h1 b 2 t tf 0.5 in. 0.8 in.
MOMENT OF INERTIA ABOUT THE z-AXIS Iweb Iweb Iflange Iflange I 1 1 t c 3 + t1 c1 32 3 29.656 in.4 1 bt 3 + btf a c1 12 f 8.07 in.
4
tf23
LOCATION OF NEUTRAL AXIS
tf 2
b
2
c c c1 c2
b1 h
h12 + t h1
Iweb + Iflange
I
37.726 in.4
FIRST MOMENT OF AREA ABOVE THE z AXIS c2 Q tc2 2 VQ tmax tmax 2221 psi ; It
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CHAPTER 5
Stresses in Beams (Basic Topics)
Built-Up Beams
Problem 5.11-1 A prefabricated wood I-beam serving as a floor joist has the cross section shown in the figure. The allowable load in shear for the glued joints between the web and the flanges is 65 lb/in. in the longitudinal direction. Determine the maximum allowable shear force Vmax for the beam.
z 0.625 in. y 0.75 in.
O
8 in.
5 in.
0.75 in.
Solution 5.11-1
Wood I-beam
All dimensions in inches. Find Vmax based upon shear in the glued joints. Allowable load in shear for the glued joints is 65 lb/in. fallow f I VQ I bh3 12 65 lb/in. Vmax (b fallow I Q t)h3 1 12 1 (4.375)(8)3 12 170.57 in.4 ;
1 (5) (9.5)3 12 Q Qflange Af df
(5)(0.75)(4.375) Vmax fallowI Q
16.406 in.3
(65 lb/in.)(170.57 in.4) 16.406 in.3
676 lb
;
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SECTION 5.11
Built-Up Beams
467
Problem 5.11-2 A welded steel girder having the cross section shown in the figure is fabricated of two 300 mm * 25 mm flange plates and a 800 mm * 16 mm web plate. The plates are joined by four fillet welds that run continuously for the length of the girder. Each weld has an allowable load in shear of 920 kN/m. Calculate the maximum allowable shear force Vmax for the girder.
z 16 mm
y 25 mm
O
800 mm
25 mm 300 mm
Solution 5.11-2 h 850 mm h1
b tf I I 300 mm 25 mm b h3 12 (b t)h13 12
9 4
800 mm 16 mm
Qflange f allow
3.094 * 106 mm3 920 kN m f 2 fallow
t
(2 welds, one either side of web) f bt f a h 2 tf b Vmax VQ I Vmax ; fI Qflange
3.236 * 10 mm
Qflange
A f df Qflange
1.924 MN
y
Problem 5.11-3 A welded steel girder having the cross section shown in the figure is fabricated of two 20 in. * 1 in. flange plates and a 60 in. * 5/16 in. web plate. The plates are joined by four longitudinal fillet welds that run continuously throughout the length of the girder. If the girder is subjected to a shear force of 280 kips, what force F (per inch of length of weld) must be resisted by each weld?
1 in.
z 5 in. 16
O
60 in.
1 in. 20 in.
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CHAPTER 5
Stresses in Beams (Basic Topics)
Solution 5.11-3 h 62 in.
b tf I I 20 in. 1 in. bh3 12 (b
h1 t
60 in. 5 in. 16 Qflange Qflange V F F
btf a
h 2
tf
b VQ I 1994 * 103 lb.in. ;
610 in3 280 k VQflange 21 1994 lb/in. f F 2F
t)h13 12
4 4
4.284 * 10 in. Af df
Qflange
Problem 5.11-4 A box beam of wood is constructed of two 260 mm * 50 mm boards and two 260 mm * 25 mm boards (see figure). The boards are nailed at a longitudinal spacing s 100 mm. If each nail has a allowable shear force F 1200 N, what is the maximum allowable shear force Vmax?
z 50 mm
y 25 mm
O 50 mm
260 mm
260 mm
25 mm
Solution 5.11-4 Wood box beam All dimensions in millimeters.
b h s F f 260 b1 260 2(50) 160 310 h1 260 nail spacing 100 mm allowable shear force for one nail shear flow between one flange and both webs 2F s 2(1200 N) 100 mm 24 kN/ m
f I 1200 N Q
VQ I
Vmax
fallow I Q 411.125 * 106 mm4 (260)(25)(142.5)
1 (bh3 12 Qflange fallowI Q 10.7 kN
b1h3) 1 Afdf
3
926.25 * 10 mm4 Vmax (24 kN/ m)(411.25 * 106 mm4) . 926.25 * 103 mm3 ;
fallow
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SECTION 5.11
Built-Up Beams
469
Problem 5.11-5 A box beam is constructed of four wood boards as shown in the figure part (a). The webs are 8 in. 1 in. and the flanges are 6 in. 1 in. boards (actual dimensions), joined by screws for which the allowable load in shear is F 250 lb per screw.
(a) Calculate the maximum permissible longitudinal spacing smax of the screws if the shear force V is 1200 lb. (b) Repeat (a) if the flanges are attached to the webs using a horizontal arrangement of screws as shown in the figure part (b).
y
Flange 1 in. Web
Flange 1 in.
z 1 in.
O 1 in.
Web 8 in. 1 in.
8 in. 1 in. 6 in. 1 in. 1 in. (b)
6 in. (a)
Solution 5.11-5 Wood box beam F 250 lb V 1200 lb
(a) Vertical screws h b I Qa f smax smax 10 in. 6 in. bh 12
3
(b) Horizontal screws h 8 in. 8 in. bh 12
3
h1 t (b
6 in. 1 in. 2t) h13 12 I Qb 233.333 in.4 21 in.3
h1 t (b
8 in. 1 in.
b I I 27 in.3 329.333 in.4 Qb f smax smax
2t) h13 12 Qa
(b VQ I
2 t) t (3.5 in.) 2F s
bt (4.5 in.) VQ I 2F S 2FI VQa 5.08 in.
2FI VQb 4.63 in. ;
;
Problem 5.11-6 Two wood box beams (beams A and B) have the same outside dimensions (200 mm * 360 mm) and the same thickness (t 20 mm) throughout, as shown |in the figure on the next page. Both beams are formed by nailing, with each nail having an allowable shear load of 250 N. The beams are designed for a shear force V 3.2 kN.
y
y
A z O 360 mm z t= 20 mm 200 mm
B O 360 mm
(a) What is the maximum longitudinal spacing SA for the t= nails in beam A? 20 mm (b) What is the maximum longitudinal spacing sB for the nails in beam B? (c) Which beam is more efficient in resisting the shear force?
200 mm
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CHAPTER 5
Stresses in Beams (Basic Topics)
Solution 5.11-6 Two wood box beams Cross-sectional dimensions are the same.
All dimensions in millimeters. b h t F V I s f f 200 360 20 allowable load per nail shear force 3.2 kN 1 (bh3 12 b1 h3) 1 250 N b1 h1 200 360 2(20) 2(20) 160 320
(a) BEAM A Q Af df (bt)a h 2 t b 1 (200)(20)a b(340) 2
680 * 103 mm3 sA 2FI VQ (2)(250 N)(340.7 * 106 mm4) (3.2 kN)(680 * 103 mm3) ;
78.3 mm (b) BEAM B Q Afdf (b
340.69 * 106 mm4
longitudinal spacing of the nails shear flow between one flange and both webs 2F s VQ I smax 2FI VQ
2t)(t)a
h 2
t
b
1 (160)(20) (340) 2 544 * 103 mm3 sB 2FI VQ (2)(250 N)(340.7 * 106 mm4) (3.2 kN)(544 * 103 mm3) ;
97.9 mm
(c) BEAM B IS MORE EFFICIENT because the shear flow on the contact surfaces is smaller and therefore fewer ; nails are needed.
Problem 5.11-7 A hollow wood beam with plywood webs has the cross-sectional dimensions shown in the figure. The plywood is attached to the flanges by means of small nails. Each nail has an allowable load in shear of 30 lb. Find the maximum allowable spacing s of the nails at cross sections where the shear force V is equal to (a) 200 lb and (b) 300 lb.
3 in. 16 3 in.
3 in. 16
y
3 in. 4 8 in. 3 in. 4
z
O
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SECTION 5.11
Built-Up Beams
471
Solution 5.11-7 Wood beam with plywood webs All dimensions in inches.
b h F s f f I Q 3.375 8.0 h1 b1 3.0 6.5 30 lb
(a) V smax
200 lb 2FI VQ 300 lb (2.77 in.) a 200 b 300 2(30 lb)(75.344 in.4) (200 lb)(8.1563 in.3) ;
allowable shear force for one nail longitudinal spacing of the nails
2.77 in. (b) V
shear flow between one flange and both webs VQ I 2F s smax b1h3) 1 Afdf 2FI VQ
By proportion, smax 1.85 in. ;
1 (bh3 12 Qflange
75.3438 in.4 (3.0)(0.75)(3.625) 8.1563 in.3
Problem 5.11-8 A beam of T cross section is formed by nailing together two boards having the dimensions shown in the figure. If the total shear force V acting on the cross section is 1500 N and each nail may carry 760 N in shear, what is the maximum allowable nail spacing s?
z
y 240 mm 60 mm C 200 mm 60 mm
Solution 5.11-8 F allow V 1500 N
h1 t 200 mm 60 mm h b
760 N 240 mm
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS I 13 1 tc + t1 h1 32 3 + I c223 t2 b 2
260 mm
A bt + h1t A 2.64 * 104 mm2 LOCATION OF NEUTRAL AXIS (z AXIS) btah1 c2 c2 c1 c1 h1 t b + th1 2 2 A 170.909 mm h c2 89.091 mm Q Q
13 bt + bt a c1 12
1.549 * 108 mm4 bt a c1 t b 2
FIRST MOMENT OF AREA OF FLANGE
8.509 * 105 mm3 VQ I F s smax 92.3 mm ;
MAXIMUM ALLOWABLE SPACING OF NAILS f smax
F allowI VQ
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CHAPTER 5
Stresses in Beams (Basic Topics)
Problem 5.11-9 The T-beam shown in the figure is fabricated by welding together two steel plates. If the allowable load for each weld is 1.8 k/in. in the longitudinal direction, what is the maximum allowable shear force V?
0.6 in.
y
5.5 in. z C
4.5 in.
0.5 in.
Solution 5.11-9 T-beam (welded) k F allow 1.8 in.
h1 t1 h A 5.5 in. 0.6 in. 6 in. bt2 + h1t1 A 5.55 in.2 b t2 4.5 in. 0.5 in.
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS I 1 1 t c 3 + t1 1 c2 311 3 + I 1 b t 3 + bt2 ac2 12 2 t223 t2 2 b 2
20.406 in.4 b t2 ac2 VQ I t2 b 2
LOCATION OF NEUTRAL AXIS (z AXIS) t2 h1 bt2 + t1 h1 a + t2 b 2 2 A 2.034 in. 3.966 in. c1 h c2
FIRST MOMENT OF AREA OF FLANGE Q Q 4.014 in.3
c2 c2 c1
MAXIMUM ALLOWABLE SHEAR FORCE f 2F 2 Fallow I Q 18.30 k ;
Vmax Vmax
Problem 5.11-10 A steel beam is built up from a W 410 * 85 wide-flange beam and two 180 mm * 9 mm cover plates (see figure). The allowable load in shear on each bolt is 9.8 kN. What is the required bolt spacing s in the longitudinal direction if the shear force V = 110kN (Note: Obtain the dimensions and moment of inertia of the W shape from Table E-1(b).)
y
180 mm 9 mm cover plates
z
W 410 O
85
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SECTION 5.11
Built-Up Beams
473
Solution 5.11-10
V Aw Iw Acp h A 110 kN F allow 9.8 kN 417 mm I W 410 * 85 10800 mm2 hw 310 * 106 mm4 (180) (9) (2) mm2 for two plates hw + (9 mm) (2) Aw + Acp A h 2 1.404 * 104 mm2 Q + Acp ac 9 mm 2 b 2
4.57 * 108 mm4 9 mm b 2
First moment of area of one flange Q 180 mm (9 mm)a c 3.451 * 105 mm3 VQ I 2F s smax 236 mm ;
LOCATION OF NEUTRAL AXIS (z AXIS) c c 217.5 mm
Maximum allowable spacing of nails f smax
Moment of inertia about the neutral axis I Iw + 180 mm (9 mm) (2) 12
3
2 Fallow I VQ
Problem 5.11-11 The three beams shown have approximately the same cross-sectional area. Beam 1 is a W 14 82 with flange plates; Beam 2 consists of a web plate with four angles; and Beam 3 is constructed of 2 C shapes with flange plates.
(a) (b) (c) (d) Which design has the largest moment capacity? Which has the largest shear capacity? Which is the most economical in bending? Which is the most economical in shear?
Assume allowable stress values are: sa 18 ksi and ta 11 ksi. The most economical beam is that having the largest capacityto-weight ratio. Neglect fabrication costs in answering (c) and (d) above. (Note: Obtain the dimensions and properties of all rolled shapes from tables in Appendix E.)
8 0.52 4 Four angles 1 66 2 8 0.52 C 15 14 0.675 0.375
50
W 14
82
4 Beam 2 Beam 3
0.375
Beam 1
Solution 5.11-11 Built-up steel beam Beam 1: properties and dimensions for W14 * 82 with flange plates
AW b1 24 in.2 8 in. t1 hw 14.3 in. Iw 88l in.4 0.52 in.
h1 tf1 AI
hw + 2t1 0.855 in. AW + 2b1t1
bf1 tw1 AI
10.1 in. 0.51 in. 32.32 in.2
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I1 I1
Iw +
b1 + t3 hw t1 2 1 2 + b1 t1 a + b2 12 2 2
Q1
b1 t1 a
h1 2 a hw 2
t1 hw b + bf1 tf1 a 2 2 tf1 b 2
2
tf1 b 2
1.338 * 103 in4 + tw1
Beam 2: properties and dimensions for L6 * 6 * 1/2 angles with web plate Aa Ia t2 A2 I2 I2 5.77 in.2 19.9 in.4 0.675 in. 4Aa + b2t2 4Ia + Aa a b2 2 ca b2 h2 1.67 in. 14 in. b2 A2
2
Q1 a b2 2 b 2 2
98.983 in.3
ha
6 in. Q2 Q2 Q3 2 Aa a b3 t3 a h2 2
3
ca b + t2
32.53 in.2 t2 b3 2 12
78.046 in. h3 2 a
ca b 4 +
t3 hc b + 2bf3 tf3 a 2 2 hc 2 2 tf3 b
2
tf3 b 2
889.627 in.4 + 2tw3 I1 tw1 Q1 I2 t2 Q2 I2 2tw3 Q3
Beam 3: properties and dimensions for C15 * 50 with flange plates Ac b3 bf3 A3 I3 I3 14.7 in.2 4 in. 3.72 in. 2Ac + 2b3 t3 Ic 2 + b3t3 3 12 t3 hc tf3 15 in. 0.65 in. A3 32.4 in. Ic h3
2
Q3
79.826 in.3
404 in.4 hc + 2t3 tw3 0.716 in.
4.448 * 103 mm2 4.964 * 103 mm2 1.14 * 104 mm2 largest value
0.375 in.
2 + b3 t3 a
hc t3 2 + b2 2 2
985.328 in.4
(a) Beam with largest moment capacity; largest section modulus controls Mmax S1 S2 S3 sallow S 2I1 h1 2I2 h2 2I3 h3 S1 S2 S3 174.449 in.3 127.09 in.3 125.121 in.3 largest value
Itw Case (3) with maximum has the largest shear Q capacity ; (c) MOST ECONOMICAL BEAM IN BENDING HAS LARGEST
BENDING CAPACITY-TO-WEIGHT RATIO
S3 A3 S1 A1
3.862 in. 5.398 in.
6
S2 A2
3.907 in.
6
Case (1) is the most economical in bending. (d) MOST ECONOMICAL BEAM IN SHEAR HAS LARGEST
SHEAR CAPACITY-TO-WEIGHT RATIO
;
case (1) with maximum S has the largest moment capacity ; (b)
BEAM WITH LARGEST SHEAR CAPACITY: LARGEST RATIO CONTROLS
Itw/Q
I1 tw1 Q1 A1 6
0.213 I3 tw3 Q3 A3
6 0.273
I2 t2 Q2 A2
0.237
Vmax
tallow I tw O
Case (3) is the most economical in shear.
;
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SECTION 5.12
Beams with Axial Loads
475
Problem 5.11-12 Two W 310 * 74 steel wide-flange beams are bolted together to form a built-up beam as shown in the figure. What is the maximum permissible bolt spacing s if the shear force V 80 kN and the allowable load in shear on each bolt is F 13.5 kN (Note: Obtain the dimensions and properties of the W shapes from Table E-1(b).)
W 310
74
W 310
74
Solution 5.11-12 V 80 kN W 310 * 74
F allow hw c 13.5 kN Iw 310 mm hw 2 b d (2) 2 Aw 9420 mm
6 2 4
FIRST MOMENT OF AREA OF FLANGE Q Aw hw 2 2F s 2Fallow I VQ smax 180 mm ; Q 1.46 * 106 mm3
310 mm hw c
163 * 10 mm
Location of neutral axis (z axis) MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS I I c Iw + Aw a
MAXIMUM ALLOWABLE SPACING OF NAILS f smax VQ I
7.786 * 108 mm4
Beams with Axial Loads
When solving the problems for Section 5.12, assume that the bending moments are not affected by the presence of lateral deflections.
P = 25 lb
Problem 5.12-1 While drilling a hole with a brace and bit, you exert a downward force P 25 lb on the handle of the brace (see figure). The diameter of the crank arm is d 7/16 in. and its lateral offset is b 4-7/8 in. Determine the maximum tensile and compressive stresses st and sc, respectively, in the crank.
7 d = in. 16
b = 4 in. 8
7
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CHAPTER 5
Stresses in Beams (Basic Topics)
Solution 5.12-1
P M d d A S
Brace and bit
25 lb (compression) Pb (25 lb)(4 7/8 in.) 121.9 lb-in. diameter 7/16 in. pd2 4 pd 32
3
MAXIMUM STRESSES st P M + A S P A M S 25 lb 0.1503 in.
2
121.9 lb-in. + 0.008221 in.3 ; 14,660 psi 14,828 psi
166 psi + 14,828 psi sc 166 psi ;
0.1503 in.2 0.008221 in.3
14,990 psi
Problem 5.12-2 An aluminum pole for a street light weights 4600 N and supports an arm that weights 660 N (see figure). The center of gravity of the arm is 1.2 m from the axis of the pole. A wind force of 300 N also acts in the ( y) direction at 9 m above the base. The outside diameter of the pole (at its base) is 225 mm, and its thickness is 18 mm. Determine the maximum tensile and compressive stresses st and sc, respectively, in the pole (at its base) due to the weights and the wind force.
W2 = 660 N
1.2 m P1 = 300 N 9m
W1 = 4600 N 18 mm z x y x 225 mm y
Solution 5.12-2 W1 4600 N
W2 P1 d2 A A 660 N 300 N d1 2t p2 1d1 4
b h d1 d222
1.2 m 9m 225 mm t 18 mm d2 4 2
Mx Mx
W2 b + P1h 3.492 * 103 N # m a Pz Mx d1 + b A I2 ; Mx d1 b I2 ; (Moment)
MAXIMUM STRESS st
I I
p 1d1 4 64
st
5.77 * 103 kPa 5770 kPa a Pz A
1.171 * 104 mm2
6.317 * 107 mm4 sc sc
AT BASE OF POLE Pz Pz Vy W1 + W2 5.26 * 10 N P1 Vy
3
6.668 * 103 6668 kPa ;
(Axial force ) 300 N (Shear force)
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SECTION 5.12
Beams with Axial Loads
477
Problem 5.12-3 A curved bar ABC having a circular axis (radius r 12 in.) is loaded by forces P 400 lb (see figure). The cross section of the bar is rectangular with height h and thickness t. If the allowable tensile stress in the bar is 12,000 psi and the height h 1.25 in., what is the minimum required thickness tmin ?
B A P 45
h C 45 r h t P
Solution 5.12-3
Curved bar
TENSILE STRESS st P M + A S
3Pr(2 12) P + 2 ht th r 12) d h r 12) d h
r e
radius of curved bar r ra1 r cos 45 12) 1 b 12
P c1 + 3(2 ht MINIMUM THICKNESS tmin
M
Pe
Pr (2 2
P c1 + 3(2 hsallow
SUBSTITUE NUMERICAL VALUES: P 400 lb s allow ; 12,000 psi r tmin 12 in. h 0.477 in. 1.25 in.
CROSS SECTION h height t thickness A ht S 12 th 6
Problem 5.12-4 A rigid frame ABC is formed by welding two steel pipes at B (see figure). Each pipe has cross-sectional area A 11.31 * 103 mm2, moment of inertia I 46.37 * 106 mm4, and outside diameter d 200 mm. Find the maximum tensile and compressive stresses st and sc, respectively, in the frame due to the load P 8.0 kN if L H 1.4 m.
B d A d L L P d H C
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CHAPTER 5
Stresses in Beams (Basic Topics)
Solution 5.12-4
Rigid frame
AXIAL FORCE: N RA sin a RAL P sin a 2 PL 2
BENDING MOMENT: M TENSILE STRESS st Load P at midpoint B REACTIONS: RA BAR AB: tan a sin a d c H L 1H2 + L2 H RC P 2 N Mc + A I
P sin a PLd + 2A 4I
SUBSTITUTE NUMERICAL VALUES P sina A st 8.0 kN L 1/12 d H 1.4 m a 45 200 mm 46.37 * 106 mm4
11.31 * 103 mm2 I (8.0 kN)(1/12) 2(11.31 * 103 mm2)
(8.0 kN)(1.4 m)(200 mm) + 4(46.37 * 106 mm4) 0.250 MPa + 12.08 MPa 11.83 MPa (tension) sc N A Mc I ; 12.08 MPa ;
diameter d/2
0.250 MPa
12.33 MPa (compression)
Problem 5.12-5 A palm tree weighing 1000 lb is inclined at an angle of 60 (see figure). The weight of the tree may be resolved into two resultant forces, a force P1 900 lb acting at a point 12 ft from the base and a force P2 100 lb acting at the top of the tree, which is 30 ft long. The diameter at the base of the tree is 14 in. Calculate the maximum tensile and compressive stresses st and sc, respectively, at the base of the tree due to its weight.
30 ft
P2 = 100 lb
12 ft
P1 = 900 lb 60
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SECTION 5.12
Beams with Axial Loads
479
Solution 5.12-5
Palm tree
M P1L1 cos 60 82,800 lb-in. N FREE-BODY DIAGRAM P1 P2 L1 L2 d A S 900 lb 100 lb 12 ft 30 ft 14 in. pd2 4 pd3 32 153.94 in.2 269.39 in.3 144 in. 360 in. (P1 + P2) sin 60 (1000 lb) sin 60 866 lb P2 L2 cos 60 [(900 lb)(144 in.) + (100 lb)(360 in.)] cos 60
MAXIMUM TENSILE STRESS st N M + A S 866 lb 153.94 in.
2
82,800 lb-in. + 269.39 in.3 ; 302 psi
5.6 psi + 307.4 psi MAXIMUM COMPRESSIVE STRESS sc 5.6 psi 307.4 psi
313 psi
;
Problem 5.12-6 A vertical pole of aluminum is fixed at the base and pulled at the top by a cable having a tensile force T (see figure). The cable is attached at the outer edge of a stiffened cover plate on top of the pole and makes an angle a 20 at the point of attachment. The pole has length L 2.5 m and a hollow circular cross section with outer diameter d2 280 mm and inner diameter d1 220 mm. The circular cover plate has diameter 1.5d2. Determine the allowable tensile force Tallow in the cable if the allowable compressive stress in the aluminum pole is 90 MPa.
1.5 d2
a L
T
d2
d1 d2
Solution 5.12-6 sallow 90 MPa
d2 t A A 280 mm d2 2 p 1d 2 42 d1 a d1 22
d1
220 mm
PN V
T cos (a) T sin (a) VL + PN a
(Axial force) (Shear force) 1.5 d2 b 2 (Moment).
20 I I
L
2.5 m d1 4 2
M
p 1d 4 64 2
2.356 * 104 mm2
1.867 * 108 mm4
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Allowable Tensile Force sc PN A M d2 I2 T cos (a) A T cos (a) a I 1.5 d2 b 2 d2 2
Tallow cos (a) + A Tallow 108.6 kN
sallow sin (a) L + cos (a) a I ; 1.5 d2 b 2 d2 2
T sin (a) L
Problem 5.12-7 Because of foundation settlement, a circular tower is leaning at an angle a to the vertical (see figure). The structural core of the tower is a circular cylinder of height h, outer diameter d2, and inner diameter d1. For simplicity in the analysis, assume that the weight of the tower is uniformly distributed along the height. Obtain a formula for the maximum permissible angle a if there is to be no tensile stress in the tower.
h
d1 d2 a
Solution 5.12-7
Leaning tower
CROSS SECTION A I W a weight of tower angle of tilt I A c p2 (d 42 p4 (d 64 2 p2 (d 64 2 d2 + d2 2 1 16 d2 2 h Wa bsin a 2 d2 ) 1 d4) 1 d2)(d2 + d2) 1 2 1
AT THE BASE OF THE TOWER N W cos a M
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481
TENSILE STRESS (EQUAL TO ZERO) st N Mc + A I Wcosa A 0
cos a A
hd2 sin a 4I
tan a
4I hd2A
d22 + d12 4hd2
d2 Wh + a sinab a b I2 2
MAXIMUM ANGLE a d22 + d12 a arctan 4hd2
;
Problem 5.12-8 A steel bar of solid circular cross section and length L 2.5 m is subjected to an axial tensile force T 24 kN and a bending moment M 3.5 kN m (see figure).
(a) Based upon an allowable stress in tension of 110 MPa, determine the required diameter d of the bar; disregard the weight of the bar itself. (b) Repeat (a) including the weight of the bar.
z
y d M z-direction T L x
Solution 5.12-8 M 3.5 kN # m
g steel sallow A 77 kN m
3
T L
24 kN 2.5 m
SOLVE NUMERICALLY FOR
d (SUBSTITUTE sallow)
d
70 mm
;
(b) INCLUDE WEIGHT OF BAR d 2 p4 d 64 Mmax M+ Agsteel L2 2
110 MPa c I
p2 d 4
AT TOP OF BEAM AT SUPPORT st sallow Mmax d T + A I2
(a) DISREGARD WEIGHT OF BAR MAX. TENSILE STRESS AT TOP OF BEAM AT SUPPORT smax T Md + A I2 4T pd
2
T Md + p2 p 42 d d 4 64 p d3
SUBSTITUTE MMAX FROM ABOVE, SOLVE FOR d NUMERICALLY d 76.5 mm ;
sallow
32 M +
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Problem 5.12-9 A cylindrical brick chimney of height H weighs w 825 lb/ft of height (see figure). The inner and outer diameters are d1 3 ft and d2 4 ft, respectively. The wind pressure against the side of the chimney is p = 10 lb/ft2 of projected area. Determine the maximum height H if there is to be no tension in the brickwork
p w H d1 d2
Solution 5.12-9
Brick Chimney
d2
I I A
p4 (d 64 2
d4) 1
p2 (d 64 2 c d2 2 qH a
d2) (d2 1 2
d2 ) 1
H
w
q
12 (d2 + d2) 1 16
AT BASE OF CHIMNEY N
V M N
W
wH
M
H b 2
1 pd H2 22
TENSILE STRESS (EQUAL TO ZERO) s1 pd2 H2 2wH Md2 N + A 2I
2 d2 + d1 2 8d2
0
or
M N
2I Ad2
p q d2 d1 W
wind pressure intensity of load outer diameter inner diameter total weight of chimney wH pd2
SOLVE FOR H
H
w(d2 + d2) 2 1 4pd2 2 4 ft d1
;
SUBSTITUTE NUMERICAL VALUES w q 825 lb/ft 10 lb/ft
2
d2 Hmax
3 ft ;
CROSS SECTION A p2 (d2 4 d2) 1
32.2 ft
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Problem 5.12-10 A flying buttress transmits a load P 25 kN, acting at an angle of 60 to the horizontal, to the top of a vertical buttress AB (see figure). The vertical buttress has height h 5.0 m and rectangular cross section of thickness t 1.5 m and width b 1.0 m (perpendicular to the plane of the figure). The stone used in the construction weighs y 26 kN/m3. What is the required weight W of the pedestal and statue above the vertical buttress (that is, above section A) to avoid any tensile stresses in the vertical buttress?
Flying buttress P W 60 A h t B B A t 2 t
h
Solution 5.12-10 Flying buttress FREE-BODY DIAGRAM OF VERTICAL BUTTRESS
CROSS SECTION A S bt 12 bt 6 (1.0 m)(1.5 m) 1.5 m2 0.375 m3 1 (1.0 m)(1.5 m)2 6
AT THE BASE N W + WB + P sin 60 W + 195 kN + (25 kN) sin 60 W + 216.651 kN M (Pcos 60) h 62.5 kN # m TENSILE STRESS (EQUAL TO ZERO) P h t b b g WB 25 kN 5.0 m 1.5 m width of buttress perpendicular to the figure 1.0 m 26 kN/m bthg 195 kN
3
(25 kN) (cos 60) (5.0 m)
st
N M + A S W + 216.651 kN 1.5 m
2
62.5 kN # m + 0.375 m3 0
0
or W
W
216.651 kN + 250 kN ;
33.3 kN
weight of vertical buttress
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Problem 5.12-11 A plain concrete wall (i.e., a wall with no steel reinforcement) rests on a secure foundation and serves as a small dam on a creek (see figure). The height of the wall is h 6.0 ft and the thickness of the wall is t 1.0 ft.
(a) Determine the maximum tensile and compressive stresses st and sc, respectively, at the base of the wall when the water level reaches the top (d h). Assume plain concrete has weight density gc 145 Ib/ft3. (b) Determine the maximum permissible depth dmax of the water if there is to be no tension in the concrete.
t
h d
Solution 5.12-11
Concrete wall
h t b gc gw d W W F
height of wall thickness of wall width of wall (perpendicular to the figure) width density of concrete weight density of water depth of water weight of wall bhtgc resultant force for the water pressure
STRESSES AT THE BASE OF THE WALL (d DEPTH OF WATER) st sc W M + A S W A M S hgc + hgc d 3gw t2 d 3gw t2 h Eq. (1) Eq.(2)
(a) STRESSES AT THE BASE WHEN d h t gc gw 6.0 ft 1.0 ft 72 in. d 12 in. 145 lb/in.3 1728 62.4 lb/in.3 1728 72 in.
MAXIMUM WATER PRESSURE = gw d F M A 1 (d)(gw d) (b) 2 d Fa b 3 bt S 13 bd gw 6 12 bt 6 12 bd gw 2
145 lb/ft3 62.4 Ib/ft3
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485
Substitute numerical values into Eqs. (1) and (2): st sc 6.042 psi + 93.600 psi 6.042 psi 93.600 psi 87.6 psi 99.6 psi ; ;
d3 dmax
(72 in.)(12 in.)2 a 28.9 in. ;
145 b 62.4
24,092 in.3
(b) MAXIMUM DEPTH FOR NO TENSION Set st = 0 in Eq. (1): hgc + d3gw t
2
0
d3
ht2 a
gc b gw
Problem 5.12-12 A circular post, a rectangular post, and a post of cruciform cross section are each compressed by loads that produce a resultant force P acting at the edge of the cross section (see figure). The diameter of the circular post and the depths of the rectangular and cruciform posts are the same.
(a) For what width b of the rectangular post will the maximum tensile stresses be the same in the circular and rectangular posts? (b) Repeat (a) for the post with cruciform cross section. (c) Under the conditions described in parts (a) and (b), which post has the largest compressive stress?
P P P
x
4
b =b 4
b 4 d d d
Load P here d =d 4
Solution 5.12-12 (a) EQUAL MAXIMUM TENSILE STRESSES
CIRCULAR POST A p2 d 4 S p3 d 32 4P pd
2
COMPRESSION sc Pd 2 RECTANGULAR POST 12 P pd
2
P A 4P pd
2
M S 16 P pd
2
20P pd 2 Pd 2 2P bd
M
Tension st P M + A S 16 P + pd
2
A TENSION st
bd
S
bd2 6 P bd
M 3P bd
P M + A S
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COMPRESSION s c
P A
M S
P bd
3P bd
4P bd
Equate compressive stresses & solve for b 12 P pd
2
Equate tensile stress expressions, solve for b 12 P pd
2
2P 3bd
3 pd
1 b
b
pd 3
;
2P bd
6 pd
1 b
b
pd 6
;
(c) THE LARGEST COMPRESSIVE STRESS substitute expressions for b above & compare compressive stresses CIRCULAR POST 20 P pd2 4P pd a bd 6 24 P pd 2
(b) CRUCIFORM CROSS SECTION A S M cbd c
3
a
bd bd 22
3
bd bd 12 + ab d 2 12 2 2 12 d Pd 3 2 a bd2 b 32 st 16P 3bd P M + A S 16 P 4P + 3bd 3bd
3 bd 2 32
sc
RECTANGULAR POST sc
TENSION
12 P 3bd M S
CRUCIFORM POST 20 P sc pd 3 d 3
20 P pd 2
COMPRESSION
sc 4P 3bd
P A
Rectangular post has the largest compressive stress ; 20 P 3bd
16 P 3bd
Problem 5.12-13 Two cables, each carrying a tensile force P 1200 lb, are bolted to a block of steel (see figure). The block has thickness t 1 in. and width b 3 in.
P t
b
P
(a) If the diameter d of the cable is 0.25 in., what are the maximum tensile and compressive stresses st and sc, respectively, in the block? (b) If the diameter of the cable is increased (without changing the force P), what happens to the maximum tensile and compressive stresses?
Solution 5.12-13
Steel block loaded by cables
P t
1200 lb 1.0 in. e
d
0.25 in. t d + 2 2 0.625 in.
b
width of block 3.0 in.
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487
CROSS SECTION OF BLOCK A bt 30 in.2 I 13 bt 12 0.25 in.4
MAXIMUM COMPRESSIVE STRESS (AT BOTTOM OF BLOCK) y sc t 2 Pey P + A I 1200 lb 3 in.2 400 psi + 1500 psi (1200 lb)(0.625 in.)( 0.25 in.4 1100 psi ; 0.5 in.) 0.5 in.
(a) MAMIMUM TENSILE STRESS (AT TOP OF BLOCK) y st t 2 0.5 in.
Pey P + A I 1200 lb 3 in.2 + (1200 lb)(0.625 in.)(0.5 in.) 0.25 in.4 1900 psi ;
(b) IF d IS INCREASED, increase the eccentricity e increases and both stresses in magnitude.
400 psi + 1500 psi
Problem 5.12-14 A bar AB supports a load P acting at the centroid of the end cross section (see figure). In the middle region of the bar the cross-sectional area is reduced by removing one-half of the bar.
(a) If the end cross sections of the bar are square with sides of length b, what are the maximum tensile and compressive stresses st and sc, respectively, at cross section mn within the reduced region? (b) If the end cross sections are circular with diameter b, what are the maximum stresses st and sc?
m B P b (b) A b b 2 n b (a)
b 2 b
b 2
Solution 5.12-14 (a) SQUARE BAR
b (b)a b 2 b Pa b 4 c
Bar with reduced cross section
(b) CIRCULAR BAR Cross section mn is a semicircle b4 96 A 1 pb2 a b 24 b4 0.1098a b 2 Pa 2b b 3p pb2 8 0.3927 b2 b2 2 b 4 1 b3 (b)a b 12 2
Cross section mn is a rectangle. A M STRESSES st sc P Mc + A I P A Mc I 2P b2 2P b2 + 6P b2 6P b2 8P b2 4P b2 ; ; I
From Appendix D, Case 10: I M 0.006860 b4 0.2122 Pb
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FOR TENSION ct 4r 3p 2b 3p 0.2122 b sc b 2 2b 3p 0.2878 b
2.546 P A
P
2
FOR COMPRESSION: cc r ct
b Mcc I
+ 6.564
P b
2
9.11
P b2
;
P 0.3927 b 2.546 P b
2 2
(0.2122 Pb)(0.2878 b) 0.006860 b4 8.903 P b
2
STRESSES st Mct P + A I P 0.3927 b2 + (0.2122 Pb)(0.2122 b) 0.006860 b4
6.36
P b2
;
Problem 5.12-15 A short column constructed of a W 12 * 35 wide-flange shape is subjected to a resultant compressive load P 12 k having its line of action at the midpoint of one flange (see figure).
(a) Determine the maximum tensile and compressive stresses st and sc, respectively, in the column. (b) Locate the neutral axis under this loading condition. (c) Recompute maximum tensile and compressive stresses if a C 10 15.3 is attached to one flange, as shown.
P = 25 k z C
y C 10 15.3 (Part c only)
2 W 12 35
1
1
2
Solution 5.12-15 Column of wide-flange shape PROPERTIES OF EACH SHAPE:
W 12 * 35 Aw hw tf Iw 10.3 in.3 12.5 in. 0.52 in. 285 in.
4
C 10 * 15.3 Ac twc xp Ic 4.48 in.2 0.24 in. 0.634 in. 2.27 in. (2-2 axis)
4
sc
P Aw
Pew c Iw w
sc
5711 psi
;
(b) NEUTRAL AXIS (W SHAPE ALONE) y0 Iw Aw ew y0 4.62 in. ;
(a) THE MAXIMUM TENSILE AND COMPRESSIVE STRESSES LOCATION OF CENTROID FOR W 12 35 ALONE cw P st hw 2 25 k cw ew 6.25 in. hw 2 tf 2 st ew 5.99 in. ;
(C) COMBINED COLUMN, W 12 * 35 with C 10 * 15.3 h h A hw + twc 12.74 in. Aw + Ac A 14.78 in.2
Pew P + c Aw Iw w
857 psi
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489
LOCATION OF CENTROID OF COMBINED SHAPE hw Aw a b + Ac (h 2 A Iw + Aw ac + Ic + Ac (h I 394.334 in.4 hw 2 b 2 xp c)2 y0 P xp) c 8.025 in. st sc 25 k e hw P Pe + c A I P A I Ae Pe (h I y0
tf 2 st c)
c 453 psi sc
e
4.215 in. ;
c I
2951 psi
; ;
6.33 in. (from centroid)
Problem 5.12-16 A short column of wide-flange shape is subjected to a compressive load that produces a resultant force P 55 kN acting at the midpoint of one flange (see figure).
(a) Determine the maximum tensile and compressive stresses st and sc, respectively, in the column. (b) Locate the neutral axis under this loading condition. (c) Recompute maximum tensile and compressive stresses if a 120 mm 10 mm cover plate is added to one flange as shown.
z
P = 55 kN
y
Cover plate (120 mm 10 mm) (Part c only) y P
C 8 mm z 12 mm 160 mm C 200 mm
Solution 5.12-16 P 55 kN
(a) MAXIMUM TENSILE AND COMPRESSIVE STRESSES FOR W
SHAPE ALONE
e st sc
d 2
tf 2
e
94 mm st sc 3.27 MPa 24.2 MPa ; ;
PROPERTIES AND DIMENSIONS FOR W SHAPE b tf Aw Aw Iw Iw 160 mm 12 mm bd (b d tw
3
P Pe d + Aw Iw 2 P Aw Pe d Iw 2
200 mm 8 mm 2 tf) 2 tf)3
2
tw) (d
(b) NEUTRAL AXIS (W SHAPE ALONE) y0 Iw Aw e y0 76.2 mm ;
5.248 * 10 mm (b tw) (d bd3 12
7
12 (c) COMBINED COLUMN-W SHAPE & COVER PLATE bp 120 mm tp 10 mm
3.761 * 10 mm4
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h h A
d
tp A 6.448 * 103 mm2
I e st sc
210 mm Aw + bp tp
4.839 * 107 mm4 tf d c e 2 P Pe + c A I P Aw Pe (h Iw st c)
74.459 mm 1.587 MPa sc ; ;
CENTROID OF COMPOSITE SECTION tp d Aw + bp tp ad + b 2 2 c A c I 119.541 mm Iw + Aw ac bp t3 p + 12 d b 2
2
20.3 MPa
NEUTRAL AXIS y0 tp 2
2
I Ae
y0
100.8 mm (from centrioid)
+ bp tp a d +
cb
Problem 5.12-17 A tension member constructed of an 1 L4 4 2 inch angle section (see Table E-4(a) in Appendix E) is subjected to a tensile load P 12.5 kips that acts through the point where the midlines of the legs intersect [see figure part (a)].
(a) Determine the maximum tensile stress st in the angle section. (b) Recompute the maximum tensile stress if two angels are used and P is applied as shown in the figure part (b).
2 3 1 P 2 (a) 3 (b) P L4 C 4 1 1 2 2L4 C 4 1 2
Solution 5.12-17
Angle section in tension
(b) TWO ANGLES: L 4 * 4 * 1/2 0.776 in. A t c e 1.315 in IL e 1.699 in. 2.258 in.
4
(a) ONE ANGLE: L 4 * 4 * 1/2 AL t c e P c1 I3 M 3.75 in.2 0.5 in. ac 1.18 in. c 12 Pe t b 12 2 c1 I3 M rmin
2AL 0.5 in 1.18 in 5.52 in.4 (2-2 axis) ac 12.5 k 2IL Pe I M 11.04 in.4 11.625 k-in. t b 2 e 0.93 in.
12.5 k
2 AL rmin
P I M
16.44 k-in.
MAXIMUM TENSILE STRESS OCCURS AT CORNER st Mc1 P + AL I3 st 15.48 ksi ;
MAXIMUM TENSILE STRESS OCCURS AT THE LOWER EDGE st P Mc + A I st 2.91 ksi ;
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491
Problem 5.12-18 A short length of a 200 * 17.1 channel is subjected to an axial compressive force P that has its line of action through the midpoint of the web of the channel [(see figure(a)].
Two L 76 y C 200 17.1 P
76 y P C
6.4 angles
(a) Determine the equation of the neutral axis under this z z C loading condition. (b) If the allowable stresses in tension and compression (a) are 76 MPa and 52 MPa respectively, find the maximum permissible load Pmax. (c) Repeat (a) and (b) if two L 76 76 6.4 angles are added to the channel as shown in the figure part (b). See Table E-3(b) in Appendix E for channel properties and Table E-4(b) for angle properties.
C 200 17.1 (b)
Solution 5.12-18
P
tw = 5.59mm bf = 57.4
sc 1 Ac e c Ic 1 ;
Pmax
67.3 kN
C 200 * 17.1 Ac Ic c2 st 2170 mm2 bf c1 c2 dc 203 mm c1 42.9 mm 52 MPa 14.5 mm 0.545 * 106 mm4 (z-axis)
(c) COMBINED COLUMN WITH 2-ANGLES L 76 * 76 * 6.4 AL cL 929 mm2 21.2 mm 4.028 * 103 mm2 h 133.4 mm IL 0.512 * 106 mm4
ALLOWABLE STRESSES 76 MPa s c tw 2 ECCENTRICITY OF THE LOAD e c1 e 11.705 mm
COMPOSITE SECTION A h Ac + 2 AL bf + 76 mm A
(a) LOCATION OF THE NEUTRAL AXIS (CHANNEL ALONE) y0 Ic Ac # e y0 21.5 mm ;
CENTROID OF COMPOSITE SECTION Ac 1bf c12 + 2 AL 1bf + cL2 c A c I I Ic + Ac 1bf 59.367 mm c1 c22 c22
(b) FIND PMAX st Pe P + c A I2 165.025 kN P Pe c A I1 P st 1 e + c2 Ac Ic
+ 2 IL + 2 AL 1bf + cL 2.845 * 106 mm4 bf tw 2 c
e bf
e
4.762 mm
57.4 mm
P sc
LOCATION OF THE NEUTRAL AXIS y0 I Ae y0 148.3 mm ;
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CHAPTER 5
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148.3 mm 7 h P Pe + c A I
133.4 mm
;
P
Thus, this composite section has no tensile stress sc
sc 1 e +c A I
Pmax
149.6 kN
;
Stress Concentrations
The problems for Section 5.13 are to be solved considering the stress-concentration factors.
M h d M
Problem 5.13-1 The beams shown in the figure are subjected to bending moments M 2100 lb-in. Each beam has a rectangular cross section with height h 1.5 in. and width b 0.375 in. (perpendicular to the plane of the figure).
(a) For the beam with a hole at midheight, determine the maximum stresses for hole diameters d 0.25, 0.50, 0.75, and 1.00 in. (b) For the beam with two identical notches (inside height h1 1.25 in .), determine the maximum stresses for notch radii R 0.05, 0.10, 0.15, and 0.20 in.
Probs. 5.13.1 through 5.13-4
(a) 2R M h h1 M
(b)
Solution 5.13-1 M 2100 lb-in. h
(a) BEAM WITH A HOLE 1 d h 2
1.5 in. b
0.375 in.
(b) BEAM WITH NOTCHES h1 1.25 in. h h1 1.5 in. 1.25 in. 1.2
Eq.(5-57): sc
6Mh b(h3 3.375 d3 ) d
3
Eq. (5-58) (1) snom 6M bh2 1 R h1 0.04 0.08 0.12 0.16 21,500 psi
50,400 1 d h 2 12Md b(h3 d3 ) 67,200 d 3.375 d h 0.1667 0.3333 0.5000 0.6667 sc Eq. (1) (psi) 15,000 15,500 17,100 d3
Eq.(5-56): sB
R (in) (2) 0.05 0.10 0.15 0.20
K (Fig. 5-50) 3.0 2.3 2.1 1.9
sm ax Ks nom sm ax ( i ps ) 65,000 49,000 45,000 41,000
d (in.) 0.25 0.50 0.75 1.00
sB Eq. (2) (psi) 17,100 28,300
sm ax (psi) 15,000 15,500 17,100 28,300
NOTE: The larger the notch radius, the smaller the stress.
NOTE: The larger the hole, the larger the stress.
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493
Problem 5.13-2 The beams shown in the figure are subjected to bending moments M 250 N # m. Each beam has a rectangular cross section with height h 44 mm and width b 10 mm (perpendicular to the plane of the figure).
(a) For the beam with a hole at midheight, determine the maximum stresses for hole diameters d 10, 16, 22 and 28 mm. (b) For the beam with two identical notches (inside height h1 40 mm ), determine the maximum stresses for notch radii R 2, 4, 6, and 8 mm.
Solution 5.13-2 M 250 N # m h
1 d h 2 sc
44 mm b
10 mm
(b) BEAM WITH NOTCHES h1 40 mm h h1 44 mm 40 mm 6M bh2 1 1.1
(a) BEAM WITH A HOLE Eq. (5-57): 6Mh b(h3 d3 ) 6.6 * 10 85,180
6
Eq. (5-58): snom MPa (1) R (mm) R h1 0.05 0.10 0.15 0.20
93.8 MPa
d3
d 1 h 2 sB
Eq. (5-56): 12Md 300 * 103d 85,180 d MPa 3 (2)
K (Fig. 5-50) 2.6 2.1 1.8 1.7
smax Ks nom smax (MPa) 240 200 170 160
b(h3
d3 )
2 4 6 8
d (mm) 10 16 22 28
d h 0.227 0.364 0.500 0.636
sB sc Eq. (2) Eq. (1) (MPa) (MPa) 78 81 89 89 133
sm ax (MPa) 78 81 89 133
NOTE: The larger the notch radius, the smaller the stress.
NOTE: The larger the hole, the larger the stress.
Problem 5.13-3 A rectangular beam with semicircular notches, as shown in part (b) of the figure, has dimensions h 0.88 in. and h1 0.80 in. The maximum allowable bending stress in the metal beam is smax 60 ksi, and the bending moment is M 600 lb-in. Determine the minimum permissible width bmin of the beam.
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Solution 5.13-3 Beam with semicircular notches h1 0.80 in. h 0.88 in.
smax h R h1 60 ksi M h1 + 2R R 0.04 in. 0.80 in. 600 lb-in. 1 (h 2 h1) 0.04 in.
smax 60 ksi
Ksnom 2.57c
Ka
6M bh2 1
b d
6(600 lb-in.) b(0.80 in.)2 ;
0.05
Solve for b: bmin L 0.24 in.
From Fig. 5-50: K L 2.57
Problem 5.13-4 A rectangular beam with semicircular notches, as shown in part (b) of the figure, has dimension h 120 mm and h1 100 mm . The maximum allowable bending stress in the plastic beam is smax 6 MPa, and the bending moment is M 150 N # m. Determine the minimum permissible width bmin of the beam.
Solution 5.13-4 Beam with semicircular notches h1 100 mm h 120 mm
smax h R h1 6 MPa M h1 + 2R R 10 mm 100 mm 150 N # m 1 (h 2 0.10 h1) 10 mm
smax 6 MPa
Ksnom 2.20 c
Ka
6M bh2 1
b d
6(150 N # m) b(100 mm)2 ;
Solve for b: bmin L 33 mm
From Fig.5-50: K L 2.20
Problem 5.13-5 A rectangular beam with notches and a hole (see figure) has dimensions h 5.5 in., h1 5 in., and width b 1.6 in. The beam is subjected to a bending moment M 130 k-in., and the maximum allowable bending stress in the material (steel) is smax 42,000 psi.
(a) What is the smallest radius Rmin that should be used in the notches? (b) What is the diameter dmax of the largest hole that should be drilled at the midheight of the beam?
2R M h1 h d M
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SECTION 5.13
Stress Concentrations
495
Solution 5.13-5 Beam with notches and a hole h 5.5 in. h1 5 in. b 1.6 in.
M 130 k-in. smax 42,000 psi
(b) LARGEST HOLE DIAMETER Assume sB 1 d 7 and use Eq. (5-56). h 2 12Md b(h3 d3) 12(130 k-in.)d (1.6 in.)[(5.5 in.)3 166.4 ; 0 d3] or
(a) MINIMUM NOTCH RADIUS h h1 snom K 5.5 in. 5 in. 6M bh2 1 smax snom 1.1 19,500 psi
42,000 psi d3 + 23.21d 42,000 psi 19,500 psi 2.15 1.1, we get dmax
Solve numerically: 4.13 in.
From Fig. 5-50, with K R L 0.090 h1 Rmin L 0.090h1
h 2.15 and h1
0.45 in.
;
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