in 5
Stresses 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).
d
R
Solution 5.4-1
Steel wire R 24 in. d 1 in. 16
R Cylinder
d
From Eq. (5-4): y emax r d2 d R d 2 2R d Substitute numerical values: emax 1 16 in. 2(24 in.) 1 16 in. 1300 10
6
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
d L
3 mm 2 r r
max
0.0024
L 2 d L (3 mm) 0.0024 3.93 m
From Eq. (5-4): emax L min y r d emax d2 L2
285
286
CHAPTER 5
Stresses in Beams (Basic Topics)
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 quartercircular 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
d
Angle equals 90 or /2 radians, r radius of curvature L L d L length of 90 bend 46 ft 552 in. 4.5 in. 2 r 4 r 2 emax r L 2 d 4L 2L emax y r d2 2L 6400 10
6
L
r r
radius
4.5 in. 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 , the curvature , and the vertical deflection at the end of the beam. Solution 5.4-4 Cantilever beam
L A C B M0
A B L M0
Assume that the deflection curve is nearly flat. Then the distance BC is the same as the length L of the beam. sin u L r 1.5 m 75 m 0.02
0
length of beam 1.5 m 0.001 max y y 75 mm emax r y 75 mm r 75 m emax 0.001 1 k 0.01333 m 1 r
L L
arcsin 0.02 0.02 rad (1 cos ) (75 m)(1 cos (0.02 rad)) 15.0 mm L NOTE: 100, which confirms that the deflection curve is nearly flat.
SECTION 5.4
Longitudinal Strains in Beams
287
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 0.25 in. Determine the longitudinal normal strain at the top surface of the strip.
M0
M0
t L -- 2 L -- 2
Solution 5.4-5
Thin strip of steel
0
The deflection curve is very flat (note that L/ and therefore is a very small angle. sin u L2 r sin u (1 L2 ( is in radians) r cos )
80)
For small angles, u
M0 M0
L -- 2
L -- 2
Substitute numerical values ( 0.25 r 1 cos 10 r
r 1
cos cos
L 2r
inches):
L
20 in. t 0.25 in.
0.2 in.
Solve numerically:
200.0 in.
NORMAL STRAIN y t 2 0.1 in. e r r 200 in.
500
10
6
(Shortening at the top surface)
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 1.2 m and the height of the bar is h 100 mm. The deflection at the midpoint is measured as 3.6 mm. What is the maximum normal strain at the top and bottom of the bar?
h P P
a
L -- 2
L -- 2
a
288
CHAPTER 5
Stresses in Beams (Basic Topics)
Solution 5.4-6
Bar of rectangular cross section
h P P
a
L -- 2
L -- 2
a
0
L
1.2 m
h
100 mm
3.6 mm
Substitute numerical values ( 0.0036 r 1 cos 0.6 r
meters):
Note that the deflection curve is nearly flat (L/ 333) and is a very small angle. sin u u L2 r
Solve numerically: NORMAL STRAIN r 1 L cos 2r e y r h2 r
50.00 m
L2 (radians) r r (1 cos u)
50 mm 50,000 mm
1000
10
6
(Elongation on top; shortening on bottom)
Normal Stresses in Beams
Problem 5.5-1 A thin strip of hard copper (E 16,400 ksi) having length L 80 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 max in the strip. (b) Does the stress increase or decrease if the thickness of the strip is increased?
3 t = -- in. 32
Solution 5.5-1 E 16,400 ksi
Copper strip bent into a circle L 80 in. t 3/32 in. Substitute numerical values: smax (16,400 ksi)(3 32 in.) 80 in. 60.4 ksi
(a) MAXIMUM BENDING STRESS L 2 r 2 r r L 2 Ey r Et L 2 Ey L
(b) CHANGE IN STRESS If the thickness t is increased, the stress
max
From Eq. (5-7): s smax 2 E(t 2) L
increases.
SECTION 5.5
Normal Stresses in Beams
289
Problem 5.5-2 A steel wire (E 200 GPa) of diameter d 1.0 mm is bent around a pulley of radius R0 400 mm (see figure).
(a) What is the maximum stress max in the wire? (b) Does the stress increase or decrease if the radius of the pulley is increased?
R0 d
Solution 5.5-2 E 200 GPa
Steel wire bent around a pulley d 1.0 mm R0 400 mm From Eq. (5-7): smax 400.5 mm Ey r (200 GPa) (0.5 mm) 400.5 mm 250 MPa
(a) MAXIMUM STRESS IN THE WIRE r y R0 d 2 d 2 400 mm 0.5 mm
(b) CHANGE IN STRESS If the radius is increased, the stress decreases.
max
0.5 mm
Problem 5.5-3 A thin, high-strength steel rule (E 30 106 psi) having thickness t 0.15 in. and length L 40 in. is bent by couples M0 into a circular arc subtending a central angle 45 (see figure). (a) What is the maximum bending stress max in the rule? (b) Does the stress increase or decrease if the central angle is increased?
L = length t M0 M0
Solution 5.5-3
L
Thin steel rule bent into an arc Substitute numerical values: smax E t L 30 106 psi 0.15 in. 40 in. 45 0.78540 rad (30 106 psi) (0.15 in.) (0.78540 rad) 2 (40 in.) 44,200 psi 44.2 ksi
(b) CHANGE IN STRESS If the angle increases. is increased, the stress
max
(a) MAXIMUM BENDING STRESS L smax r Ey r r L E(t 2) L radians Et 2L
290
CHAPTER 5
Stresses in Beams (Basic Topics)
Problem 5.5-4 A simply supported wood beam AB with span length L 3.5 m carries a uniform load of intensity q 6.4 kN/m (see figure). Calculate the maximum bending stress max due to the load q if the beam has a rectangular cross section with width b 140 mm and height h 240 mm.
q A B h
L
b
Solution 5.5-4 L b Mmax smax
Simple beam with uniform load Substitute numerical values: smax 3(6.4 kN m)(3.5 m) 2 4(140 mm)(240 mm) 2 7.29 MPa
3.5 m q 6.4 kN/m 140 mm h 240 mm qL 8
2
S
bh 6 3qL2 4bh2
2
Mmax S
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 three steel plates so as to form an I-shaped cross section (see figure) having section modulus S 3600 in3. What is the maximum bending stress max in a girder due to the uniform load?
Solution 5.5-5
Bridge girder
q
L S Mmax
L
180 ft 3600 in.3 qL2 8 Mmax S
q
1.6 k/ft
smax smax
qL2 8S 21.6 ksi
(1.6 k ft)(180 ft) 2 (12 in. ft) 8(3600 in.3 )
SECTION 5.5
Normal Stresses in Beams
291
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 R is b 200 mm. Calculate the maximum bending stress max in the axle if P 47 kN.
P A d R b L R
P B d
b
Solution 5.5-6
Freight-car axle MAXIMUM BENDING STRESS smax Mmax S 32Pb d3
Diameter d 80 mm Distance b 200 mm Load P 47 kN M max Pb S d3 32
Substitute numerical values: smax 32(47 kN)(200 mm) (80 mm) 3 187 MPa
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
h b
P
q
P
8 in. h 1.5 in. 3 lb/ft P 90 lb Pd qL2 2
d
8.0 ft 135.4 lb-ft
L
9.5 ft
Mmax
720 lb-ft 855.4 lb-ft
10,264 lb-in.
d L
d L
bh2 3.0 in3. 6 M 10,264 lb-in. smax S 3.0 in.3 S
3420 psi
292
CHAPTER 5
Stresses in Beams (Basic Topics)
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 46 m and an I-shaped cross section with dimensions as shown in the figure. The load on each girder (during construction) is assumed to be 11.0 kN/m, which includes the weight of the girder. Determine the maximum bending stress in a girder due to this load.
50 mm
2400 mm 25 mm
600 mm
Solution 5.5-8
Bridge girder Mmax qL2 2 Mmax c I b1h3 1 12 1 (11.0 kN m)(46 m) 2 2 c h 2 1200 mm 11,638 kN m
q
tf
smax
h1 tw L h2
I
bh3 12
b
1 1 (0.6 m)(2.4 m) 3 (0.575 m)(2.3 m) 3 12 12 0.6912 m4 0.5830 m4 0.1082 m4 smax Mmax c I (11,638 kN m)(1.2 m) 0.1082 m4 129 MPa
L q b tf h1 b1
46 m 11.0 kN/m 600 mm h 2400 mm 50 mm tw 25 mm h 2tf 2300 mm b tw 575 mm
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 8.8 k, and if the distance from the line of action of that force to point B is 14 ft, what is the maximum bending stress in the beam due to the pumping force?
A
B
C 0.875 in.
0.625 in.
20.0 in.
8.0 in.
SECTION 5.5
Normal Stresses in Beams
293
Solution 5.5-9
Beam in an oil-well pump Mmax
tf
PL (8.8 k)(14 ft) 123,200 lb-ft 1,478,400 lb-in. Mmax c I b1h3 1 12 1 (7.375 in.)(18.25 in.) 3 12 1,597.7 in.4 c h 2 10.0 in.
smax
P L h1 tw h2
I
bh3 12
b
1 (8.0 in.)(20.0 in.) 3 12 5,333.3 in.4
3,735.7 in.4 (1.4784
L P b tf h1 b1
14 ft 8.8 k 8.0 in. h 20.0 in. 0.875 in. tw 0.625 in. h 2tf 18.25 in. b tw 7.375 in.
smax
Mmax c I
106 lb-in.)(10.0 in.) 1,597.7 in.4 9.25 ksi
9250 psi
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 max 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 DATA P L q
Railroad tie (or sleeper) 250 mm Substitute numerical values: M1
3
175 kN b 300 mm h 1500 mm a 500 mm 2P L 2a S bh2 6 3.125 10
17,500 N m 21,875 N m
M2
21,875 N m
m3
Mmax
BENDING-MOMENT DIAGRAM
M1 0 M2 M1
MAXIMUM BENDING STRESS smax Mmax 5 21,875 N m 3.125 10 3 m3
7.0 MPa
(Tension on top; compression on bottom)
M1 M2
q L 2 2 P
qa2 2
Pa2 L 2a a
2
L P (2a 4
L 2a 2 L)
a
PL 2
2
PL 2
294
CHAPTER 5
Stresses in Beams (Basic Topics)
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
q t
a
s L
a
d1 d2
L s
36 ft 11 ft
432 in. 132 in. A
d2 d1 4
6.0 in. d2 2t (d 2 2 d 2) 1
t
0.25 in. 5.5 in. 4.5160 in.2
I q
(d 4 d 4 ) 18.699 in.4 1 64 2 A (0.053 lb/in.3)(4.5160 in.2)
0.23935 lb/in.
0.053 lb/in.3 a (L s)/2
150 in.
BENDING-MOMENT DIAGRAM
0 M1 M2 M1
MAXIMUM BENDING STRESS smax smax Mmax c I c d2 2 3.0 in. 432 psi
M1 M2 Mmax
2,692.7 lb-in.
qa2 2 qL L 4 2
2,692.7 lb-in. s 2,171.4 lb-in.
(2,692.7 lb-in.)(3.0 in.) 18.699 in.4
(Tension on top)
SECTION 5.5
Normal Stresses in Beams
295
Problem A 5.5-12 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 max in the beams, assuming that the weight density of water is 9.81 kN/m3.
h t
A
B
Solution 5.5-12
Vertical wood beam
t A A
MAXIMUM BENDING MOMENT
q0 q q0 ( x ) L B
h RA
x L
RA
B q0
q0 L 6
h t
2.0 m 120 mm 9.81 kN/m3 (water)
Let b width of beam perpendicular to the plane of the figure Let q0 q0 gbh maximum intensity of distributed load S bt 6
2
q0 x3 6L q0 Lx q0 x3 6 6L dM q0L q0 x2 L 0 x dx 6 2L 3 Substitute x L 3 into the equation for M: M RAx
Mmax
For the vertical wood beam: L Maximum bending stress smax Mmax S 2q0 h2
q0 L L 6 3
q0 L3 6L 3 3
q0 L2
9
3
h; Mmax
q0 h2
9
3
2gh3 3 3 t2
3
3 bt 2
SUBSTITUTE NUMERICAL VALUES:
max
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 Mmax/S 2.10 MPa max
296
CHAPTER 5
Stresses in Beams (Basic Topics)
Problem 5.5-13 Determine the maximum tensile stress t (due to pure bending by positive bending moments M) for beams having cross sections as follows (see figure): (a) a semicircle of diameter d, and (b) an isosceles trapezoid with bases b1 b and b2 4b/3, and altitude h.
b1 C d (a) C b2 (b) h
Solution 5.5-13 (a) SEMICIRCLE
Maximum tensile stress (b) TRAPEZOID
C c d c b2 b1 C h
From Appendix D, Case 10: IC c st (9 4r 3 Mc IC
2
64)r4
(9
72 2d 3 768M (9 2 64)d3
64)d 4 1152
2
4b 3 From Appendix D, Case 8: b1 b b2 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
30.93
M d3
Problem 5.5-14 Determine the maximum bending stress max (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 60. (Hint: Use the formulas given in Appendix D, Cases 9 and 15.)
C
d
SECTION 5.5
Normal Stresses in Beams
297
Solution 5.5-14
Circular core From Appendix D, Cases 9 and 15:
y
C
y
Iy r d 2
r4 4
r4 2
ab r2 b
d
2 a
2ab3 r4
MAXIMUM BENDING STRESS smax smax For smax Mc Iy
3
c
r sin b
d sin b 2
64M sin b d (4b sin 4b) 60 /3 rad:
3
radians Iy d 64
4
d4 64 d4 64 d4 64
d 32 2
4
radians b b b b
r sin
sin b cos b
d4 32 2 d4 32 2
(sin b cos b)(1
2 sin b cos3 b 2 cos2b)
b
r cos
576M (8 3 9)d
10.96
M d3
d4 32 2
d4 (4b 128
sin 4b)
1 sin 4b 4
1 sin 2b ( cos 2b) 2
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 max 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
P x A d
Wheel loads on a beam
P
Substitute x into the equation for M: L d P S 24 ft 288 in. 5 ft 60 in. 3k 16.2 in.3 Mmax P L 2L d 2 2
B
MAXIMUM BENDING STRESS smax Mmax S P L 2LS
RA
L
MAXIMUM BENDING MOMENT RA M dM dx P (L L RAx x) P (L L dx 4x) x d) P (2L L d 2x)
Substitute numerical values: smax
d 2 2 30 in.) 2
3k (288 in. 2(288 in.)(16.2 in.3 ) 21.4 ksi
P (2Lx L d
2x2 ) 0 x L 2 d 4
P (2L L
298
CHAPTER 5
Stresses in Beams (Basic Topics)
Problem 5.5-16 Determine the maximum tensile stress t and maximum compressive stress c due to the load P acting on the simple beam AB (see figure). Data are as follows: P 5.4 kN, L 3.0 m, d 1.2 m, b 75 mm, t 25 mm, h 100 mm, and h1 75 mm.
t P A d B h h1
L
b
Solution 5.5-16
Simple beam of T-section
t P A d B c2 RA L RB b
c1 C
h1
h
P b d
5.4 kN 75 mm 1.2 m
L t h
3.0 m 25 mm 100 mm h1 75 mm
MAXIMUM BENDING MOMENT Mmax RA(L d) RB(d) 3888 N m
MAXIMUM TENSILE STRESS st Mmax c2 IC 43.9 MPa MAXIMUM COMPRESSIVE STRESS sc Mmax c1 IC 73.2 MPa (3888 N m)(0.0625 m) 3.3203 106 mm4
PROPERTIES OF THE CROSS SECTION A c1 IC 3750 mm2 c2 37.5 mm
(3888 N m)(0.0375 m) 3.3203 106 mm4
62.5 mm 3.3203
106 mm4
REACTIONS OF THE BEAM RA 2.16 kN RB 3.24 kN
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 t and maximum compressive stress c if the cross section has the dimensions indicated and the moment of inertia about the z axis (the neutral axis) is I 2.81 in.4 (Note: The uniform load represents the weight of the beam.)
200 lb 20 lb/ft B 5.0 ft y z C 3.0 ft
A
0.606 in. 2.133 in.
SECTION 5.5
Normal Stresses in Beams
299
Solution 5.5-17
Cantilever beam (channel section)
200 lb
I
2.81 in.4
c1
0.606 in.
20 lb/ft B
5.0 ft
Mmax
(200 lb)(5.0 ft)
A 3.0 ft 8.0 ft
1000 lb-ft 640 lb-ft 19,680 lb-in.
(20 lb ft)(8.0 ft) 1640 lb-ft
c2
2.133 in.
8.0 ft 2
y z C
0.606 in. 2.133 in.
MAXIMUM TENSILE STRESS Mc1 (19,680 lb-in.)(0.606 in.) st I 2.81 in.4 4,240 psi MAXIMUM COMPRESSIVE STRESS sc Mc2 I (19,680 lb-in.)(2.133 in.) 2.81 in.4
14,940 psi
Problem 5.5-18 A cantilever beam AB of triangular cross section has length L 0.8 m, width b 80 mm, and height h 120 mm (see figure). The beam is made of brass weighing 85 kN/m3. (a) Determine the maximum tensile stress t and maximum compressive stress c due to the beam's own weight. (b) If the width b is doubled, what happens to the stresses? (c) If the height h is doubled, what happens to the stresses?
A B
b h L
Solution 5.5-18
Triangular beam
q y b z L h/3 2h h 3
C
L
0.8 m b 85 kN/m3 g bh 2
80 mm
h
120 mm
Compressive stress:
c
2
t t c
Substitute numerical values:
(a) MAXIMUM STRESSES q Iz gA IC Mmax h 3 Mc1 Iz c2 qL2 2 2h 3 3gL h
2
1.36 MPa 2.72 MPa
gbhL2 4
(b) WIDTH b IS DOUBLED No change in stresses. (c) HEIGHT h IS DOUBLED Stresses are reduced by half.
bh3 36
c1
Tensile stress: st
300
CHAPTER 5
Stresses in Beams (Basic Topics)
Problem 5.5-19 A beam ABC with an overhang from B to C supports a uniform load of 160 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 5.14 in.4 Calculate the maximum tensile stress t and maximum compressive stress c due to the uniform load.
160 lb/ft A B 10 ft C
5 ft
y z C
0.674 in. 2.496 in.
Solution 5.5-19
Beam with an overhang
q 160 lb/ft y 0.674 in. 2.496 in.
A L 10 ft
B b 5 ft
C
z C
AT CROSS SECTION OF MAXIMUM POSITIVE
M1 0 3.75 ft M2
BENDING MOMENT
st sc
M1c2 Iz M1c1 Iz
(13,500 lb-in.)(2.496 in.) 5.14 in.4 (13,500 lb-in.)(0.674 in.) 5.14 in.4
6,560 psi 1,770 psi
Iz c1 RA M1 M2
5.14 in.4 0.674 in. c2 2.496 in. 600 lb RB 1800 lb 1125 lb-ft 13,500 lb-in. 2000 lb-ft 24,000 lb-in.
AT CROSS SECTION OF MAXIMUM NEGATIVE
BENDING MOMENT
st sc
M2c1 Iz M2c2 Iz
(24,000 lb-in.)(0.674 in.) 5.14 in.4 (24,000 lb-in.)(2.496 in.) 5.14 in.4
3,150 psi 11,650 psi
MAXIMUM STRESSES 6,560 psi t c
11,650 psi
Problem 5.5-20 A frame ABC travels horizontally with an acceleration a0 (see figure). Obtain a formula for the maximum stress max in the vertical arm AB, which has length L, thickness t, and mass density .
L
A t a0 = acceleration B C
SECTION 5.5
Normal Stresses in Beamss
301
Solution 5.5-20 L t
Accelerating frame TYPICAL UNITS FOR USE
IN THE PRECEDING EQUATION
length of vertical arm thickness of vertical arm 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
t
SI UNITS: kg/m3 L meters (m) a0 t m/s2 meters (m)
max
N s2/m4
N/m2 (pascals) slug/ft3 ft/s2 t lb-s2/ft4 ft
L
q q
rbta0 bt 2 6
Mmax smax
qL2 2 Mmax S
rbta0 L2 2 3rL2a0 t
USCS UNITS: L
max
ft
a0
S
lb/ft2 (Divide by 144 to obtain psi)
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 1/2 in. Determine the maximum tensile and compressive stresses in the beam.
P = 625 lb L1 = 4 ft q = 80 lb/ft
1
t = -- in. 2 t = -- in. 2
1
h= 3 in.
L2 = 8 ft
L3 = 5 ft
b = 2 -- in. 2
1
Solution 5.5-21
P L1 A
Beam of T-section
q B C qL2 3 2
BENDING-MOMENT DIAGRAM
M1 RA L1 9,000 lb in.
L2 RA
RB
L3
BENDING MOMENT
M2
12,000 lb in.
AT CROSS SECTION OF MAXIMUM POSITIVE 5 ft 60 in. st M1c2 IC 4,320 psi sc M1c1 IC 8,640 psi
L1 P
4 ft 48 in. L2 8 ft 96 in. L3 625 lb q 80 lb/ft 6.6667 lb/in.
t c1 C h
t b
c2
PROPERTIES OF THE CROSS SECTION b 2.5 in. h 3.0 in. t 0.5 in. A bt (h t)t 2.50 in.2 c1 2.0 in. c2 1.0 in. IC 25 4 in. 12 2.0833 in.4
AT CROSS SECTION OF MAXIMUM NEGATIVE
BENDING MOMENT
st
M2c1 IC
11,520 psi
sc
M2c2 IC
5,760 psi
REACTIONS RA 187.5 lb (upward) RB 837.5 lb (upward)
MAXIMUM STRESSES 11,520 psi t
c
8,640 psi
302
CHAPTER 5
Stresses in Beams (Basic Topics)
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
y c1 z c2 B C y B
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS (THE z AXIS) All dimensions in millimeters. Rectangle: Iz Ic Ad 2 1 (25)(50) 3 (25)(50)(25 12 260,420 Hole: Iz Ic Ad 2 878 261,300 mm4 24.162) 2
24.162) 2
MAXIMUM BENDING MOMENT M PL (600 N)(0.4 m) 240 N m
PROPERTIES OF THE CROSS SECTION A1 area of rectangle (25 mm)(50 mm) A2 area of hole (10 mm) 78.54 mm 4 area of cross section A1 A2 1171.5 mm2
2 2
1250 mm2
490.87 Cross-section: I 261,300 14,460
(10) 4 (78.54)(37.5 64 13,972 14,460 mm4 246,800 mm4
A
STRESS AT THE TOP OF THE BEAM Mc1 (240 N m)(25.838 mm) s1 I 246,800 mm4
25.1 MPa (tension) STRESS AT THE TOP OF THE HOLE My s2 y c1 7.5 mm 18.338 mm I s2 (240 N m)(18.338 mm) 246,800 mm4
Using line B-B as reference axis: Ai yi y A1(25 mm) A2(37.5 mm) 28,305 mm3
3 a Ai yi 28,305 mm A 1171.5 mm2 Distances to the centroid C:
24.162 mm
17.8 MPa (tension)
c2 c1
y
24.162 mm c2 25.838 mm
50 mm
STRESS AT THE BOTTOM OF THE BEAM Mc2 (240 N m)(24.162 mm) s3 I 246,800 mm4
23.5 MPa (compression)
SECTION 5.5
Normal Stresses in Beams
303
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 max in the wood beams versus the depth d of the water above the lower support at B. Plot the stress max (psi) as the ordinate and the depth d (ft) as the abscissa. (Note: The weight density 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
A t
Vertical wood beam in a dam 6 ft 2.5 in. 62.4 lb/ft3 Let b width of beam (perpendicular to the figure) Let q0 intensity of load at depth d q0 bd h t MAXIMUM BENDING STRESS 1 2 Section modulus: S bt 6 smax q0 smax Mmax S bd gd 3 1 t2
h d B
q0
6 q0 d 2 B 1 6 bt 2 d L
d L
ANALYSIS OF BEAM
q0
L RA
A
C
B
RB x0 MC
RA L
d
RB
h 6 ft q0 d 2 6L q0 d 3 6 d d B 3L RA (L
d L d) q0 d 6
2
SUBSTITUTE NUMERICAL VALUES: d depth of water (ft) (Max. d h 6 ft) L h 6 ft 62.4 lb/ft3 t 2.5 in. psi max smax (62.4)d 3 1 (2.5) 2 d 6 9d d d 9B 18 d
max
2d d 3LB 3L
2d d R 3LB 3L
0.1849d 3 (54
1
2d)
V
RA
d L
d (ft) 0 1 2 3 4 5 6
(psi)
0
C
x0 RB
0 9 59 171 347 573 830
M
Mmax MC
1000 750
830 psi
0
max (psi)
Mmax
q0 d 2 1 6
d L
2d d 3LB 3L
500 250 0 1 2 3
d (ft)
4
5
6