Compression, 1
Tension, and Shear
Normal Stress and Strain
Problem 1.2-1 A solid circular post ABC (see figure) supports a load P1 2500 lb acting at the top. A second load P2 is uniformly distributed around the shelf at B. The diameters of the upper and lower parts of the post are dAB 1.25 in. and dBC 2.25 in., respectively. (a) Calculate the normal stress AB in the upper part of the post. (b) If it is desired that the lower part of the post have the same compressive stress as the upper part, what should be the magnitude of the load P2?
C A
P1
dAB P2 B dBC
Solution 1.2-1 P1 2500 lb dAB dBC 1.25 in. 2.25 in.
Circular post in compression ALTERNATE SOLUTION FOR PART (b) sBC sAB P1 P2 P1 P2 2 ABC 4 dBC P1 P1 sBC 2 AAB 4 dAB P1 2 or P2 dAB
(a) NORMAL STRESS IN PART AB P1 2500 lb sAB 2040 psi AAB 4 (1.25 in.) 2 (b) LOAD P FOR EQUAL STRESSES 2 sBC P1 P2 ABC
AB
P1 P2 2 dBC
2500 lb P2 2 4 (2.25 in.) 2040 psi 5600 lb
P1 A
P2
dBC 1.8 dAB 2.24 P1 5600 lb
P1 B
sAB dBC 2 dAB
1R
Solve for P2: P2
P2 B
C
1
2
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-2 Calculate the compressive stress c in the circular piston rod (see figure) when a force P 40 N is applied to the brake pedal. Assume that the line of action of the force P is parallel to the piston rod, which has diameter 5 mm. Also, the other dimensions shown in the figure (50 mm and 225 mm) are measured perpendicular to the line of action of the force P.
50 mm 5 mm 225 mm P = 40 N Piston rod
Solution 1.2-2
Free-body diagram of brake pedal
50 mm A
EQUILIBRIUM OF BRAKE PEDAL MA 0 275 mm 50 mm F(50 mm) (40 N) P(275 mm) 275 50 0 220 N 5 mm)
F
225 mm P = 40 N
F
COMPRESSIVE STRESS IN PISTON ROD (d F d compressive force in piston rod diameter of piston rod 5 mm sc F A 220 N (5 mm) 2 4 11.2 MPa
P
Problem 1.2-3 A steel rod 110 ft long hangs inside a tall tower and holds a 200-pound weight at its lower end (see figure). If the diameter of the circular rod is 1/4 inch, calculate the maximum normal stress max in the rod, taking into account the weight of the rod itself. (Obtain the weight density of steel from Table H-1, Appendix H.)
1 -- in. 4
110 ft
200 lb
SECTION 1.2
Normal Stress and Strain
3
Solution 1.2-3
Long steel rod in tension P 200 lb 110 ft
1
smax gL 490 lb/ft3 P A
max
W A
P
gL
d L
L d
/4 in.
Weight density: W Weight of rod (Volume) AL
374.3 psi 200 lb 2 4 (0.25 in.) 374 psi
(490 lb ft3 )(110 ft) 4074 psi
P A
4074 psi
1 ft2 144 in.2 4448 psi
Rounding, we get
max
4450 psi
P = 200 lb
Problem 1.2-4 A circular aluminum tube of length L 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. (a) If the measured strain is 550 10 6, what is the shortening of the bar? (b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?
Strain gage P L = 400 mm P
Solution 1.2-4
Aluminum tube in compression
Strain gage P P
e L d2 d1
550
10
6
(b) COMPRESSIVE LOAD P 40 MPa A
2 [d 2 d1 ] 4 2 863.9 mm2
400 mm 60 mm 50 mm
OF THE BAR
4
[ (60 mm) 2
(50 mm) 2 ]
(a) SHORTENING eL
P
A
(40 MPa)(863.9 mm2)
(550
10 6)(400 mm)
34.6 kN
0.220 mm
4
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-5 The cross section of a concrete pier that is loaded uniformly in compression is shown in the figure. (a) Determine the average compressive stress c in the concrete if the load is equal to 2500 k. (b) Determine the coordinates x and y of the point where the resultant load must act in order to produce uniform normal stress.
y 20 in.
16 in. 48 in. 16 in. 16 in. O 20 in. 16 in. x
Solution 1.2-5
y
Concrete pier in compression (a) AVERAGE COMPRESSIVE STRESS
16 in.
c
P sc
2500 k P A 2500 k 1472 in.2 1.70 ksi
x 48 in.
C y
16 in. 1 16 in.
2 3 4 x
(b) COORDINATES OF CENTROID C From symmetry, y x x 1 (48 in.) 2 24 in.
O
20 in.
16 in.
USE THE FOLLOWING AREAS: A1 A2 A3 A (48 in.)(20 in.) A4 960 in.2 128 in.2
xi Ai (see Chapter 12, Eq. 12-7a) A 1 (x A 2x2 A2 x3 A3 ) A 1 1 1 [ (10 in.)(960 in.2 ) 1472 in.2 2(25.333 in.)(128 in.2) (28 in.)(256 in.2)]
1 (16 in.)(16 in.) 2
(16 in.)(16 in.) A1 A2 (960 A3
256 in.2 A4 256 128) in.2
128
15.8 in.
1472 in.2
Problem 1.2-6 A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). The cable has an effective cross-sectional area of 490 mm2, and the angle of the incline is 30. Calculate the tensile stress t in the cable.
Cable
SECTION 1.2
Normal Stress and Strain
5
Solution 1.2-6
Car on inclined track TENSILE STRESS IN THE CABLE W T Weight of car Tensile force in cable Angle of incline
R2
FREE-BODY DIAGRAM OF CAR
W
st
T A
W sin A
SUBSTITUTE NUMERICAL VALUES: W A st 130 kN 490 mm2 (130 kN)(sin 30 ) 490 mm2 133 MPa 30
A
R1
Effective area of cable
R1, R2
Wheel reactions (no friction force between wheels and rails)
EQUILIBRIUM IN THE INCLINED DIRECTION FT T 0 Q b T W sin 0
W sin
Problem 1.2-7 Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle 34 to the horizontal and wire BC is at an angle 48. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals 0.001 in.) Determine the tensile stresses AB and BC in the two wires.
C A
B
Solution 1.2-7 Two steel wires supporting a lamp FREE-BODY DIAGRAM OF POINT B
TAB TBC
SUBSTITUTE NUMERICAL VALUES: TAB(0.82904) 34 d 30 mils d2 4 48 0.030 in. 10
6
TBC(0.66913) TBC(0.74314)
0 18 0
TAB(0.55919)
SOLVE THE EQUATIONS: in.2 TAB 12.163 lb TBC 15.069 lb
y W = 18 lb 0 x
A
706.9
TENSILE STRESSES IN THE WIRES sAB TAB A TBC A 17,200 psi 21,300 psi
EQUATIONS OF EQUILIBRIUM Fx Fy 0 0 TAB cos TAB sin TBC cos TBC sin 0 W 0
sBC
6
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-8 A long retaining wall is braced by wood shores set at an angle of 30 and supported by concrete thrust blocks, as shown in the first part of the figure. The shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are idealized as shown in the second part of the figure. Note that the base of the wall and both ends of the shores are assumed to be pinned. The pressure of the soil against the wall is assumed to be triangularly distributed, and the resultant force acting on a 3-meter length of the wall is F 190 kN. If each shore has a 150 mm 150 mm square cross section, what is the compressive stress c in the shores?
Soil
Retaining wall Concrete Shore thrust block 30
B F 1.5 m A 30 0.5 m 4.0 m C
Solution 1.2-8
Retaining wall braced by wood shores
Wall B F 1.5 m A 4.0 m 30 0.5 m C Shore
F A A
190 kN area of one shore (150 mm)(150 mm) 22,500 mm2 0.0225 m2
FREE-BODY DIAGRAM OF WALL AND SHORE
B F 1.5 m 30 A AH AV CV CH
30
SUMMATION OF MOMENTS ABOUT POINT A MA 0 0
F(1.5 m) CV (4.0 m) CH (0.5 m) or (190 kN)(1.5 m) C 117.14 kN C(sin 30 )(4.0 m)
C
C(cos 30 )(0.5 m)
0
C CH CV CH CV
compressive force in wood shore horizontal component of C vertical component of C C cos 30 C sin 30
COMPRESSIVE STRESS IN THE SHORES sc C A 117.14 kN 0.0225 m2 5.21 MPa
SECTION 1.2
Normal Stress and Strain
7
Problem 1.2-9 A loading crane consisting of a steel girder ABC supported by a cable BD is subjected to a load P (see figure). The cable has an effective cross-sectional area A 0.471 in2. The dimensions of the crane are H 9 ft, L1 12 ft, and L2 4 ft. (a) If the load P 9000 lb, what is the average tensile stress in the cable? (b) If the cable stretches by 0.382 in., what is the average strain?
D
Cable H Girder A L1 B L2 P C
Solution 1.2-9
Loading crane with girder and cable EQUILIBRIUM
D
MA
0 (9000 lb)(16 ft) 0
TV (12 ft)
H
TV TH TV
A L1 B L2 P = 9000 lb C
TH
12,000 lb L1 12 ft H 9 ft 12 TH TV 9 16,000 lb
H L2 A P
9 ft 4 ft 0.471
L1 A in.2
12 ft effective area of cable
(12,000 lb)
2 TH 2 TV
12 9 (16,000 lb) 2 (12,000 lb) 2
TENSILE FORCE IN CABLE T
9000 lb
20,000 lb FREE-BODY DIAGRAM OF GIRDER
T TV
(a) AVERAGE TENSILE STRESS IN CABLE s T A 20,000 lb 0.471 in.2 42,500 psi
A 12 ft
TH
B 4 ft
C
(b) AVERAGE STRAIN IN CABLE L length of cable stretch of cable e L 0.382 in. (15 ft)(12 in. ft) L H2 L2 1 15 ft
P = 9000 lb
0.382 in. 2120 10
6
T P
tensile force in cable 9000 lb
8
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-10 Solve the preceding problem if the load P 32 kN; the cable has effective cross-sectional area A 481 mm2; the dimensions of the crane are H 1.6 m, L1 3.0 m, and L2 1.5 m; and the cable stretches by 5.1 mm. Figure is with Prob. 1.2-9. Solution 1.2-10
D
Loading crane with girder and cable
H
H
1.6 m 1.5 m 481 mm2 A P
L1 32 kN
3.0 m
L2 A
A L1 B L2 P = 32 kN C
effective area of cable
FREE-BODY DIAGRAM OF GIRDER
T TV T = tensile force in cable C
1.5 m
TENSILE FORCE IN CABLE T
2 TH
TV2
(90 kN) 2
(48 kN) 2
A 3.0 m
TH
B
102 kN (a) AVERAGE TENSILE STRESS IN CABLE
P = 32 kN
EQUILIBRIUM MA 0 (32 kN)(4.5 m) 0
s
T A
102 kN 481 mm2
212 MPa
(b) AVERAGE STRAIN IN CABLE L L length of cable H2 L2 1 3.4 m
TV (3.0 m) TV TH TV
48 kN L1 3.0 m H 1.6 m 3.0 TH TV 1.6 3.0 TH (48 kN) 1.6 90 kN Problem 1.2-11 A reinforced concrete slab 8.0 ft square and 9.0 in. thick is lifted by four cables attached to the corners, as shown in the figure. The cables are attached to a hook at a point 5.0 ft above the top of the slab. Each cable has an effective cross-sectional area A 0.12 in2. Determine the tensile stress t in the cables due to the weight of the concrete slab. (See Table H-1, Appendix H, for the weight density of reinforced concrete.)
stretch of cable 5.1 mm e L 5.1 mm 3.4 m 1500 10
6
Cables
Reinforced concrete slab
SECTION 1.2
Normal Stress and Strain
9
Solution 1.2-11
W
Reinforced concrete slab supported by four cables T tensile force in a cable
Cable AB:
A H Cable
TV T TV
t
B L L
EQUILIBRIUM Fvert W 4TV W 4
T
H LAB H H
2
L2 2
(Eq. 1)
0 0 (Eq. 2)
H L t
height of hook above slab length of side of square slab thickness of slab weight density of reinforced concrete
TV
W D
weight of slab
L2t L 2
length of diagonal of slab
T T
COMBINE EQS. (1) & (2): H H W 4
2
L 2
2
W 4 W 4 1 L2 2H 2
H 2 L2 2 H
DIMENSIONS OF CABLE AB
A LAB H B D= L 2 2
TENSILE STRESS IN A CABLE LAB B A L2 2 st effective cross-sectional area of a cable T A W 4A 1 L2 2H2
t:
length of cable H2
SUBSTITUTE NUMERICAL VALUES AND OBTAIN H 5.0 ft 150 W lb/ft3 7200 lb 1 L2 2H2 22,600 psi L A 8.0 ft 0.12 in.2 t
9.0 in.
0.75 ft
FREE-BODY DIAGRAM OF HOOK AT POINT A
W A T TH
L2t W 4A
st
TV T T
T
T
Problem 1.2-12 A round bar ACB of length 2L (see figure) rotates about an axis through the midpoint C with constant angular speed (radians per second). The material of the bar has weight density . (a) Derive a formula for the tensile stress x in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress max?
A L
C
x L
B
10
CHAPTER 1
Tension, Compression, and Shear
Solution 1.2-12
D C x L
Rotating Bar
dM B
Consider an element of mass dM at distance from the midpoint C. The variable ranges from x to L. dM dF dF Fx
D
d
g g A dj Inertia force (centrifugal force) of element of mass dM g (dM)(j 2 ) g A 2jdj
B L
angular speed (rad/s) A cross-sectional area weight density g g mass density
dF
x
g A 2jdj g
gA 2 2 (L 2g
x 2)
(a) TENSILE STRESS IN BAR AT DISTANCE x sx Fx A g 2 2 (L 2g x 2) --
We wish to find the axial force Fx in the bar at Section D, distance x from the midpoint C. The force Fx equals the inertia force of the part of the rotating bar from D to B.
(b) MAXIMUM TENSILE STRESS x 0 smax g 2L2 -- 2g
Mechanical Properties of Materials
Problem 1.3-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon. (a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea water from Table H-1, Appendix H.)
Solution 1.3-1
Hanging wire of length L W
S
total weight of steel wire weight density of steel 490 lb/ft3
Lmax
smax gS
40,000 psi (144 in.2 ft2 ) 490 lb ft3
L
W
11,800 ft (b) WIRE HANGING IN SEA WATER F F Lmax tensile force at top of wire (gS gW ) AL smax F A (gS gW )L
weight density of sea water 63.8 lb/ft3
A
max
cross-sectional area of wire 40 ksi (yield strength)
(a) WIRE HANGING IN AIR W smax
S AL
smax gS gW 40,000 psi (144 in.2 ft2 ) (490 63.8)lb ft3
W A
gSL
13,500 ft
SECTION 1.3
Mechanical Properties of Materials
11
Problem 1.3-2 Imagine that a long wire of tungsten hangs vertically from a high-altitude balloon. (a) What is the greatest length (meters) it can have without breaking if the ultimate strength (or breaking strength) is 1500 MPa? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of tungsten and sea water from Table H-1, Appendix H.) Solution 1.3-2 Hanging wire of length L W
T
total weight of tungsten wire weight density of tungsten 190
W
(b) WIRE HANGING IN SEA WATER F F smax Lmax tensile force at top of wire (
T W)AL
L
kN/m3
weight density of sea water 10.0 kN/m3
F A
(gT
gW )L
A
max
cross-sectional area of wire 1500 MPa (breaking strength)
smax gT gW 1500 MPa (190 10.0) kN m3 8300 m
(a) WIRE HANGING IN AIR W smax Lmax
T AL
W A smax gT
gTL 1500 MPa 190 kN m3
7900 m
Problem 1.3-3 Three different materials, designated A, B, and C, are tested in tension using test specimens having diameters of 0.505 in. and gage lengths of 2.0 in. (see figure). At failure, the distances between the gage marks are found to be 2.13, 2.48, and 2.78 in., respectively. Also, at the failure cross sections the diameters are found to be 0.484, 0.398, and 0.253 in., respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile.
P
Gage length
P
12
CHAPTER 1
Tension, Compression, and Shear
Solution 1.3-3
P
Tensile tests of three materials
P 2.0 in
0.505 in
Percent reduction in area
A0
1
A1 A0
(100)
Percent elongation L0 2.0 in.
L1 L0
L0
(100)
Percent elongation where L1 is in inches.
L1 2.0
1 (100)
L1 L0
(Eq. 1)
1 100
d0 A1 A0
initial diameter d1
Percent reduction in area B1
d1 2 d0
final diameter
A1 (100) A0
d0
0.505 in.
where d1 is in inches.
Material A B C L1 (in.) 2.13 2.48 2.78 d1 (in.) 0.484 0.398 0.253 % Elongation (Eq. 1) 6.5% 24.0% 39.0%
d1 2 R (100) 0.505
% Reduction (Eq. 2) 8.1% 37.9% 74.9%
(Eq. 2)
Brittle or Ductile? Brittle Ductile Ductile
Problem 1.3-4 The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus, the strength-to-weight ratio RS/W for a material in tension is defined as RS/W
in which is the characteristic stress and is the weight density. Note that the ratio has units of length. Using the ultimate stress U as the strength parameter, calculate the strength-to-weight ratio (in units of meters) for each of the following materials: aluminum alloy 6061-T6, Douglas fir (in bending), nylon, structural steel ASTM-A572, and a titanium alloy. (Obtain the material properties from Tables H-1 and H-3 of Appendix H. When a range of values is given in a table, use the average value.)
Solution 1.3-4
Strength-to-weight ratio
U
The ultimate stress U for each material is obtained from Table H-3, Appendix H, and the weight density is obtained from Table H-1. The strength-to-weight ratio (meters) is RS W sU (MPa) (103 ) g(kN m3 )
U,
(MPa) Aluminum alloy 6061-T6 Douglas fir Nylon Structural steel ASTM-A572 Titanium alloy 310 65 60 500 1050
(kN/m3) 26.0 5.1 9.8 77.0 44.0
RS/W (m) 11.9 12.7 6.1 6.5 23.9 103 103 103 103 103
Values of
, and RS/W are listed in the table.
Titanium has a high strength-to-weight ratio, which is why it is used in space vehicles and high-performance airplanes. Aluminum is higher than steel, which makes it desirable for commercial aircraft. Some woods are also higher than steel, and nylon is about the same as steel.
SECTION 1.3
Mechanical Properties of Materials
13
Problem 1.3-5 A symmetrical framework consisting of three pinconnected bars is loaded by a force P (see figure). The angle between the inclined bars and the horizontal is 48. The axial strain in the middle bar is measured as 0.0713. Determine the tensile stress in the outer bars if they are constructed of aluminum alloy having the stress-strain diagram shown in Fig. 1-13. (Express the stress in USCS units.)
A
B
C
D P
Solution 1.3-5
Symmetrical framework L length of bar BD distance BC L cot L2
D P
A
B
C
L1
L(cot 48 )
0.9004 L
length of bar CD L csc L(csc 48 ) 1.3456 L eBDL
Elongation of bar BD eBDL L3 L3 0.0713 L distance CE L2 1 (L
distance DE
Aluminum alloy 48 eBD 0.0713
B C
eBD L) 2 L2 (1 0.0713) 2
Use stress-strain diagram of Figure 1-13
(0.9004L) 2 1.3994 L
elongation of bar CD L3
L L2 L3 D
BDL
L2
0.0538L
Strain in bar CD 0.0538L 1.3456L 0.0400
L2
From the stress-strain diagram of Figure 1-13: s
E
31 ksi
14
CHAPTER 1
Tension, Compression, and Shear
Problem 1.3-6 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table. Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.2% offset. Is the material ductile or brittle?
P P
STRESS-STRAIN DATA FOR PROBLEM 1.3-6
Stress (MPa) 8.0 17.5 25.6 31.1 39.8 44.0 48.2 53.9 58.1 62.0 62.1
Strain 0.0032 0.0073 0.0111 0.0129 0.0163 0.0184 0.0209 0.0260 0.0331 0.0429 Fracture
Solution 1.3-6
Tensile test of a plastic
PL
Using the stress-strain data given in the problem statement, plot the stress-strain curve:
proportional limit
PL
47 MPa 2.4 GPa
Modulus of elasticity (slope)
Y
yield stress at 0.2% offset 53 MPa
60
Stress (MPa)
Y PL
Y
40 slope 40 MPa = 2.4 GPa 0.017 20 0.2% offset 0
Material is brittle, because the strain after the proportional limit is exceeded is relatively small. --
0.01
0.02 0.03 Strain
0.04
Problem 1.3-7 The data shown in the accompanying table were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for Prob. 1.3-3). At fracture, the elongation between the gage marks was 0.12 in. and the minimum diameter was 0.42 in. Plot the conventional stress-strain curve for the steel and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in., and percent reduction in area.
TENSILE-TEST DATA FOR PROBLEM 1.3-7
Load (lb) 1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600
Elongation (in.) 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture
SECTION 1.3
Mechanical Properties of Materials
15
Solution 1.3-7 d0 A0 0.505 in.
2 d0 4
Tensile test of high-strength steel L0 2.00 in. ENLARGEMENT OF PART OF THE STRESS-STRAIN CURVE
Stress (psi) 70,000
PL YP
0.200 in.2
CONVENTIONAL STRESS AND STRAIN s P A0 e L0 Elongation (in.)
0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture
YP
69,000 psi (0.1% offset) 65,000 psi
PL
Load P (lb)
1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600
Stress (psi)
5,000 10,000 30,000 50,000 60,000 64,500 67,000 68,000 69,000 70,000 72,000 76,000 84,000 92,000 100,000 112,000 113,000
60,000
Strain e
0.00010 0.00030 0.00100 0.00165 0.00195 0.00215 0.00235 0.00270 0.00315 0.00450 0.00510 0.00650 0.01150 0.01680 0.02535 0.05540
50,000 0 0.0020
0.1% pffset 50,000 psi Slope 0.00165 30 106 psi 0.0040 Strain
RESULTS Proportional limit 65,000 psi 30 106 psi
Modulus of elasticity (slope) Yield stress at 0.1% offset
69,000 psi
Ultimate stress (maximum stress) 113,000 psi Percent elongation in 2.00 in. L1 L0 L0 (100) 6%
STRESS-STRAIN DIAGRAM
0.12 in. (100) 2.00 in.
150,000 Stress (psi) 100,000
Percent reduction in area A0 A0 A1 (100)
0.200 in.2 4 (0.42 in.) 2 (100) 0.200 in.2 31%
50,000
0
0.0200
0.0400 Strain
0.0600
16
CHAPTER 1
Tension, Compression, and Shear
Elasticity, Plasticity, and Creep
Problem 1.4-1 A bar made of structural steel having the stressstrain diagram shown in the figure has a length of 48 in. The yield stress of the steel is 42 ksi and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 30 103 ksi. The bar is loaded axially until it elongates 0.20 in., and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.)
(ksi) 60 40 20 0 0
0.002
0.004
0.006
Solution 1.4-1
Steel bar in tension ELASTIC RECOVERY eE
Y
A
B
eE
sB ksi Slope
30
42 103 ksi
0.00140
E
RESIDUAL STRAIN eR eR eB eE 0.00417 0.00140 0.00277 PERMANENT SET
0
R
B
L
48 in.
Y
Yield stress Slope 30
42 ksi
eRL
(0.00277)(48 in.) 0.13 in.
103 ksi
0.20 in. STRESS AND STRAIN AT POINT B
B Y
Final length of bar is 0.13 in. greater than its original length.
42 ksi 0.20 in. 48 in. 0.00417
eB
L
Problem 1.4-2 A bar of length 2.0 m is made of a structural steel having the stress-strain diagram shown in the figure. The yield stress of the steel is 250 MPa and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 200 GPa. The bar is loaded axially until it elongates 6.5 mm, and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.)
(MPa) 300 200 100 0 0
0.002
0.004
0.006
SECTION 1.4
Elasticity, Plasticity, and Creep
17
Solution 1.4-2
Steel bar in tension ELASTIC RECOVERY eE L 2.0 m 2000 mm
Y
Y
A
B
Yield stress Slope
E
250 MPa
eE
sB Slope
250 MPa 200 GPa
0.00125
200 GPa
RESIDUAL STRAIN eR eR eB eE 0.00325 0.00125 0.00200 Permanent set eRL (0.00200)(2000 mm)
6.5 mm
0
R
B
STRESS AND STRAIN AT POINT B
B Y
4.0 mm Final length of bar is 4.0 mm greater than its original length.
250 MPa 6.5 mm 2000 mm 0.00325
eB
L
Problem 1.4-3 An aluminum bar has length L 4 ft and diameter d 1.0 in. The stress-strain curve for the aluminum is shown in Fig. 1-13 of Section 1.3. The initial straight-line part of the curve has a slope (modulus of elasticity) of 10 106 psi. The bar is loaded by tensile forces P 24 k and then unloaded.
(a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.)
Solution 1.4-3
Aluminum bar in tension
B
STRESS AND STRAIN AT POINT B sB P A 24 k 2 4 (1.0 in.) 31 ksi 0.04
B
A
From Fig. 1-13: eB
E
ELASTIC RECOVERY eE eE sB Slope 31 ksi 10 106 psi 0.0031
0
R
B
L d P
4 ft 1.0 in. 24 k
48 in.
RESIDUAL STRAIN eR eR eB eE 0.04 0.0031 0.037 (Note: The accuracy in this result is very poor because eB is approximate.) 106 psi. (a) PERMANENT SET eRL (0.037)(48 in.) 1.8 in. (b) PROPORTIONAL LIMIT WHEN RELOADED
B
See Fig. 1-13 for stress-strain diagram Slope from O to A is 10
31 ksi
18
CHAPTER 1
Tension, Compression, and Shear
Problem 1.4-4 A circular bar of magnesium alloy is 800 mm long. The stress-strain diagram for the material is shown in the figure. The bar is loaded in tension to an elongation of 5.6 mm, and then the load is removed. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.)
200 (MPa) 100
0
0
0.005
0.010
Solution 1.4-4
( (
PL )2 PL )1
Magnesium bar in tension
B
Slope
(sPL ) 1 eA
88 MPa 0.002
44 GPa
A
STRESS AND STRAIN AT POINT B eB L 5.6 mm 800 mm 0.007
B
From -e diagram:
E
(
PL)2
170 MPa
0
R
B
ELASTIC RECOVERY eE eE sB Slope (sPL ) 2 Slope 170 MPa 44 GPa 0.00386
L
800 mm 5.6 mm
(
PL )1
initial proportional limit 88 MPa (from stress-strain diagram)
RESIDUAL STRAIN eR eR eB eE 0.007 0.00386 0.00314 (a) PERMANENT SET eRL (0.00314)(800 mm) 2.51 mm
(
PL )2
proportional limit when the bar is reloaded
INITIAL SLOPE OF STRESS-STRAIN CURVE From -e diagram: At point A: (
PL )1
88 MPa 0.002
(b) PROPORTIONAL LIMIT WHEN RELOADED (
PL)2 B
eA
170 MPa
Problem 1.4-5 A wire of length L 4 ft and diameter d 0.125 in. is stretched by tensile forces P 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: 18,000 0 0.03 ( ksi) 1 300 in which is nondimensional and square inch (ksi). has units of kips per
(a) Construct a stress-strain diagram for the material. (b) Determine the elongation of the wire due to the forces P. (c) If the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?
SECTION 1.5
Linear Elasticity, Hooke's Law, and Poisson's Ratio
19
Solution 1.4-5 L P 4 ft 600 lb
Wire stretched by forces P 0.125 in.
ALTERNATIVE FORM OF THE STRESS-STRAIN RELATIONSHIP Solve Eq. (1) for e in terms of : s e 0 s 54 ksi (s ksi) (Eq. 2) 18,000 300s This equation may also be used when plotting the stress-strain diagram.
48 in. d
COPPER ALLOY s 18,000e 1 300e 0 e 0.03 (s ksi) (Eq. 1)
(b) ELONGATION s
= 54 ksi B
OF THE WIRE
(a) STRESS-STRAIN DIAGRAM (From Eq. 1)
60
B
P A
600 lb (0.125 in.) 2 4
48,900 psi
48.9 ksi
From Eq. (2) or from the stress-strain diagram: e 0.0147 eL (0.0147)(48 in.) 0.71 in.
40 (ksi) 20
B
STRESS AND STRAIN AT POINT B (see diagram) 48.9 ksi eB 0.0147
E
=
B- R
ELASTIC RECOVERY eE
0.03
0
R
B
0.01
0.02
eE
sB Slope
48.9 ksi 18,000 ksi
0.00272
INITIAL SLOPE OF STRESS-STRAIN CURVE Take the derivative of ds de (1 with respect to e: RESIDUAL STRAIN eR eR eB eE 0.0147 eR L 0.0027 0.0120
300e)(18,000) (18,000e)(300) (1 300e) 2
(c) Permanent set
(0.0120)(48 in.) 0.58 in.
18,000 (1 300e) 2 At e 0, ds de 18,000 ksi
(d) Proportional limit when reloaded
B
B
Initial slope 18,000 ksi
49 ksi
Linear Elasticity, Hooke's Law, and Poisson's Ratio
When solving the problems for Section 1.5, assume that the material behaves linearly elastically. Problem 1.5-1 A high-strength steel bar used in a large crane has diameter d 2.00 in. (see figure). The steel has modulus of elasticity E 29 106 psi and Poisson's ratio 0.29. Because of clearance requirements, the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load Pmax that is permitted?
d P P
20
CHAPTER 1
Tension, Compression, and Shear
Solution 1.5-1 STEEL BAR d E LATERAL STRAIN e d d
Steel bar in compression 2.00 in. 29 Max. d 0.001 in. 0.29 AXIAL STRESS Ee (29 106 psi)( 0.001724) 50.00 ksi (compression) Assume that the yield stress for the high-strength steel is greater than 50 ksi. Therefore, Hooke's law is valid. MAXIMUM COMPRESSIVE LOAD 0.001724 Pmax sA 157 k (50.00 ksi) 4
(2.00
106 psi
0.001 in. 2.00 in.
0.0005
AXIAL STRAIN e 0.0005 e n 0.29 (shortening)
in.) 2
Problem 1.5-2 A round bar of 10 mm diameter is made of aluminum alloy 7075-T6 (see figure). When the bar is stretched by axial forces P, its diameter decreases by 0.016 mm. Find the magnitude of the load P. (Obtain the material properties from Appendix H.)
P 7075-T6
d = 10 mm
P
Solution 1.5-2 d 10 mm
Aluminum bar in tension d 0.016 mm AXIAL STRESS Ee (72 GPa)(0.004848)
(Decrease in diameter) 7075-T6 From Table H-2: E 72 GPa
Y
349.1 MPa (Tension) 0.33 480 MPa Because <
Y,
Hooke's law is valid.
(10
From Table H-3: Yield stress LATERAL STRAIN e d d 0.016 mm 10 mm
LOAD P (TENSILE FORCE) P sA
0.0016
(349.1 MPa) 27.4 kN
4
mm) 2
AXIAL STRAIN e 0.0016 0.33 0.004848 (Elongation) e n
Problem 1.5-3 A nylon bar having diameter d1 3.50 in. is placed inside a steel tube having inner diameter d2 3.51 in. (see figure). The nylon bar is then compressed by an axial force P. At what value of the force P will the space between the nylon bar and the steel tube be closed? (For nylon, assume E 400 ksi and 0.4.)
Steel tube d1 d2 Nylon bar
SECTION 1.5
Linear Elasticity, Hooke's Law, and Poisson's Ratio
21
Solution 1.5-3
Nylon bar inside steel tube AXIAL STRAIN
d1 d2
e n (Shortening) e AXIAL STRESS
0.002857 0.4
0.007143
COMPRESSION d1 3.50 in. d2 3.51 in. Nylon: E 400 ksi 0.4 d1 0.01 in.
Ee
(400 ksi)( 0.007143) 2.857 ksi
(Compressive stress) Assume that the yield stress is greater than Hooke's law is valid. FORCE P (COMPRESSION) P sA (2.857 ksi) 27.5 k 4
(3.50
and
LATERAL STRAIN e e d1 (Increase in diameter) d1 0.01 in. 3.50 in. 0.002857
in.) 2
Problem 1.5-4 A prismatic bar of circular cross section is loaded by tensile forces P (see figure). The bar has length L 1.5 m and diameter d 30 mm. It is made of aluminum alloy with modulus 1 of elasticity E 75 GPa and Poisson's ratio /3. If the bar elongates by 3.6 mm, what is the decrease in diameter d? What is the magnitude of the load P?
P L
d
P
Solution 1.5-4 L E 1.5 m 75 GPa
Aluminum bar in tension d 30 mm
1
DECREASE IN DIAMETER d ed (0.0008)(30 mm) 0.024 mm
/3
3.6 mm (elongation) AXIAL STRAIN e L 3.6 mm 1.5 m 0.0024
AXIAL STRESS Ee (75 GPa)(0.0024) 180 MPa (This stress is less than the yield stress, so Hooke's law is valid.) LOAD P (TENSION) P sA (180 MPa)
(30
LATERAL STRAIN e ne ( 1 )(0.0024) 3 0.0008 (Minus means decrease in diameter) mm) 2
4
127 kN
22
CHAPTER 1
Tension, Compression, and Shear
Problem 1.5-5 A bar of monel metal (length L 8 in., diameter d 0.25 in.) is loaded axially by a tensile force P 1500 lb (see figure from Prob. 1.5-4). Using the data in
Table H-2, Appendix H, determine the increase in length of the bar and the percent decrease in its cross-sectional area.
Solution 1.5-5 L 8 in.
Bar of monel metal in tension d 0.25 in. 25,000 ksi P 1500 lb 0.32 Original area: A0 DECREASE IN CROSS-SECTIONAL AREA d2 4
From Table H-2: E AXIAL STRESS P 1500 lb s A 4 (0.25 in.) 2
30,560 psi
Final area: A1 A1 4 4 (d [d2 d) 2 2dd (d) 2 ]
Assume is less than the proportional limit, so that Hooke's law is valid. AXIAL STRAIN e s E 30,560 psi 25,000 ksi 0.001222
Decrease in area: A A A0 4 A1 d)
(d)(2d
INCREASE IN LENGTH eL (0.001222)(8 in.) 0.00978 in.
PERCENT DECREASE IN AREA Percent A (100) A0 (d)(2d d2 d) (100)
LATERAL STRAIN e ne (0.32)(0.001222) 0.0003910
(0.0000978)(0.4999) (100) (0.25) 2 0.078%
DECREASE IN DIAMETER d ed (0.0003910)(0.25 in.) 0.0000978 in.
Problem 1.5-6 A tensile test is peformed on a brass specimen 10 mm in diameter using a gage length of 50 mm (see figure). When the tensile load P reaches a value of 20 kN, the distance between the gage marks has increased by 0.122 mm. (a) What is the modulus of elasticity E of the brass? (b) If the diameter decreases by 0.00830 mm, what is Poisson's ratio?
10 mm 50 mm P P
SECTION 1.5
Linear Elasticity, Hooke's Law, and Poisson's Ratio
23
Solution 1.5-6 d P 10 mm 20 kN
Brass specimen in tension Gage length L 0.122 mm 50 mm d 0.00830 mm E (a) MODULUS OF ELASTICITY s e 254.6 MPa 0.002440 104 GPa
AXIAL STRESS P 20 kN 254.6 MPa A 4 (10 mm) 2 Assume is below the proportional limit so that Hooke's law is valid. s AXIAL STRAIN e 0.122 mm 50 mm 0.002440
(b) POISSON'S RATIO e d n e ed d ed ed 0.00830 mm (0.002440)(10 mm) 0.34
L
Problem 1.5-7 A hollow steel cylinder is compressed by a force P (see figure). The cylinder has inner diameter d1 3.9 in., outer diameter d2 4.5 in., and modulus of elasticity E 30,000 ksi. When the force P increases from zero to 40 k, the outer diameter of the cylinder increases by 455 10 6 in. (a) Determine the increase in the inner diameter. (b) Determine the increase in the wall thickness. (c) Determine Poisson's ratio for the steel.
P
d1 d2
Solution 1.5-7 d1 d2 t E P d2 3.9 in. 4.5 in. 0.3 in.
Hollow steel cylinder (c) POISSON'S RATIO
d1 d2 t
Axial stress: s A [d2 d2 ] 1 4 2 3.9584 in.2 P A
Y;
P A 4 [ (4.5 in.) 2 (3.9 in.) 2 ]
30,000 ksi 40 k (compression) 455 10
6 in.
(increase) s
LATERAL STRAIN e d2 d2 455 10 6 in. 4.5 in. 0.0001011 (
40 k 3.9584 in.2 Hooke's law is valid) 10.105 ksi 30,000 ksi 0.0001011 0.000337
10.105 ksi (compression) Axial strain:
(a) INCREASE IN INNER DIAMETER d1 ed1 (0.0001011)(3.9 in.) 394 10
6
e
s E e e 0.30
in. n
0.000337
(b) INCREASE IN WALL THICKNESS t et (0.0001011)(0.3 in.) 30 10
6
in.
24
CHAPTER 1
Tension, Compression, and Shear
Problem 1.5-8 A steel bar of length 2.5 m with a square cross section 100 mm on each side is subjected to an axial tensile force of 1300 kN (see figure). Assume that E 200 GPa and v 0.3. Determine the increase in volume of the bar.
100 mm
100 mm 1300 kN
1300 kN 2.5 m
Solution 1.5-8
Square bar in tension DECREASE IN SIDE DIMENSION e b ne eb 195 (195 10
6
Find increase in volume. Length: L Side: b Force: P E 2.5 m 100 mm 1300 kN 0.3 2500 mm
10 6 )(100 mm)
0.0195 mm FINAL DIMENSIONS L1 b1 L b L b 2501.625 mm 99.9805 mm
200 GPa
AXIAL STRESS s s P A P b2 130 MPa
1300 kN (100 mm) 2
FINAL VOLUME V1 L1b2 1 25,006,490 mm3
Stress is less than the yield stress, so Hooke's law is valid. AXIAL STRAIN e s E 130 MPa 200 GPa 650 10
6
INITIAL VOLUME V Lb2 25,000,000 mm3
INCREASE IN VOLUME V V1 V 6490 mm3
INCREASE IN LENGTH L eL (650 10 6 )(2500 mm) 1.625 mm
SECTION 1.6
Shear Stress and Strain
25
Shear Stress and Strain
Problem 1.6-1 An angle bracket having thickness t 0.5 in. is attached to the flange of a column by two 5/8-inch diameter bolts (see figure). A uniformly distributed load acts on the top face of the bracket with a pressure p 300 psi. The top face of the bracket has length L 6 in. and width b 2.5 in. Determine the average bearing pressure b between the angle bracket and the bolts and the average shear stress aver in the bolts. (Disregard friction between the bracket and the column.)
p b
L
t
Solution 1.6-1
Angle bracket bolted to a column p pressure acting on top of the bracket 300 psi
b
F
F
resultant force acting on the bracket pbL (300 psi) (2.5 in.) (6.0 in.) 4.50 k
L
BEARING PRESSURE BETWEEN BRACKET AND BOLTS Ab bearing area of one bolt dt (0.625 in.) (0.5 in.) 4.50 k 2(0.3125 in.2 ) 0.3125 in.2 7.20 ksi
t
sb
F 2Ab
Two bolts d t b L 0.625 in. thickness of angle 2.5 in. 6.0 in. 0.5 in.
AVERAGE SHEAR STRESS IN THE BOLTS As Shear area of one bolt 4 taver d2 F 2As 4 (0.625 in.) 2 4.50 k 2(0.3068 in.2 ) 0.3068 in.2 7.33 ksi
26
CHAPTER 1
Tension, Compression, and Shear
Problem 1.6-2 Three steel plates, each 16 mm thick, are joined by two 20-mm diameter rivets as shown in the figure. (a) If the load P 50 kN, what is the largest bearing stress acting on the rivets? (b) If the ultimate shear stress for the rivets is 180 MPa, what force Pult is required to cause the rivets to fail in shear? (Disregard friction between the plates.)
P/2 P/2
P
P
P
Solution 1.6-2
P/2 P/2
Three plates joined by two rivets
P t
sb
P 2Ab
P 2dt
50 kN 2(20 mm)(16 mm)
78.1 MPa (b) ULTIMATE LOAD IN SHEAR
P
P
Shear force on two rivets Shear force on one rivet t d P
ULT
P 2 P 4
thickness of plates diameter of rivets 50 kN
16 mm 20 mm Let A
180 MPa (for shear in the rivets)
cross-sectional area of one rivet P4 P P Shear stress t d2 A d2 4( 4 ) 2 or, P d At the ultimate load: PULT d 2tULT 226 kN (20 mm) 2 (180 MPa)
(a) MAXIMUM BEARING STRESS ON THE RIVETS Maximum stress occurs at the middle plate. Ab bearing area for one rivet dt
Problem 1.6-3 A bolted connection between a vertical column and a diagonal brace is shown in the figure. The connection consists of three 5/8-in. bolts that join two 1/4-in. end plates welded to the brace and a 5/8-in. gusset plate welded to the column. The compressive load P carried by the brace equals 8.0 k. Determine the following quantities: (a) The average shear stress aver in the bolts, and (b) The average bearing stress b between the gusset plate and the bolts. (Disregard friction between the plates.)
P
Column Brace
End plates for brace Gusset plate
SECTION 1.6
Shear Stress and Strain
27
Solution 1.6-3
Diagonal brace
P
(a) AVERAGE SHEAR STRESS IN THE BOLTS A cross-sectional area of one bolt d2 4 V 0.3068 in.2
End plates
P Gusset plate
1 P P 3 2 6 V P taver A 6A
shear force acting on one bolt
8.0 k 6(0.3068 in.2 )
4350 psi 3 bolts in double shear P d t1
5
(b) AVERAGE BEARING STRESS AGAINST GUSSET PLATE 8.0 k 0.625 in. Ab F bearing area of one bolt t1d (0.625 in.)(0.625 in.) 0.3906 in.2
5
compressive force in brace diameter of bolts /8 in.
thickness of gusset plate /8 in. 0.625 in.
t2
1
thickness of end plates /4 in. 0.25 in. sb
bearing force acting on gusset plate from one bolt P 3 P 8.0 k 6830 psi 3Ab 3(0.3906 in.2 )
Problem 1.6-4 A hollow box beam ABC of length L is supported at end A by a 20-mm diameter pin that passes through the beam and its supporting pedestals (see figure). The roller support at B is located at distance L/3 from end A. (a) Determine the average shear stress in the pin due to a load P equal to 10 kN. (b) Determine the average bearing stress between the pin and the box beam if the wall thickness of the beam is equal to 12 mm.
P Box beam A B C
L -- 3
2L -- 3
Box beam Pin at support A
28
CHAPTER 1
Tension, Compression, and Shear
Solution 1.6-4
Hollow box beam
P
P d t
10 kN diameter of pin 20 mm 12 mm
A
B
C
wall thickness of box beam
(a) AVERAGE SHEAR STRESS IN PIN
L -- 3 R = 2P 2P 2L -- 3
Double shear taver 2 d2 4 2P 4P d2 31.8 MPa
(b) AVERAGE BEARING STRESS ON PIN
R =P 2 R =P 2
sb
2P 2(dt)
P dt
41.7 MPa
Problem 1.6-5 The connection shown in the figure consists of five steel plates, each 3/16 in. thick, joined by a single 1/4-in. diameter bolt. The total load transferred between the plates is 1200 lb, distributed among the plates as shown. (a) Calculate the largest shear stress in the bolt, disregarding friction between the plates. (b) Calculate the largest bearing stress acting against the bolt.
360 lb 480 lb 360 lb
600 lb 600 lb
Solution 1.6-5 d t
Plates joined by a bolt
1
diameter of bolt thickness of plates
/4 in.
3
/16 in.
(a) MAXIMUM SHEAR STRESS IN BOLT Vmax 4Vmax tmax 7330 psi d2 d2 4 (b) MAXIMUM BEARING STRESS
FREE-BODY DIAGRAM OF BOLT
360 lb 480 lb 360 lb A B B A A B B A 600 lb 600 lb
Fmax Fmax sb
maximum force applied by a plate against the bolt 600 lb Fmax dt 12,800 psi
Section A Section B Vmax
A: V B: V
360 lb 240 lb
max. shear force in bolt 360 lb
SECTION 1.6
Shear Stress and Strain
29
Problem 1.6-6 A steel plate of dimensions 2.5 1.2 0.1 m is hoisted by a cable sling that has a clevis at each end (see figure). The pins through the clevises are 18 mm in diameter and are located 2.0 m apart. Each half of the cable is at an angle of 32 to the vertical. For these conditions, determine the average shear stress aver in the pins and the average bearing stress b between the steel plate and the pins.
P
Cable sling 32 32 Clevis
2.0 m Steel plate (2.5 1.2 0.1 m)
Solution 1.6-6
Steel plate hoisted by a sling 1.2 0.1 m 0.300 m3 TENSILE FORCE T IN CABLE Fvertical T cos 32 18 mm T 0 W 2 0 23.10 kN 2 cos 32 13.62 kN
Dimensions of plate: 2.5 Volume of plate: V
(2.5) (1.2) (0.1) m 77.0 kN/m3 23.10 kN
Weight density of steel: Weight of plate: W d t V
diameter of pin through clevis thickness of plate 0.1 m
100 mm
W 2 cos 32
FREE-BODY DIAGRAMS OF SLING AND PIN
P=W T 32 Pin W 2 32 32
SHEAR STRESS IN THE PINS (DOUBLE SHEAR) taver T 2Apin 13.62 kN 2( 4 )(18 mm) 2 26.8 MPa
H Cable
BEARING STRESS BETWEEN PLATE AND PINS Ab bearing area td T td 13.62 kN (100 mm)(18 mm)
sb
7.57 MPa
W 2 H 2.0 m H W 2
30
CHAPTER 1
Tension, Compression, and Shear
Problem 1.6-7 A special-purpose bolt of shank diameter d 0.50 in. passes through a hole in a steel plate (see figure). The hexagonal head of the bolt bears directly against the steel plate. The radius of the circumscribed circle for the hexagon is r 0.40 in. (which means that each side of the hexagon has length 0.40 in.). Also, the thickness t of the bolt head is 0.25 in. and the tensile force P in the bolt is 1000 lb. (a) Determine the average bearing stress b between the hexagonal head of the bolt and the plate. (b) Determine the average shear stress aver in the head of the bolt.
Steel plate d 2r t P
Solution 1.6-7
Bolt in tension d r
d
0.50 in. 0.40 in. 0.25 in. 1000 lb
(a) BEARING STRESS BETWEEN BOLT HEAD AND PLATE Ab Ab bearing area area of hexagon minus area of bolt 3r2 3 d2 2 4 3 (0.40 in.) 2 ( 3) (0.50 in.) 2 2 4 0.4157 in.2 0.1963 in.2 0.2194 in.2 P Ab 1000 lb 0.2194 in.2 4560 psi
2r
P
t P
t
Ab Area of one equilateral triangle
r
r2 4
3
sb
Area of hexagon
2r
(b) SHEAR STRESS IN HEAD OF BOLT As taver shear area As P As P dt dt 1000 lb (0.50 in.)(0.25 in.)
3r2 3 2
2550 psi
Problem 1.6-8 An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force V during a static loading test (see figure). The pad has dimensions a 150 mm and b 250 mm, and the elastomer has thickness t 50 mm. When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0 mm with respect to the bottom plate. What is the shear modulus of elasticity G of the chloroprene?
b a V
t
SECTION 1.6
Shear Stress and Strain
31
Solution 1.6-8
d = 8.0 mm V
Bearing pad subjected to shear taver
t = 50 mm
V ab d t t g
12 kN (150 mm)(250 mm) 8.0 mm 50 mm 0.32 MPa 0.16 0.16 2.0 MPa
0.32 MPa
gaver G
b = 250 mm
V
12 kN 150 mm 250 mm 8.0 mm
Width of pad: a Length of pad: b d
Problem 1.6-9 A joint between two concrete slabs A and B is filled with a flexible epoxy that bonds securely to the concrete (see figure). The height of the joint is h 4.0 in., its length is L 40 in., and its thickness is t 0.5 in. Under the action of shear forces V, the slabs displace vertically through the distance d 0.002 in. relative to each other. (a) What is the average shear strain aver in the epoxy? (b) What is the magnitude of the forces V if the shear modulus of elasticity G for the epoxy is 140 ksi?
h
A
L
B
t d A h V V B
t
Solution 1.6-9
Epoxy joint between concrete slabs
d
(a) AVERAGE SHEAR STRAIN
B h
A V V t
gaver
d t
0.004
(b) SHEAR FORCES V Average shear stress :
aver
G
aver
h t L d G
4.0 in. 0.5 in. 40 in. 0.002 in. 140 ksi
V
aver(hL)
G
aver(hL)
(140 ksi)(0.004)(4.0 in.)(40 in.) 89.6 k