01 01 chapgere[1]
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The document describes several problems involving the calculation of normal stress and strain in structural elements like posts, rods, wires, and beams. The problems involve circular and rectangular cross-sections under compression, tension or a combination of forces. Diagrams are provided and the geometry, forces and material properties are used to calculate stress, strain or unknown forces through equilibrium equations. Solutions show the relevant equations and step-by-step workings to arrive at the final numerical answers.
![Solution 1.2-3 Long steel rod in tension
Problem 1.2-4 A circular aluminum tube of length L ϭ 400
mm is loaded in compression by forces P (see figure). The out-
side 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 ⑀ ϭ 550 ϫ 10Ϫ6, 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?
Solution 1.2-4 Aluminum tube in compression
SECTION 1.2 Normal Stress and Strain 3
ϭ 374.3 psi
maxϭ 374 psi ϩ 4074 psi ϭ 4448 psi
Rounding, we get
max ϭ 4450 psi
P
A
ϭ
200 lb
4 (0.25 in.)2 ϭ 4074 psi
gL ϭ (490 lbրft3
)(110 ft)¢
1
144
ft2
in.2 ≤
smax ϭ
W ϩ P
A
ϭ gL ϩ
P
A
d
L
P = 200 lb
P ϭ 200 lb
L ϭ 110 ft
d ϭ 1
⁄4 in.
Weight density: ␥ ϭ 490 lb/ft3
W ϭ Weight of rod
ϭ ␥(Volume)
ϭ ␥AL
Strain gage
L = 400 mm
PP
e ϭ 550 ϫ 10Ϫ6
L ϭ 400 mm
d2 ϭ 60 mm
d1 ϭ 50 mm
(a) SHORTENING ␦ OF THE BAR
␦ ϭ eL ϭ (550 ϫ 10Ϫ6)(400 mm)
ϭ 0.220 mm
(b) COMPRESSIVE LOAD P
ϭ 40 MPa
ϭ 863.9mm2
P ϭ A ϭ (40 MPa)(863.9 mm2)
ϭ 34.6 kN
A ϭ
4
[d2
2
Ϫ d1
2
] ϭ
4
[(60 mm)2
Ϫ (50 mm)2
]
Strain gage
PP](https://image.slidesharecdn.com/01-01chapgere1-130611230425-phpapp02/85/01-01-chapgere-1-3-320.jpg)
![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 xෆ and yෆ of the point where the
resultant load must act in order to produce uniform normal
stress.
Solution 1.2-5 Concrete pier in compression
4 CHAPTER 1 Tension, Compression, and Shear
O 16 in.20 in.
48 in.
y
x
16 in.
16 in.
16 in.
x
C
y
1
2
3
4
USE THE FOLLOWING AREAS:
A1 ϭ (48 in.)(20 in.) ϭ 960 in.2
A3 ϭ (16 in.)(16 in.) ϭ 256 in.2
A ϭ A1 ϩ A2 ϩ A3 ϩ A4
ϭ (960 ϩ 128 ϩ 256 ϩ 128) in.2
ϭ 1472 in.2
A2 ϭ A4 ϭ
1
2
(16 in.)(16 in.) ϭ 128 in.2
(a) AVERAGE COMPRESSIVE STRESS c
P ϭ 2500k
(b) COORDINATES OF CENTROID C
(see Chapter 12, Eq. 12-7a)
ϩ 2(25.333 in.)(128 in.2)
ϩ(28 in.)(256 in.2)]
ϭ15.8 in.
ϭ
1
1472 in.2 [(10 in.)(960 in.2
)
x ϭ
1
A
(x1 A1 ϩ 2x2 A2 ϩ x3 A3)
x ϭ
© xi Ai
A
From symmetry, y ϭ
1
2
(48 in.) ϭ 24 in.
sc ϭ
P
A
ϭ
2500 k
1472 in.2 ϭ 1.70 ksi
O 16 in.20 in.
48 in.
y
x
16 in.
16 in.
16 in.
20 in.
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
␣](https://image.slidesharecdn.com/01-01chapgere1-130611230425-phpapp02/85/01-01-chapgere-1-4-320.jpg)
![Problem 1.5-5 A bar of monel metal (length L ϭ 8 in.,
diameter d ϭ 0.25 in.) is loaded axially by a tensile force
P ϭ 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.
22 CHAPTER 1 Tension, Compression, and Shear
Solution 1.5-5 Bar of monel metal in tension
L ϭ 8in. d ϭ 0.25in. P ϭ 1500lb
From Table H-2: E ϭ 25,000 ksi ϭ 0.32
AXIAL STRESS
Assume is less than the proportional limit, so that
Hooke’s law is valid.
AXIAL STRAIN
INCREASE IN LENGTH
␦ ϭ e L ϭ (0.001222)(8in.) ϭ 0.00978in.
LATERAL STRAIN
DECREASE IN DIAMETER
ϭ 0.0000978 in.
¢d ϭ Ϳe¿d Ϳ ϭ (0.0003910)(0.25 in.)
ϭ Ϫ0.0003910
e¿ ϭ Ϫ ne ϭ Ϫ(0.32)(0.001222)
e ϭ
s
E
ϭ
30,560 psi
25,000 ksi
ϭ 0.001222
ϭ 30,560 psis ϭ
P
A
ϭ
1500 lb
4 (0.25 in.)2
DECREASE IN CROSS-SECTIONAL AREA
Original area:
Final area:
Decrease in area:
⌬A ϭ A0 Ϫ A1
PERCENT DECREASE IN AREA
ϭ 0.078%
ϭ
(0.0000978)(0.4999)
(0.25)2 (100)
Percent ϭ
¢A
A0
(100) ϭ
(¢d)(2d Ϫ ¢d)
d2 (100)
¢A ϭ
4
(¢d)(2d Ϫ ¢d)
A1 ϭ
4
[d2
Ϫ 2d¢d ϩ (¢d)2
]
A1 ϭ
4
(d Ϫ ¢d)2
A0 ϭ
d2
4
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
PP
50 mm](https://image.slidesharecdn.com/01-01chapgere1-130611230425-phpapp02/85/01-01-chapgere-1-22-320.jpg)
![Problem 1.5-7 A hollow steel cylinder is compressed by a force P
(see figure). The cylinder has inner diameter d1 ϭ 3.9 in., outer diameter
d2 ϭ 4.5 in., and modulus of elasticity E ϭ 30,000 ksi. When the force P
increases from zero to 40 k, the outer diameter of the cylinder increases
by 455 ϫ 10Ϫ6 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.
SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 23
Solution 1.5-6 Brass specimen in tension
d ϭ 10mm Gage length L ϭ 50mm
P ϭ 20kN ␦ ϭ 0.122mm ⌬d ϭ 0.00830mm
AXIAL STRESS
Assume is below the proportional limit so that
Hooke’s law is valid.
AXIAL STRAIN
e ϭ
␦
L
ϭ
0.122 mm
50 mm
ϭ 0.002440
s ϭ
P
A
ϭ
20 kN
4 (10 mm)2 ϭ 254.6 MPa
(a) MODULUS OF ELASTICITY
(b) POISSON’S RATIO
eЈ ϭ e
⌬d ϭ eЈd ϭ ed
n ϭ
¢d
ed
ϭ
0.00830 mm
(0.002440)(10 mm)
ϭ 0.34
E ϭ
s
e
ϭ
254.6 MPa
0.002440
ϭ 104 GPa
P
d1
d2
Solution 1.5-7 Hollow steel cylinder
d1 ϭ 3.9in.
d2 ϭ 4.5in.
t ϭ 0.3in.
E ϭ 30,000ksi
P ϭ 40k (compression)
⌬d2 ϭ 455 ϫ 10Ϫ6 in. (increase)
LATERAL STRAIN
(a) INCREASE IN INNER DIAMETER
(b) INCREASE IN WALL THICKNESS
ϭ 30 ϫ 10Ϫ6
in.
¢t ϭ e¿t ϭ (0.0001011)(0.3 in.)
ϭ 394 ϫ 10Ϫ6
in.
¢d1 ϭ e¿d1 ϭ (0.0001011)(3.9 in.)
e¿ ϭ
¢d2
d2
ϭ
455 ϫ 10Ϫ6
in.
4.5 in.
ϭ 0.0001011
(c) POISSON’S RATIO
Axial stress:
(compression)
( Ͻ Y; Hooke’s law is valid)
Axial strain:
ϭ 0.30
n ϭ
e¿
e
ϭ
0.0001011
0.000337
ϭ 0.000337
e ϭ
s
E
ϭ
10.105 ksi
30,000 ksi
ϭ 10.105 ksi
s ϭ
P
A
ϭ
40 k
3.9584 in.2
ϭ 3.9584 in.2
A ϭ
4
[d2
2
Ϫ d1
2
] ϭ
4
[(4.5 in.)2
Ϫ (3.9 in.)2
]
s ϭ
P
Ad2d1
t](https://image.slidesharecdn.com/01-01chapgere1-130611230425-phpapp02/85/01-01-chapgere-1-23-320.jpg)
01 01 chapgere[1]
- 1. Normal Stress and Strain Problem 1.2-1 A solid circular post ABC (see figure) supports a load P1 ϭ 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 ϭ 1.25 in. and dBC ϭ 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? Solution 1.2-1 Circular post in compression 1 Tension, Compression, and Shear 1 A B C P1 dAB dBC P2 P1 ϭ 2500 lb dAB ϭ 1.25 in. dBC ϭ 2.25 in. (a) NORMAL STRESS IN PART AB (b) LOAD P2 FOR EQUAL STRESSES ϭAB ϭ 2040 psi Solve for P2: P2 ϭ 5600 lb sBC ϭ P1 ϩ P2 ABC ϭ 2500 lb ϩ P2 4 (2.25 in.)2 sAB ϭ P1 AAB ϭ 2500 lb 4 (1.25 in.)2 ϭ 2040 psi ALTERNATE SOLUTION FOR PART (b) ∴ P2 ϭ 2.24P1 ϭ 5600 lb dBC dAB ϭ 1.8 P1 ϩ P2 dBC 2 ϭ P1 dAB 2 or P2 ϭ P1 B¢ dBC dAB ≤ 2 Ϫ 1R sAB ϭ P1 AAB ϭ P1 4 dAB 2 sBC ϭ sAB sBC ϭ P1 ϩ P2 ABC ϭ P1 ϩ P2 4 dBC 2 A B C P1 P2
- 2. Piston rod 5 mm 50 mm 225 mm P = 40 N Problem 1.2-2 Calculate the compressive stress c in the circular piston rod (see figure) when a force P ϭ 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 perpen- dicular to the line of action of the force P. Solution 1.2-2 Free-body diagram of brake pedal 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, calcu- late 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.) 2 CHAPTER 1 Tension, Compression, and Shear F ϭ compressive force in piston rod d ϭ diameter of piston rod ϭ 5 mm EQUILIBRIUM OF BRAKE PEDAL F(50 mm) Ϫ P(275 mm) ϭ 0 COMPRESSIVE STRESS IN PISTON ROD (d ϭ 5 mm) sc ϭ F A ϭ 220 N 4 (5 mm)2 ϭ 11.2 MPa F ϭP¢ 275 mm 50 mm ≤ ϭ (40 N) ¢ 275 50 ≤ ϭ 220 N ©MA ϭ 0 ە ۔ 50 mm 225 mm P = 40 N F A in. 200 lb 110 ft 1 4 —
- 3. Solution 1.2-3 Long steel rod in tension Problem 1.2-4 A circular aluminum tube of length L ϭ 400 mm is loaded in compression by forces P (see figure). The out- side 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 ⑀ ϭ 550 ϫ 10Ϫ6, 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? Solution 1.2-4 Aluminum tube in compression SECTION 1.2 Normal Stress and Strain 3 ϭ 374.3 psi maxϭ 374 psi ϩ 4074 psi ϭ 4448 psi Rounding, we get max ϭ 4450 psi P A ϭ 200 lb 4 (0.25 in.)2 ϭ 4074 psi gL ϭ (490 lbրft3 )(110 ft)¢ 1 144 ft2 in.2 ≤ smax ϭ W ϩ P A ϭ gL ϩ P A d L P = 200 lb P ϭ 200 lb L ϭ 110 ft d ϭ 1 ⁄4 in. Weight density: ␥ ϭ 490 lb/ft3 W ϭ Weight of rod ϭ ␥(Volume) ϭ ␥AL Strain gage L = 400 mm PP e ϭ 550 ϫ 10Ϫ6 L ϭ 400 mm d2 ϭ 60 mm d1 ϭ 50 mm (a) SHORTENING ␦ OF THE BAR ␦ ϭ eL ϭ (550 ϫ 10Ϫ6)(400 mm) ϭ 0.220 mm (b) COMPRESSIVE LOAD P ϭ 40 MPa ϭ 863.9mm2 P ϭ A ϭ (40 MPa)(863.9 mm2) ϭ 34.6 kN A ϭ 4 [d2 2 Ϫ d1 2 ] ϭ 4 [(60 mm)2 Ϫ (50 mm)2 ] Strain gage PP
- 4. 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 xෆ and yෆ of the point where the resultant load must act in order to produce uniform normal stress. Solution 1.2-5 Concrete pier in compression 4 CHAPTER 1 Tension, Compression, and Shear O 16 in.20 in. 48 in. y x 16 in. 16 in. 16 in. x C y 1 2 3 4 USE THE FOLLOWING AREAS: A1 ϭ (48 in.)(20 in.) ϭ 960 in.2 A3 ϭ (16 in.)(16 in.) ϭ 256 in.2 A ϭ A1 ϩ A2 ϩ A3 ϩ A4 ϭ (960 ϩ 128 ϩ 256 ϩ 128) in.2 ϭ 1472 in.2 A2 ϭ A4 ϭ 1 2 (16 in.)(16 in.) ϭ 128 in.2 (a) AVERAGE COMPRESSIVE STRESS c P ϭ 2500k (b) COORDINATES OF CENTROID C (see Chapter 12, Eq. 12-7a) ϩ 2(25.333 in.)(128 in.2) ϩ(28 in.)(256 in.2)] ϭ15.8 in. ϭ 1 1472 in.2 [(10 in.)(960 in.2 ) x ϭ 1 A (x1 A1 ϩ 2x2 A2 ϩ x3 A3) x ϭ © xi Ai A From symmetry, y ϭ 1 2 (48 in.) ϭ 24 in. sc ϭ P A ϭ 2500 k 1472 in.2 ϭ 1.70 ksi O 16 in.20 in. 48 in. y x 16 in. 16 in. 16 in. 20 in. 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 ␣
- 5. SECTION 1.2 Normal Stress and Strain 5 Solution 1.2-6 Car on inclined track FREE-BODY DIAGRAM OF CAR W ϭ Weight of car T ϭ Tensile force in cable ␣ ϭ Angle of incline A ϭ Effective area of cable R1, R2 ϭ Wheel reactions (no friction force between wheels and rails) EQUILIBRIUM IN THE INCLINED DIRECTION T ϭ W sin ␣ ©FT ϭ 0 Qϩ bϪ T Ϫ W sin ␣ ϭ 0 TENSILE STRESS IN THE CABLE SUBSTITUTE NUMERICAL VALUES: W ϭ 130 kN ␣ ϭ 30Њ A ϭ 490 mm2 ϭ 133 MPa st ϭ (130 kN)(sin 30Њ) 490 mm2 st ϭ T A ϭ W sin ␣ A ␣ W R1 R2 Problem 1.2-7 Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle ␣ ϭ 34° to the horizontal and wire BC is at an angle  ϭ 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 B A ␣  Solution 1.2-7 Two steel wires supporting a lamp FREE-BODY DIAGRAM OF POINT B SUBSTITUTE NUMERICAL VALUES: ϪTAB(0.82904) ϩ TBC(0.66913) ϭ 0 TAB(0.55919) ϩ TBC(0.74314) Ϫ 18 ϭ 0 SOLVE THE EQUATIONS: TAB ϭ 12.163 lb TBC ϭ 15.069 lb TENSILE STRESSES IN THE WIRES sBC ϭ TBC A ϭ 21,300 psi sAB ϭ TAB A ϭ 17,200 psi TAB TBC W = 18 lb y x0 ␣  EQUATIONS OF EQUILIBRIUM ⌺Fx ϭ 0 Ϫ TAB cos ␣ ϩ TBC cos  ϭ 0 ⌺Fy ϭ 0 TAB sin ␣ ϩ TBC sin  Ϫ W ϭ 0 ␣ ϭ 34Њ  ϭ 48Њ d ϭ 30 mils ϭ 0.030 in. A ϭ d2 4 ϭ 706.9 ϫ 10Ϫ6 in.2
- 6. 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 ϭ 190 kN. If each shore has a 150 mm ϫ 150 mm square cross sec- tion, what is the compressive stress c in the shores? 6 CHAPTER 1 Tension, Compression, and Shear Retaining wall Shore 30° 30° Concrete thrust block Soil 4.0 m 0.5 m F B C A 1.5 m Solution 1.2-8 Retaining wall braced by wood shores Wall Shore 30° 4.0 m F B C A 1.5 m 0.5 m F ϭ 190 kN A ϭ area of one shore A ϭ (150 mm)(150 mm) ϭ 22,500 mm2 ϭ 0.0225 m2 FREE-BODY DIAGRAM OF WALL AND SHORE C ϭ compressive force in wood shore CH ϭ horizontal component of C CV ϭ vertical component of C CH ϭ C cos 30Њ CV ϭ C sin 30Њ SUMMATION OF MOMENTS ABOUT POINT A ϪF(1.5 m)ϩCV (4.0 m)ϩCH (0.5 m) ϭ 0 or Ϫ (190 kN)(1.5 m) ϩ C(sin 30Њ)(4.0 m) ϩ C(cos 30Њ)(0.5 m) ϭ 0 ∴ C ϭ 117.14 kN COMPRESSIVE STRESS IN THE SHORES ϭ5.21 MPa sc ϭ C A ϭ 117.14 kN 0.0225 m2 ©MA ϭ 0 ە ۔ 30°F B A 1.5 m C 30° AV AH CV CH
- 7. Solution 1.2-9 Loading crane with girder and cable 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 ϭ 0.471 in2. The dimensions of the crane are H ϭ 9 ft, L1 ϭ 12 ft, and L2 ϭ 4 ft. (a) If the load P ϭ 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 H C P BA Girder Cable L2L1 T ϭ tensile force in cable P ϭ 9000 lb D H C P = 9000 lb BA L2L1 C P = 9000 lb BA 12 ft 4 ft TH T TV H ϭ 9 ft L1 ϭ 12 ft L2 ϭ 4 ft A ϭ effective area of cable A ϭ 0.471 in.2 P ϭ 9000 lb FREE-BODY DIAGRAM OF GIRDER EQUILIBRIUM TV (12 ft) Ϫ (9000 lb)(16 ft) ϭ 0 TV ϭ 12,000lb ϭ 16,000 lb TENSILE FORCE IN CABLE ϭ 20,000lb (a) AVERAGE TENSILE STRESS IN CABLE (b) AVERAGE STRAIN IN CABLE ␦ ϭ stretch of cable ␦ ϭ 0.382 in. e ϭ ␦ L ϭ 0.382 in. (15 ft)(12 in.րft) ϭ 2120 ϫ 10Ϫ6 L ϭ length of cable L ϭ ͙H 2 ϩ L1 2 ϭ 15 ft s ϭ T A ϭ 20,000 lb 0.471 in.2 ϭ 42,500 psi T ϭ ͙TH 2 ϩ TV 2 ϭ ͙(16,000 lb)2 ϩ (12,000 lb)2 TH ϭ (12,000 lb)¢ 12 9 ≤ ∴ TH ϭ TV ¢ 12 9 ≤ TH TV ϭ L1 H ϭ 12 ft 9 ft ©MA ϭ 0 ە ۔
- 8. Problem 1.2-10 Solve the preceding problem if the load P ϭ 32 kN; the cable has effective cross-sectional area A ϭ 481 mm2; the dimensions of the crane are H ϭ 1.6 m, L1 ϭ 3.0 m, and L2 ϭ 1.5 m; and the cable stretches by 5.1 mm. Figure is with Prob. 1.2-9. 8 CHAPTER 1 Tension, Compression, and Shear Solution 1.2-10 Loading crane with girder and cable FREE-BODY DIAGRAM OF GIRDER EQUILIBRIUM TV (3.0 m) Ϫ (32 kN)(4.5 m) ϭ 0 TV ϭ 48 kN ∴ ϭ 90 kN TH ϭ (48 kN)¢ 3.0 1.6 ≤ TH ϭ TV ¢ 3.0 1.6 ≤ TH TV ϭ L1 H ϭ 3.0 m 1.6 m ©MA ϭ 0 ۔ە TENSILE FORCE IN CABLE (a) AVERAGE TENSILE STRESS IN CABLE (b) AVERAGE STRAIN IN CABLE L ϭ length of cable ␦ ϭ stretch of cable ␦ ϭ 5.1 mm e ϭ ␦ L ϭ 5.1 mm 3.4 m ϭ 1500 ϫ 10Ϫ6 L ϭ ͙H2 ϩ L1 2 ϭ 3.4 m s ϭ T A ϭ 102 kN 481 mm2 ϭ 212 MPa ϭ 102 kN T ϭ ͙TH 2 ϩ TV 2 ϭ ͙(90 kN)2 ϩ (48 kN)2 D H C P = 32 kN BA L2L1 CBA 3.0 m 1.5 m TH T TV P = 32 kN T = tensile force in cable 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 ϭ 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.) Cables Reinforced concrete slab H ϭ 1.6 m L1ϭ 3.0 m L2 ϭ 1.5 m A ϭ effective area of cable A ϭ 481mm2 P ϭ 32 kN
- 9. Solution 1.2-11 Reinforced concrete slab supported by four cables SECTION 1.2 Normal Stress and Strain 9 H ϭ height of hook above slab L ϭ length of side of square slab t ϭ thickness of slab ␥ ϭ weight density of reinforced concrete W ϭ weight of slab ϭ ␥L2t D ϭ length of diagonal of slab DIMENSIONS OF CABLE AB ϭ L͙2 Cable B L L A W H t T T T T A W TH T TV B A H LAB D 2 L 2 = LAB ϭ length of cable ϭ B H2 ϩ L2 2 T ϭ tensile force in a cable Cable AB: (Eq. 1) EQUILIBRIUM ⌺Fvert ϭ 0 ↑ϩ↓Ϫ W Ϫ 4TV ϭ 0 (Eq. 2) COMBINE EQS. (1) & (2): TENSILE STRESS IN A CABLE A ϭ effective cross-sectional area of a cable SUBSTITUTE NUMERICAL VALUES AND OBTAIN t: H ϭ 5.0 ft L ϭ 8.0 ft t ϭ 9.0 in. ϭ 0.75 ft ␥ ϭ 150 lb/ft3 A ϭ 0.12 in.2 W ϭ ␥L2t ϭ 7200 lb st ϭ W 4A ͙1 ϩ L2 ր2H2 ϭ 22,600 psi st ϭ T A ϭ W 4A ͙1 ϩ L2 ր2H2 T ϭ W 4 ͙H2 ϩ L2 ր2 H ϭ W 4 ͙1 ϩ L2 ր2H 2 T ¢ H ͙H2 ϩ L2 ր2 ≤ ϭ W 4 TV ϭ W 4 TV ϭ T ¢ H ͙H2 ϩ L2 ր2 ≤ TV T ϭ H LAB 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? CA B L L x FREE-BODY DIAGRAM OF HOOK AT POINT A
- 10. Solution 1.3-1 Hanging wire of length L Solution 1.2-12 Rotating Bar 10 CHAPTER 1 Tension, Compression, and Shear ϭ angular speed (rad/s) A ϭ cross-sectional area ␥ ϭ weight density ϭ mass density g g Consider an element of mass dM at distance from the midpoint C. The variable ranges from x to L. dF ϭ Inertia force (centrifugal force) of element of mass dM (a) TENSILE STRESS IN BAR AT DISTANCE x (b) MAXIMUM TENSILE STRESS x ϭ 0 smax ϭ g2 L2 2g — sx ϭ Fx A ϭ g2 2g (L2 Ϫ x2 ) — Fx ϭ Ύ B D dF ϭ Ύ L x g g A2 jdj ϭ gA2 2g (L2 Ϫ x2 ) dF ϭ (dM)(j2 ) ϭ g g A2 jdj dM ϭ g g A dj C B L x dM d D 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. 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.) W ϭ total weight of steel wire ␥S ϭ weight density of steel ϭ 490lb/ft3 ␥W ϭ weight density of sea water ϭ 63.8lb/ft3 (b) WIRE HANGING IN SEA WATER F ϭ tensile force at top of wire ϭ 13,500 ft ϭ 40,000 psi (490 Ϫ 63.8)lbրft3 (144 in.2 րft2 ) Lmax ϭ smax gS Ϫ gW F ϭ (gS Ϫ gW) AL smax ϭ F A ϭ (gS Ϫ gW)L ϭ 11,800 ft Lmax ϭ smax gS ϭ 40,000 psi 490 lbրft3 (144 in.2 րft2 ) L A ϭ cross-sectional area of wire max ϭ 40 ksi (yield strength) (a) WIRE HANGING IN AIR W ϭ ␥SAL smax ϭ W A ϭ gSL
- 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 SECTION 1.3 Mechanical Properties of Materials 11 W ϭ total weight of tungsten wire ␥T ϭ weight density of tungsten ϭ 190 kN/m3 ␥W ϭ weight density of sea water ϭ 10.0 kN/m3 A ϭ cross-sectional area of wire max ϭ 1500 MPa (breaking strength) (a) WIRE HANGING IN AIR W ϭ ␥TAL ϭ 7900 m Lmax ϭ smax gT ϭ 1500 MPa 190 kNրm3 smax ϭ W A ϭ gTL (b) WIRE HANGING IN SEA WATER F ϭ tensile force at top of wire F ϭ (␥TϪ␥W)AL ϭ 8300 m ϭ 1500 MPa (190 Ϫ 10.0) kNրm3 Lmax ϭ smax gT Ϫ gW smax ϭ F A ϭ (gT Ϫ gW)L L 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. Gage lengthP P
- 12. 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 character- istic 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 ϭ ᎏ ␥ ᎏ 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.) 12 CHAPTER 1 Tension, Compression, and Shear Percent elongation L0 ϭ 2.0 in. Percent elongation (Eq. 1) where L1 is in inches. ϭ ¢ L1 2.0 Ϫ 1≤(100) ϭ L1 Ϫ L0 L0 (100) ϭ ¢ L1 L0 Ϫ 1≤100 d0 ϭ initial diameter d1 ϭ final diameter Percent reduction in area (Eq. 2) where d1 is in inches. ϭ B1 Ϫ ¢ d1 0.505 ≤ 2 R (100) A1 A0 ϭ ¢ d1 d0 ≤ 2 d0 ϭ 0.505 in. ϭ ¢1 Ϫ A1 A0 ≤(100) Percent reduction in area ϭ A0 Ϫ A1 A0 (100) 2.0 in PP 0.505 in L1 d1 % Elongation % Reduction Brittle or Material (in.) (in.) (Eq. 1) (Eq. 2) Ductile? A 2.13 0.484 6.5% 8.1% Brittle B 2.48 0.398 24.0% 37.9% Ductile C 2.78 0.253 39.0% 74.9% Ductile Solution 1.3-4 Strength-to-weight ratio 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 Values of U, ␥, and RS/W are listed in the table. RSրW ϭ sU(MPa) g(kNրm3 ) (103 ) U ␥ RS/W (MPa) (kN/m3) (m) Aluminum alloy 310 26.0 11.9 ϫ 103 6061-T6 Douglas fir 65 5.1 12.7 ϫ 103 Nylon 60 9.8 6.1 ϫ 103 Structural steel 500 77.0 6.5 ϫ 103 ASTM-A572 Titanium alloy 1050 44.0 23.9 ϫ 103 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. Solution 1.3-3 Tensile tests of three materials
- 13. SECTION 1.3 Mechanical Properties of Materials 13 Problem 1.3-5 A symmetrical framework consisting of three pin- connected bars is loaded by a force P (see figure). The angle between the inclined bars and the horizontal is ␣ ϭ 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.) P D A B C ␣ Solution 1.3-5 Symmetrical framework Aluminum alloy ␣ ϭ 48Њ eBD ϭ 0.0713 Use stress-strain diagram of Figure 1-13 L ϭ length of bar BD L1 ϭ distance BC ϭ L cot ␣ ϭ L(cot 48Њ) ϭ 0.9004 L L2 ϭ length of bar CD ϭ L csc ␣ ϭ L(csc 48Њ) ϭ 1.3456 L Elongation of bar BD ϭ distance DE ϭ eBDL eBDL ϭ 0.0713 L L3 ϭ distance CE ␦ ϭ elongation of bar CD ␦ ϭ L3 Ϫ L2 ϭ 0.0538L Strain in bar CD From the stress-strain diagram of Figure 1-13: s Ϸ 31 ksi ϭ ␦ L2 ϭ 0.0538L 1.3456L ϭ 0.0400 ϭ 1.3994 L ϭ ͙(0.9004L)2 ϩ L2 (1 ϩ 0.0713)2 L3 ϭ ͙L1 2 ϩ (L ϩ eBD L)2 P D A B C ␣ E ␣ D B C L L2 L3 ⑀BDL
- 14. 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. 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) Strain 8.0 0.0032 17.5 0.0073 25.6 0.0111 31.1 0.0129 39.8 0.0163 44.0 0.0184 48.2 0.0209 53.9 0.0260 58.1 0.0331 62.0 0.0429 62.1 Fracture Solution 1.3-6 Tensile test of a plastic Using the stress-strain data given in the problem statement, plot the stress-strain curve: PL ϭ proportional limit PL Ϸ 47 MPa Modulus of elasticity (slope) Ϸ 2.4 GPa Y ϭ yield stress at 0.2% offset Y Ϸ 53 MPa Material is brittle, because the strain after the proportional limit is exceeded is relatively small. — 0 20 40 60 Stress (MPa) 0.01 0.02 0.03 0.04 Strain 0.2% offset slope ≈ 40 MPa 0.017 = 2.4 GPa Y PL TENSILE-TEST DATA FOR PROBLEM 1.3-7 Load (lb) Elongation (in.) 1,000 0.0002 2,000 0.0006 6,000 0.0019 10,000 0.0033 12,000 0.0039 12,900 0.0043 13,400 0.0047 13,600 0.0054 13,800 0.0063 14,000 0.0090 14,400 0.0102 15,200 0.0130 16,800 0.0230 18,400 0.0336 20,000 0.0507 22,400 0.1108 22,600 Fracture
- 15. SECTION 1.3 Mechanical Properties of Materials 15 Solution 1.3-7 Tensile test of high-strength steel d0 ϭ 0.505in. L0 ϭ 2.00in. CONVENTIONAL STRESS AND STRAIN Load P Elongation ␦ Stress (lb) (in.) (psi) Strain e 1,000 0.0002 5,000 0.00010 2,000 0.0006 10,000 0.00030 6,000 0.0019 30,000 0.00100 10,000 0.0033 50,000 0.00165 12,000 0.0039 60,000 0.00195 12,900 0.0043 64,500 0.00215 13,400 0.0047 67,000 0.00235 13,600 0.0054 68,000 0.00270 13,800 0.0063 69,000 0.00315 14,000 0.0090 70,000 0.00450 14,400 0.0102 72,000 0.00510 15,200 0.0130 76,000 0.00650 16,800 0.0230 84,000 0.01150 18,400 0.0336 92,000 0.01680 20,000 0.0507 100,000 0.02535 22,400 0.1108 112,000 0.05540 22,600 Fracture 113,000 s ϭ P A0 e ϭ ␦ L0 A0 ϭ d0 2 4 ϭ 0.200 in.2 150,000 100,000 50,000 0 0.0200 0.0400 0.0600 Stress (psi) Strain 70,000 60,000 50,000 0 0.0020 0.0040 Stress (psi) Strain Slope ≈ ≈ 30 × 106 psi 50,000 psi 0.00165 0.1% pffset YP ≈ 69,000 psi PL ≈ 65,000 psi (0.1% offset) YP PL STRESS-STRAIN DIAGRAM ENLARGEMENT OF PART OF THE STRESS-STRAIN CURVE RESULTS Proportional limit Ϸ 65,000psi Modulus of elasticity (slope) Ϸ 30 ϫ 106 psi Yield stress at 0.1% offset Ϸ 69,000psi Ultimate stress (maximum stress) Ϸ 113,000psi Percent elongation in 2.00in. Percent reduction in area ϭ 31% ϭ 0.200 in.2 Ϫ 4 (0.42 in.)2 0.200 in.2 (100) ϭ A0 Ϫ A1 A0 (100) ϭ 0.12 in. 2.00 in. (100) ϭ 6% ϭ L1 Ϫ L0 L0 (100)
- 16. 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.) 16 CHAPTER 1 Tension, Compression, and Shear Elasticity, Plasticity, and Creep Problem 1.4-1 A bar made of structural steel having the stress- strain 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 ϫ 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) 0 60 40 20 0.0020 0.0060.004 ⑀ Solution 1.4-1 Steel bar in tension Y A B 0 ⑀R ⑀B ⑀ ⑀E L ϭ 48in. Yield stress Y ϭ 42 ksi Slope ϭ 30 ϫ 103 ksi ␦ ϭ 0.20 in. STRESS AND STRAIN AT POINT B B ϭ Y ϭ 42 ksi eB ϭ ␦ L ϭ 0.20 in. 48 in. ϭ 0.00417 ELASTIC RECOVERY eE RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.00417Ϫ0.00140 ϭ 0.00277 PERMANENT SET eRL ϭ (0.00277)(48in.) ϭ 0.13in. Final length of bar is 0.13in. greater than its original length. eE ϭ sB Slope ϭ 42 ksi 30 ϫ 103 ksi ϭ 0.00140 (MPa) 0 300 200 100 0.0020 0.0060.004 ⑀
- 17. SECTION 1.4 Elasticity, Plasticity, and Creep 17 Solution 1.4-2 Steel bar in tension Y A B 0 ⑀R ⑀B ⑀ ⑀E L ϭ 2.0m ϭ 2000 mm Yield stress Y ϭ 250 MPa Slope ϭ 200 GPa ␦ ϭ 6.5 mm ELASTIC RECOVERY eE RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.00325Ϫ0.00125 ϭ0.00200 Permanent set ϭ eRL ϭ (0.00200)(2000mm) ϭ 4.0mm Final length of bar is 4.0mm greater than its original length. eE ϭ sB Slope ϭ 250 MPa 200 GPa ϭ 0.00125 Problem 1.4-3 An aluminum bar has length L ϭ 4 ft and diameter d ϭ 1.0 in. The stress-strain curve for the alu- minum is shown in Fig. 1-13 of Section 1.3. The initial straight-line part of the curve has a slope (modulus of elas- ticity) of 10 ϫ 106 psi. The bar is loaded by tensile forces P ϭ 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 A B 0 ⑀R ⑀B ⑀ ⑀E L ϭ 4ft ϭ 48in. d ϭ 1.0in. P ϭ 24k See Fig. 1-13 for stress-strain diagram Slope from O to A is 10 ϫ 106 psi. STRESS AND STRAIN AT POINT B From Fig. 1-13: eB Ϸ 0.04 ELASTIC RECOVERY eE RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.04 Ϫ 0.0031 ϭ 0.037 (Note: The accuracy in this result is very poor because eB is approximate.) (a) PERMANENT SET eRL ϭ (0.037)(48in.) Ϸ 1.8in. (b) PROPORTIONAL LIMIT WHEN RELOADED B ϭ 31ksi eE ϭ sB Slope ϭ 31 ksi 10 ϫ 106 psi ϭ 0.0031 sB ϭ P A ϭ 24 k 4 (1.0 in.)2 ϭ 31 ksi STRESS AND STRAIN AT POINT B B ϭ Y ϭ 250 MPa eB ϭ ␦ L ϭ 6.5 mm 2000 mm ϭ 0.00325
- 18. Solution 1.4-4 Magnesium bar in tension 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.) (MPa) 0 0 200 100 0.005 0.010 ⑀ (PL)2 (PL)1 A B 0 ⑀R ⑀B ⑀ ⑀E L ϭ 800mm ␦ ϭ 5.6mm (PL)1 ϭ initial proportional limit ϭ 88 MPa (from stress-strain diagram) (PL)2 ϭ proportional limit when the bar is reloaded INITIAL SLOPE OF STRESS-STRAIN CURVE From -e diagram: At point A: (PL)1 ϭ 88 MPa eA ϭ 0.002 Problem 1.4-5 A wire of length L ϭ 4 ft and diameter d ϭ 0.125 in. is stretched by tensile forces P ϭ 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: ϭ ᎏ 1 1 ϩ 8,0 3 0 0 0 0 ⑀ ⑀ ᎏ 0 Յ ⑀ Յ 0.03 ( ϭ ksi) in which ⑀ is nondimensional and has units of kips per square inch (ksi). (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? STRESS AND STRAIN AT POINT B From -e diagram: B ϭ (PL)2 ϭ 170 MPa ELASTIC RECOVERY eE RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.007 Ϫ 0.00386 ϭ 0.00314 (a) PERMANENT SET eRL ϭ (0.00314)(800 mm) ϭ 2.51 mm (b) PROPORTIONAL LIMIT WHEN RELOADED (PL)2 ϭ B ϭ 170 MPa eE ϭ sB Slope ϭ (sPL)2 Slope ϭ 170 MPa 44 GPa ϭ 0.00386 eB ϭ ␦ L ϭ 5.6 mm 800 mm ϭ 0.007 Slope ϭ (sPL)1 eA ϭ 88 MPa 0.002 ϭ 44 GPa
- 19. Solution 1.4-5 Wire stretched by forces P SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 19 L ϭ 4 ft ϭ 48 in. d ϭ 0.125 in. P ϭ 600 lb COPPER ALLOY (Eq. 1) (a) STRESS-STRAIN DIAGRAM (From Eq. 1) 0 Յ e Յ 0.03 (s ϭ ksi)s ϭ 18,000e 1 ϩ 300e ALTERNATIVE FORM OF THE STRESS-STRAIN RELATIONSHIP Solve Eq. (1) for e in terms of : (Eq. 2) This equation may also be used when plotting the stress-strain diagram. (b) ELONGATION ␦ OF THE WIRE From Eq. (2) or from the stress-strain diagram: e ϭ 0.0147 ␦ ϭ eL ϭ (0.0147)(48in.) ϭ 0.71in. STRESS AND STRAIN AT POINT B (see diagram) B ϭ 48.9ksi eB ϭ 0.0147 ELASTIC RECOVERY eE RESIDUAL STRAIN eR eR ϭ eB Ϫ eE ϭ 0.0147 Ϫ 0.0027 ϭ 0.0120 (c) Permanent set ϭ eRL ϭ (0.0120)(48in.) ϭ 0.58in. (d) Proportional limit when reloaded ϭ B Bϭ49ksi eE ϭ sB Slope ϭ 48.9 ksi 18,000 ksi ϭ 0.00272 s ϭ P A ϭ 600 lb 4 (0.125 in.)2 ϭ 48,900 psi ϭ 48.9 ksi 0 Յ s Յ 54 ksi (s ϭ ksi)e ϭ s 18,000 Ϫ 300s 60 40 20 0 0.01 0.02 0.03 ⑀ ⑀R ⑀B ⑀E = ⑀B − ⑀R = 54 ksi B (ksi) B INITIAL SLOPE OF STRESS-STRAIN CURVE Take the derivative of with respect to e: ∴ Initial slopeϭ18,000ksi At e ϭ 0, ds de ϭ 18,000 ksi ϭ 18,000 (1 ϩ 300e)2 ds de ϭ (1 ϩ 300e)(18,000) Ϫ (18,000e)(300) (1 ϩ 300e)2 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 ϭ 2.00 in. (see figure). The steel has modulus of elasticity E ϭ 29 ϫ 106 psi and Poisson’s ratio ϭ 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? P d P
- 20. Solution 1.5-1 Steel bar in compression 20 CHAPTER 1 Tension, Compression, and Shear STEEL BAR d ϭ 2.00in. Max. ⌬d ϭ 0.001in. E ϭ 29 ϫ 106 psi ϭ 0.29 LATERAL STRAIN AXIAL STRAIN (shortening) e ϭ Ϫ e¿ n ϭ Ϫ 0.0005 0.29 ϭ Ϫ 0.001724 e¿ ϭ ¢d d ϭ 0.001 in. 2.00 in. ϭ 0.0005 AXIAL STRESS ϭ Ee ϭ (29 ϫ 106 psi)(Ϫ0.001724) ϭϪ50.00ksi (compression) Assume that the yield stress for the high-strength steel is greater than 50ksi. Therefore, Hooke’s law is valid. MAXIMUM COMPRESSIVE LOAD ϭ157k Pmax ϭ sA ϭ (50.00 ksi)¢ 4 ≤(2.00 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 prop- erties from Appendix H.) d = 10 mm 7075-T6 PP Solution 1.5-2 Aluminum bar in tension d ϭ 10mm ⌬d ϭ 0.016mm (Decrease in diameter) 7075-T6 From Table H-2: E ϭ 72GPa ϭ 0.33 From Table H-3: Yield stress Y ϭ 480 MPa LATERAL STRAIN AXIAL STRAIN ϭ 0.004848 (Elongation) e ϭ Ϫ Ϫe¿ n ϭ 0.0016 0.33 e¿ ϭ ¢d d ϭ Ϫ0.016 mm 10 mm ϭ Ϫ0.0016 AXIAL STRESS ϭ Ee ϭ (72GPa)(0.004848) ϭ 349.1 MPa (Tension) Because < Y, Hooke’s law is valid. LOAD P (TENSILE FORCE) ϭ 27.4kN P ϭ sA ϭ (349.1 MPa)¢ 4 ≤(10 mm)2 Problem 1.5-3 A nylon bar having diameter d1 ϭ 3.50 in. is placed inside a steel tube having inner diameter d2 ϭ 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 ϭ 400 ksi and ϭ 0.4.) d2d1 Steel tube Nylon bar
- 21. Solution 1.5-3 Nylon bar inside steel tube SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 21 COMPRESSION d1ϭ3.50 in. ⌬d1 ϭ 0.01in. d2ϭ3.51in. Nylon: E ϭ 400 ksi ϭ 0.4 LATERAL STRAIN e¿ ϭ 0.01 in. 3.50 in. ϭ 0.002857 e¿ ϭ ¢d1 d1 (Increase in diameter) AXIAL STRAIN (Shortening) AXIAL STRESS ϭ Ee ϭ (400ksi)(Ϫ0.007143) ϭϪ2.857ksi (Compressive stress) Assume that the yield stress is greater than and Hooke’s law is valid. FORCE P (COMPRESSION) ϭ 27.5k P ϭ sA ϭ (2.857 ksi)¢ 4 ≤(3.50 in.)2 e ϭ Ϫ e¿ n ϭ Ϫ 0.002857 0.4 ϭ Ϫ0.007143 d2d1 Problem 1.5-4 A prismatic bar of circular cross section is loaded by tensile forces P (see figure). The bar has length L ϭ 1.5 m and diameter d ϭ 30 mm. It is made of aluminum alloy with modulus of elasticity E ϭ 75 GPa and Poisson’s ratio ϭ 1 ⁄3. If the bar elongates by 3.6 mm, what is the decrease in diameter ⌬d? What is the magnitude of the load P? d PP L Solution 1.5-4 Aluminum bar in tension L ϭ 1.5m d ϭ 30mm E ϭ 75GPa ϭ 1 ⁄3 ␦ ϭ 3.6mm (elongation) AXIAL STRAIN LATERAL STRAIN ϭϪ0.0008 (Minus means decrease in diameter) e¿ ϭ Ϫ ne ϭ Ϫ(1 3)(0.0024) e ϭ ␦ L ϭ 3.6 mm 1.5 m ϭ 0.0024 DECREASE IN DIAMETER ⌬d ϭ eЈd ϭ (0.0008)(30mm) ϭ 0.024mm AXIAL STRESS ϭ Ee ϭ (75GPa)(0.0024) ϭ180 MPa (This stress is less than the yield stress, so Hooke’s law is valid.) LOAD P (TENSION) ϭ127kN P ϭ sA ϭ (180 MPa)¢ 4 ≤(30 mm)2
- 22. Problem 1.5-5 A bar of monel metal (length L ϭ 8 in., diameter d ϭ 0.25 in.) is loaded axially by a tensile force P ϭ 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. 22 CHAPTER 1 Tension, Compression, and Shear Solution 1.5-5 Bar of monel metal in tension L ϭ 8in. d ϭ 0.25in. P ϭ 1500lb From Table H-2: E ϭ 25,000 ksi ϭ 0.32 AXIAL STRESS Assume is less than the proportional limit, so that Hooke’s law is valid. AXIAL STRAIN INCREASE IN LENGTH ␦ ϭ e L ϭ (0.001222)(8in.) ϭ 0.00978in. LATERAL STRAIN DECREASE IN DIAMETER ϭ 0.0000978 in. ¢d ϭ Ϳe¿d Ϳ ϭ (0.0003910)(0.25 in.) ϭ Ϫ0.0003910 e¿ ϭ Ϫ ne ϭ Ϫ(0.32)(0.001222) e ϭ s E ϭ 30,560 psi 25,000 ksi ϭ 0.001222 ϭ 30,560 psis ϭ P A ϭ 1500 lb 4 (0.25 in.)2 DECREASE IN CROSS-SECTIONAL AREA Original area: Final area: Decrease in area: ⌬A ϭ A0 Ϫ A1 PERCENT DECREASE IN AREA ϭ 0.078% ϭ (0.0000978)(0.4999) (0.25)2 (100) Percent ϭ ¢A A0 (100) ϭ (¢d)(2d Ϫ ¢d) d2 (100) ¢A ϭ 4 (¢d)(2d Ϫ ¢d) A1 ϭ 4 [d2 Ϫ 2d¢d ϩ (¢d)2 ] A1 ϭ 4 (d Ϫ ¢d)2 A0 ϭ d2 4 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 PP 50 mm
- 23. Problem 1.5-7 A hollow steel cylinder is compressed by a force P (see figure). The cylinder has inner diameter d1 ϭ 3.9 in., outer diameter d2 ϭ 4.5 in., and modulus of elasticity E ϭ 30,000 ksi. When the force P increases from zero to 40 k, the outer diameter of the cylinder increases by 455 ϫ 10Ϫ6 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. SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 23 Solution 1.5-6 Brass specimen in tension d ϭ 10mm Gage length L ϭ 50mm P ϭ 20kN ␦ ϭ 0.122mm ⌬d ϭ 0.00830mm AXIAL STRESS Assume is below the proportional limit so that Hooke’s law is valid. AXIAL STRAIN e ϭ ␦ L ϭ 0.122 mm 50 mm ϭ 0.002440 s ϭ P A ϭ 20 kN 4 (10 mm)2 ϭ 254.6 MPa (a) MODULUS OF ELASTICITY (b) POISSON’S RATIO eЈ ϭ e ⌬d ϭ eЈd ϭ ed n ϭ ¢d ed ϭ 0.00830 mm (0.002440)(10 mm) ϭ 0.34 E ϭ s e ϭ 254.6 MPa 0.002440 ϭ 104 GPa P d1 d2 Solution 1.5-7 Hollow steel cylinder d1 ϭ 3.9in. d2 ϭ 4.5in. t ϭ 0.3in. E ϭ 30,000ksi P ϭ 40k (compression) ⌬d2 ϭ 455 ϫ 10Ϫ6 in. (increase) LATERAL STRAIN (a) INCREASE IN INNER DIAMETER (b) INCREASE IN WALL THICKNESS ϭ 30 ϫ 10Ϫ6 in. ¢t ϭ e¿t ϭ (0.0001011)(0.3 in.) ϭ 394 ϫ 10Ϫ6 in. ¢d1 ϭ e¿d1 ϭ (0.0001011)(3.9 in.) e¿ ϭ ¢d2 d2 ϭ 455 ϫ 10Ϫ6 in. 4.5 in. ϭ 0.0001011 (c) POISSON’S RATIO Axial stress: (compression) ( Ͻ Y; Hooke’s law is valid) Axial strain: ϭ 0.30 n ϭ e¿ e ϭ 0.0001011 0.000337 ϭ 0.000337 e ϭ s E ϭ 10.105 ksi 30,000 ksi ϭ 10.105 ksi s ϭ P A ϭ 40 k 3.9584 in.2 ϭ 3.9584 in.2 A ϭ 4 [d2 2 Ϫ d1 2 ] ϭ 4 [(4.5 in.)2 Ϫ (3.9 in.)2 ] s ϭ P Ad2d1 t
- 24. 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 ϭ 200 GPa and v ϭ 0.3. Determine the increase in volume of the bar. 24 CHAPTER 1 Tension, Compression, and Shear 100 mm100 mm 1300 kN 2.5 m 1300 kN Solution 1.5-8 Square bar in tension Find increase in volume. Length: L ϭ 2.5m ϭ 2500mm Side: b ϭ 100mm Force: P ϭ 1300kN E ϭ 200GPa ϭ 0.3 AXIAL STRESS Stress is less than the yield stress, so Hooke’s law is valid. AXIAL STRAIN INCREASE IN LENGTH ϭ 1.625 mm ¢L ϭ eL ϭ (650 ϫ 10Ϫ6 )(2500 mm) ϭ 650 ϫ 10Ϫ6 e ϭ s E ϭ 130 MPa 200 GPa s ϭ 1300 kN (100 mm)2 ϭ 130 MPa s ϭ P A ϭ P b2 DECREASE IN SIDE DIMENSION FINAL DIMENSIONS FINAL VOLUME INITIAL VOLUME V ϭ Lb2 ϭ 25,000,000 mm3 INCREASE IN VOLUME ⌬V ϭ V1ϪV ϭ 6490 mm3 V1 ϭ L1b1 2 ϭ 25,006,490 mm3 b1 ϭ b Ϫ ¢b ϭ 99.9805 mm L1 ϭ L ϩ ¢L ϭ 2501.625 mm ϭ 0.0195 mm ¢b ϭ e¿b ϭ (195 ϫ 10Ϫ6 )(100 mm) e¿ ϭ ne ϭ 195 ϫ 10Ϫ6
- 25. Solution 1.6-1 Angle bracket bolted to a column SECTION 1.6 Shear Stress and Strain 25 Shear Stress and Strain Problem 1.6-1 An angle bracket having thickness t ϭ 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 ϭ 300 psi. The top face of the bracket has length L ϭ 6 in. and width b ϭ 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.) t L p b Two bolts d ϭ 0.625in. t ϭ thickness of angle ϭ 0.5in. b ϭ 2.5in. L ϭ 6.0in. p ϭ pressure acting on top of the bracket ϭ 300psi F ϭ resultant force acting on the bracket ϭ pbL ϭ (300psi) (2.5in.) (6.0in.) ϭ 4.50k BEARING PRESSURE BETWEEN BRACKET AND BOLTS Ab ϭ bearing area of one bolt ϭ dt ϭ (0.625in.) (0.5in.) ϭ 0.3125in.2 AVERAGE SHEAR STRESS IN THE BOLTS As ϭ Shear area of one bolt taver ϭ F 2As ϭ 4.50 k 2(0.3068 in.2 ) ϭ 7.33 ksi ϭ 4 d2 ϭ 4 (0.625 in.)2 ϭ 0.3068 in.2 sb ϭ F 2Ab ϭ 4.50 k 2(0.3125 in.2 ) ϭ 7.20 ksit L F b
- 26. Solution 1.6-2 Three plates joined by two rivets 26 CHAPTER 1 Tension, Compression, and Shear t ϭ thickness of plates ϭ 16mm d ϭ diameter of rivets ϭ 20mm P ϭ 50kN ULT ϭ 180 MPa (for shear in the rivets) (a) MAXIMUM BEARING STRESS ON THE RIVETS Maximum stress occurs at the middle plate. ϭ dt Ab ϭ bearing area for one rivet (b) ULTIMATE LOAD IN SHEAR Let A ϭ cross-sectional area of one rivet or, P ϭ d2 At the ultimate load: ϭ 226 kN PULT ϭ d2 tULT ϭ (20 mm)2 (180 MPa) Shear stress t ϭ Pր4 A ϭ P 4(d 2 4 ) ϭ P d2 Shear force on one rivet ϭ P 4 Shear force on two rivets ϭ P 2 ϭ 78.1 MPa sb ϭ P 2Ab ϭ P 2dt ϭ 50 kN 2(20 mm)(16 mm) P P/2 P/2 P P t 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 End plates for brace Gusset plate Brace 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 ϭ 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 P/2 P/2 P P
- 27. Solution 1.6-3 Diagonal brace SECTION 1.6 Shear Stress and Strain 27 3 bolts in double shear P ϭ compressive force in brace ϭ 8.0k d ϭ diameter of bolts ϭ 5 ⁄8 in. ϭ 0.625in. t1 ϭ thickness of gusset plate ϭ 5 ⁄8 in. ϭ 0.625in. t2 ϭ thickness of end plates ϭ 1 ⁄4 in. ϭ 0.25in. (a) AVERAGE SHEAR STRESS IN THE BOLTS A ϭ cross-sectional area of one bolt V ϭ shear force acting on one bolt (b) AVERAGE BEARING STRESS AGAINST GUSSET PLATE Ab ϭ bearing area of one bolt ϭ t1d ϭ (0.625in.)(0.625in.) ϭ 0.3906in.2 F ϭ bearing force acting on gusset plate from one bolt sb ϭ P 3Ab ϭ 8.0 k 3(0.3906 in.2 ) ϭ 6830 psi ϭ P 3 ϭ 4350 psi taver ϭ V A ϭ P 6A ϭ 8.0 k 6(0.3068 in.2 ) ϭ 1 3 ¢ P 2 ≤ ϭ P 6 ϭ d2 4 ϭ 0.3068 in.2 Gusset plate End plates P P 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. — — Box beam Box beam P CA Pin at support A B 2L 3 L 3
- 28. Solution 1.6-4 Hollow box beam 28 CHAPTER 1 Tension, Compression, and Shear P ϭ 10kN d ϭ diameter of pin ϭ 20mm t ϭ wall thickness of box beam ϭ 12mm (a) AVERAGE SHEAR STRESS IN PIN Double shear (b) AVERAGE BEARING STRESS ON PIN sb ϭ 2P 2(dt) ϭ P dt ϭ 41.7 MPa taver ϭ 2P 2¢ 4 d2 ≤ ϭ 4P d2 ϭ 31.8 MPa — — P CA B 2L 3 L 3 2P R = 2P R 2 = P R 2 = P 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 Plates joined by a bolt d ϭ diameter of bolt ϭ 1 ⁄4 in. t ϭ thickness of plates ϭ 3 ⁄16 in. FREE-BODY DIAGRAM OF BOLT (a) MAXIMUM SHEAR STRESS IN BOLT (b) MAXIMUM BEARING STRESS Fmax ϭ maximum force applied by a plate against the bolt Fmax ϭ 600lb sb ϭ Fmax dt ϭ 12,800 psi tmax ϭ Vmax d2 4 ϭ 4Vmax d2 ϭ 7330 psi A B B A A B B A 360 lb 480 lb 360 lb 600 lb 600 lb Section A Ϫ A: V ϭ 360lb Section B Ϫ B: V ϭ 240lb Vmax ϭ max. shear force in bolt ϭ 360lb
- 29. Problem 1.6-6 A steel plate of dimensions 2.5 ϫ 1.2 ϫ 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. SECTION 1.6 Shear Stress and Strain 29 2.0 m Steel plate (2.5 × 1.2 × 0.1 m) Cable sling 32°32° Clevis P Solution 1.6-6 Steel plate hoisted by a sling Dimensions of plate: 2.5 ϫ 1.2 ϫ 0.1m Volume of plate: V ϭ (2.5) (1.2) (0.1) m ϭ 0.300m3 Weight density of steel: ␥ ϭ 77.0 kN/m3 Weight of plate: W ϭ ␥V ϭ 23.10 kN d ϭ diameter of pin through clevis ϭ18mm t ϭ thickness of plate ϭ 0.1m ϭ 100mm FREE-BODY DIAGRAMS OF SLING AND PIN TENSILE FORCE T IN CABLE ⌺Fvertical ϭ 0 ↑ϩ ↓Ϫ SHEAR STRESS IN THE PINS (DOUBLE SHEAR) ϭ 26.8 MPa BEARING STRESS BETWEEN PLATE AND PINS Ab ϭ bearing area ϭ td ϭ 7.57 MPa sb ϭ T td ϭ 13.62 kN (100 mm)(18 mm) taver ϭ T 2Apin ϭ 13.62 kN 2( 4 )(18 mm)2 T ϭ W 2 cos 32Њ ϭ 23.10 kN 2 cos 32Њ ϭ 13.62 kN T cos 32Њ Ϫ W 2 ϭ 0 2.0 m Cable 32°32° P = W H H W 2 W 2 W 2 T 32° H Pin
- 30. Problem 1.6-7 A special-purpose bolt of shank diameter d ϭ 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 circum- scribed circle for the hexagon is r ϭ 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. 30 CHAPTER 1 Tension, Compression, and Shear P d t Steel plate 2r Solution 1.6-7 Bolt in tension d ϭ 0.50 in. r ϭ 0.40 in. t ϭ 0.25 in. P ϭ 1000 lb (a) BEARING STRESS BETWEEN BOLT HEAD AND PLATE Abϭ bearing area Abϭ area of hexagon minus area of bolt Ab ϭ ϭ 0.4157 in.2Ϫ0.1963 in.2 ϭ 0.2194 in.2 (b) SHEAR STRESS IN HEAD OF BOLT As ϭ shear area As ϭ dt ϭ 2550psi taver ϭ P As ϭ P dt ϭ 1000 lb (0.50 in.)(0.25 in.) sb ϭ P Ab ϭ 1000 lb 0.2194 in.2 ϭ 4560 psi 3 2 (0.40 in.)2 (͙3) Ϫ ¢ 4 ≤(0.50 in.)2 ϭ 3r2 ͙3 2 Ϫ d2 4 P d t 2r 2r r Area of one equilateral triangle Area of hexagon ϭ 3r2 ͙3 2 ϭ r2 ͙3 4 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 ϭ 150 mm and b ϭ 250 mm, and the elastomer has thickness t ϭ 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? a b V t
- 31. Solution 1.6-8 Bearing pad subjected to shear SECTION 1.6 Shear Stress and Strain 31 V ϭ 12kN Width of pad: a ϭ 150mm Length of pad: b ϭ 250mm d ϭ 8.0mm G ϭ t g ϭ 0.32 MPa 0.16 ϭ 2.0 MPa gaver ϭ d t ϭ 8.0 mm 50 mm ϭ 0.16 taver ϭ V ab ϭ 12 kN (150 mm)(250 mm) ϭ 0.32 MPa V t = 50 mm d = 8.0 mm ␣ b = 250 mm 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 ϭ 4.0 in., its length is L ϭ 40 in., and its thickness is t ϭ 0.5 in. Under the action of shear forces V, the slabs displace vertically through the distance d ϭ 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? t h A BL t d h V V A B Solution 1.6-9 Epoxy joint between concrete slabs h ϭ 4.0in. t ϭ 0.5in. L ϭ 40in. d ϭ 0.002in. G ϭ 140ksi (a) AVERAGE SHEAR STRAIN (b) SHEAR FORCES V Average shear stress : aver ϭ G␥aver V ϭ aver(hL) ϭ G␥aver(hL) ϭ (140ksi)(0.004)(4.0in.)(40in.) ϭ 89.6k gaver ϭ d t ϭ 0.004 t d h V V A B