Physics, 6th Edition
Chapter 13. Elasticity
Chapter 13. Elasticity Elastic Properties of Matter 13-1. When a mass of 500 g is hung from a spring, the spring stretches 3 cm. What is the spring constant? [ m = 0.500 kg; x = 0.03 m, F = W = mg ] F = -kx;
F (0.50 kg)(9.8 m/s 2 ) ; k= = x 0.03 m
k = 163 N/m
13-2. What will be the increase in stretch for the spring of Problem 13-1 if an additional 500-g mass is hung blow the first? [ F = W = mg ] ∆x =
∆F (0.500 kg)(9.8 m/s 2 ) ; = k 163 N/m
∆x = 3.00 cm
13-3. The spring constant for a certain spring is found to be 3000 N/m. What force is required to compress the spring for a distance of 5 cm? F = kx = (3000 N/m)(0.05 m);
F = 150 N
13-4. A 6-in. spring has a 4-lb weight hung from one end, causing the new length to be 6.5 in. What is the spring constant? What is the strain? [ ∆x = 6.5 in. – 6.0 in. = 0.50 in. ] k=
F (4 lb) = ; x 0.5 in.
Strain =
k = 8.00 lb/in.
∆L 0.50 in. = ; L 6.00 in.
Strain = 0.0833
13-5. A coil spring 12 cm long is used to support a 1.8-kg mass producing a strain of 0.10. How far did the spring stretch? What is the spring constant? Strain =
∆L ; L k=
∆L = Lo ( strain) = (12.0 cm)(0.10) ; ∆F (1.8 kg)(9.8 m/s 2 ) = ; ∆L 0.0120 m
174
∆L = 1.20 cm
k = 1470 N/m
Physics, 6th Edition
Chapter 13. Elasticity
13-6. For the coil of Problem 13-5, what total mass should be hung if an elongation of 4 cm is desired? F = mg = kx;
m=
kx (1470 N/m)(0.04 m) = ; g 9.80 m/s 2
m = 6.00 kg
Young’s Modulus 13-7. A 60-kg weight is suspended by means of a cable having a diameter of 9 mm. What is the stress?
[ F = mg = (60 kg)(9.8 m/s2); F = 588 N; D = 0.009 m ]
π D 2 π (0.009 m) 2 A= = = 6.36 x 10-5 m 2 ; 4 4 Stress =
F 588 N = ; A 0.00707 m 2
Stress = 9.24 x 106 Pa
13-8. A 50-cm length of wire is stretched to a new length of 50.01 cm. What is the strain? ∆L = 50.01 – 50 cm;
Strain =
∆L 0.01 cm = ; L 50 cm
Strain = 2.00 x 10-4
13-9. A 12-m rod receives a compressional strain of -0.0004. What is the new length of the rod? Strain =
∆L ; L
∆L = Lo ( strain) = (12.0 m)(-0.0004) ;
∆L = -0.00480 m
L = Lo + ∆L = 12.000 m – 0.00480 m; L = 11.995 m 13-10. Young’s modulus for a certain rod is 4 x 1011 Pa. What strain will be produced by a tensile stress of 420 Mpa? Y=
Stress ; Strain
Strain =
Stress 420 x 106 Pa = Y 4 x 1011Pa
Strain = 1.05 x 10-3
175
Physics, 6th Edition
Chapter 13. Elasticity
13-11. A 500-kg mass is hung from the end of a 2-m length of metal wire 1 mm in diameter. If the wire stretches by 1.40 cm, what are the stress and strain? What is Young’s modulus for this metal? [ F = mg = (500 kg)(9.8 m/s2); F = 4900 N; D = 0.001 m; ∆L = 0.014 m ] A= Stress =
F 4900 N = ; A 7.85 x 10-7 m 2
Strain =
Y=
π D 2 π (0.001 m) 2 = = 7.85 x 10-7 m 2 4 4
∆L 0.014 m = ; L0 2.00 m
Stress 6.24 x 109 Pa ; = Strain 7 x 10-3
Stress = 6.24 x 109 Pa
Strain = 7.00 x 10-3
Y = 8.91 x 1011 Pa
13-12. A 16 ft steel girder with a cross-sectional area of 10 in.2 supports a compressional load of 20 tons. What is the decrease in length of the girder? [ Y = 30 x 106 Pa; 1 ton = 2000 lb ] Y=
FL ; A ∆L
∆L =
FL (−40, 000 lb)(16 ft)(12 in/ft) = ; ∆L = - 0.0256 in. YA (30 x 106 lb/in.2 )(10 in.2 )
13-13. How much will a 60 cm length of brass wire, 1.2 mm in diameter, elongate when a 3-kg mass is hung from an end? [ Y = 89.6 x 109 Pa; D = 0.0012 m; Lo = 0.60 m; m = 3 kg ] A=
π D 2 π (0.0012 m) 2 = = 1.13 x 10-6 m 2 ; 4 4
F = (3 kg)(9.8 m/s2) = 29.4 N;
∆L =
FL (29.4 N)(0.60 m) = ; YA (89.6 x 109 )(1.13 x 10-6 m 2 )
176
Y=
FL ; A ∆L
∆L =1.74 x 10-4 m
Physics, 6th Edition
Chapter 13. Elasticity
*13-14. A wire of cross-section 4 mm2 is stretched 0.1 mm by a certain weight. How far will a wire of the same material and length stretch if is cross-sectional area is 8 mm2 and the same weight is attached? Y=
FL FL = ; A1 ∆L1 A2 ∆L2
A1 ∆L1 = A2 ∆L2
A1∆L1 (4 mm 2 )(0.10 mm) ∆L2 = = ; A2 (8 mm 2 )
∆L2 = 0.0500 mm
13-15. A wire 15 ft long and 0.1 in.2 in cross-section is found to increase its length by 0.01 ft under a tension of 2000 lb. What is Young’s modulus for this wire? Can you identify the material? Y=
FL (2000 lb)(15 ft) = ; A ∆L (0.10 in.2 )(0.01 ft)
Y = 30 x 106 lb/in.2 , steel
Shear Modulus 13-16. A shearing force of 40,000 N is applied to the top of a cube that is 30 cm on a side. What is the shearing stress? [ A = (0.30 m)(0.30 m) = 0.09 m2 ] Stress =
F 40, 000 N = ; A 0.09 m 2
Stress = 4.44 x 105 Pa
*13-17. If the cube in Problem 13-16 is made of copper, what will be the lateral displacement of the upper surface of the cube? S=
F/A ; φ
φ=
d ; l
φ=
F / A 4.44 x 105 Pa = ; S 42.3 x 109 Pa
φ = 1.05 x 10-5 rad
d = lφ = (0.30 m)( 1.05 x 10-5 rad);
177
d = 3.15 µm
Physics, 6th Edition
Chapter 13. Elasticity
13-18. A shearing force of 26,000 N is distributed uniformly over the cross-section of a pin 1.3 cm in diameter. What is the shearing stress. [A = πD2/4 ] A=
π D 2 π (0.0130 m) 2 = = 1.33 x 10-4 m 2 ; 4 4
Stress =
F 26, 000 N = ; A 1.33 x 10-4 m 2
Stress = 1.96x 108 Pa
13-19. An aluminum rod 20 mm in diameter projects 4.0 cm from the wall. The end of the bolt is subjected to a shearing force of 48,000 N. Compute the downward deflection.
π D 2 π (0.020 m) 2 A= = = 3.14 x 10-4 m 2 ; 4 4 S=
d=
F / A F / A Fl = = ; φ d /l Ad
(48, 000 N)(0.04 m) ; (23.7 x 109 Pa)(3.14 x 10-4 m 2 )
l = 0.04 m; F = 48000 N
d=
Fl SA
d = 2.58 x 10-4 m
13-20. A steel rod projects 1.0 in. above a floor and is 0.5 in. in diameter. The shearing force F is 6000 lb and the shear modulus is 11.6 x 106 lb/in.2. What is the shearing stress and what is horizontal deflection? A=
π D 2 π (0.50 in.) 2 = = 0.196 in.2 ; l = 1.0 in.; F = 6000 lb 4 4 Stress =
d=
F 6, 000 lb = ; A 0.196 in.2
Stress = 3.06 x 104 lb/in.2
Fl (6000 lb)(1.0 in.) = ; SA (30 x 106 lb/in.2 )(0.196 in.2 )
178
d = 1.02 x 10-3 in.
Physics, 6th Edition
Chapter 13. Elasticity
13-21. A 1500-kg load is supported at the end of a 5-m aluminum beam as shown in Fig. 13-9. The beam has a cross-sectional area of 26 cm2 and the shear modulus is 23,700 MPa. What is the shearing stress and the downward deflection of the beam? A = 26 cm2 (10-4 m2/cm2) = 2.60 x 10-3 m2; S = 23.7 x 109 Pa; l = 5 m Stress =
φ=
F (1500 kg)(9.8 m/s 2 ) ; = A 2.60 x 10-3 m 2 F / A 5.65 x 106 Pa = ; S 23.7 x 109 Pa
S=
F/A ; φ
d ; l
d = lφ = (5.0 m)(2.39 x 10-4 rad) ;
φ=
Stress = 5.65 x 106 Pa
φ = 2.39 x 10-4 rad
d = 1.19 mm
13-22. A steel plate 0.5 in. thick has an ultimate shearing strength of 50,000 lb/in.2. What force must be applied to punch a ¼-in. hole through the plate? A=
π D 2 π (0.25 in.) 2 = = 0.0491 in.2 ; 4 4
Stress =
F = (50,000 lb/in.2)(0.0491 in.2);
F = 50, 000 lb/in.2 A
F = 2454 lb
Bulk Modulus 13-23. A pressure of 3 x 108 Pa is applied to a block of volume 0.500 m3. If the volume decreases by 0.004 m3, what is the bulk modulus? What is the compressibility? B=
−P −(3 x 108 Pa) ; = ∆V / V −0.004 m3 / 0.500 m3 k=
1 1 = ; B 37.5 x 109 Pa
179
B = 37.5 x 109 Pa
k = 2.67 x 10-11 Pa-1
Physics, 6th Edition
Chapter 13. Elasticity
*13-24. The bulk modulus for a certain grade of oil is 2.8 x 1010 Pa. What pressure is required to decrease its volume by a factor of 1.2 percent? [ ∆V/V = -1.2% = -0.012 ] B=
(
−P ; ∆V / V
-P = B ∆V
V
) = 2.8 x 10
10
Pa ( -0.012 ) ;
P = 3.36 x 108 Pa
*13-25. A solid brass sphere (B = 35,000 Mpa) of volume 0.8 m3 is dropped into the ocean to a depth where the water pressure is 20 Mpa greater than it is at the surface. What is the change in [ P = 20 x 106 Pa ]
volume of the sphere? B=
−P ; ∆V / V
∆V =
-PV −(20 x 106 Pa)(0.8 m3 ) ; = B 35 x 109 Pa
∆V = -4.57 x 10-4 m3.
13-26. A certain fluid compresses 0.40 percent under a pressure of 6 MPa. What is the compressibility of this fluid? [ ∆V/V = 0.04% = 0.0004 ] k=
∆V / V −0.0004 = ; −P −6 x 106 Pa
k = 6.67 x 10-11 Pa-1
13-27. What is the fractional decrease in the volume of water when it is subjected to a pressure of 15 MPa? B=
−P ; ∆V / V
∆V -P −(15 x 106 Pa) ; = = V B 2.10 x 109 Pa
∆V/V = -7.14 x 10-3
Challenge Problems 13-28. A 10-m steel wire, 2.5 mm in diameter, stretches a distance of 0.56 mm when a load is attached to its end. What was the mass of the load? A=
π D 2 π (0.0025 m) 2 = = 4.91 x 10-6 m 2 ; F = mg; 4 4
180
Y=
FL mgL = ; A ∆L A∆L
Physics, 6th Edition
Chapter 13. Elasticity
m=
13-28. (Cont.)
YA ∆L (207 x 109 Pa)(4.91 x 10-6 m)(0.00056 m) = ; gL (9.8 m/s 2 )(10 m)
m = 5.81 kg
13-29. A shearing force of 3000 N is applied to the upper surface of a copper cube 40 mm on a side. If S = 4.2 x 1010 Pa., what is the shearing angle? [A = (0.04 m)2 = 1.6 x 10-3 m2. ] S=
F/A ; φ
φ=
F 3000 N = ; -3 2 AS (1.6 x 10 m )(4.2 x 1010 Pa)
φ = 4.46 x 10-5 rad
13-30. A solid cylindrical steel column is 6 m long and 8 cm in diameter. What is the decrease in length if the column supports a 90,000-kg load? A=
π D 2 π (0.08 m) 2 = = 5.03 x 10-3 m 2 ; F = W = mg; L = 6.00 m 4 4 F = (90,000 kg)(9.8 m/s2) = 8.82 x 105 N;
∆L =
FL (8.82 x 105 N)(6.0 m) = ; YA (207 x 109 )(5.03 x 10-3m 2 )
Y=
FL ; A ∆L
∆L = -5.08 x 10-3 m
13-31. A piston, 8 cm in diameter, exerts a force of 2000 N on 1 liter of benzene. What is the decrease in volume of the benzene? A=
F 2000 N π D 2 π (0.08 m) 2 = 3.98 x 105 Pa = = 5.03 x 10-3 m 2 ; P = = -3 2 A 5.03 x 10 m 4 4
−P B= ; ∆V / V
-PV −(3.98 x 105 Pa)(0.001 m3 ) ; ∆V = = B 1.05 x 109 Pa
∆V = -3.79 x 10-7 m3.
13-32. How much will a 600-mm length of brass wire, 1.2 mm in diameter, stretch when a 4-kg mass is hung from its end? A=
π D 2 π (0.0012 m) 2 = = 1.13 x 10-6 m 2 ; F = (4 kg)(9.8 m/s2) = 39.2 N 4 4 181
Physics, 6th Edition
Chapter 13. Elasticity ∆L =
13-32. (Cont.)
FL (39.2 N)(0.60 m) = ; YA (89.6 x 109 )(1.13 x 10-6 m 2 )
∆L =2.32 x 10-4 m
13-33. A solid cylindrical steel column is 12 feet tall and 6 in. in diameter. What load does it support if its decrease in length is -0.0255 in.? A=
F=
π D 2 π (6 in.) 2 = = 28.3 in.2 ; 4 4
Y=
FL ; A ∆L
L = 12 ft = 144 in.
YA ∆L (30 x 106 lb/in.2 )(28.3 in.2 )(0.0255 in.) ; = L (144 in.)
F = 1.50 x 105 lb
13-34. Compute the volume contraction of mercury if its original volume of 1600 cm3 is subjected to a pressure of 400,000 Pa. [ 1600 cm3 = 1.6 x 10-3 m3 ] B=
−P ; ∆V / V
∆V =
− PV −(400, 000 Pa)(1.6 x 10-3 m3 ) ; = B 27.0 x 109 Pa
∆V = -2.37 x 10-8 m3.
*13-35. What is the minimum diameter of a brass rod if it is to undergo a 400 N tension without exceeding the elastic limit? F = 379 x 106 Pa; A D2 =
[ Elastic limit = 379 x 106 Pa ] 4F π D2 F 2 D = ; A= = π (379 x 106 Pa) 4 379 x 106 Pa
4(400 N) = 1.34 x 10-6 m 2 ; 6 π (379 x 10 Pa)
D = 1.16 mm
13-36. A cubical metal block 40 cm on a side is given a shearing force of 400,000 N at the top edge. What is the shear modulus if the upper edge deflects a distance of 0.0143 mm.
φ=
d 0.0143 mm = = 3.575 x 10-5 ; l 400 mm
S=
A = (0.40 m)2 = 0.160 m2
F/A 400, 000 N = ; φ (3.575 x 10-3 )(0.160 m 2 )
182
S = 6.99 x 1010 Pa
Physics, 6th Edition
Chapter 13. Elasticity
13-37. A steel piano wire has an ultimate strength of about 35,000 lb/in.2. How large a load can a 0.5-in.-diameter steel wire hold without breaking?
π D 2 π (0.5 in.) 2 A= = = 0.196 in.2 ; 4 4
F = 35, 000 lb/in.2 A
F = (35,000 lb/in.2)(0.196 in.2);
F = 6870 lb
Critical Thinking Questions *13-38. A metal wire increases its length by 2 mm when subjected to tensile force. What elongation can be expected from this same force if the diameter of the wire was reduced to one-half of its initial value? Suppose the metal wire maintains its diameter, but doubles its length. What elongation would be expected for the same load? Y=
FL ; A ∆L
∆L =
D2 ∆L2 = ∆L1 12 D2
FL 4 FL 4 FL = ; ∆L D 2 = ; 2 AY π D Y πY
∆L1 D12 = ∆L2 D22
2
2 mm D ; = (2 mm) = 4 2D
Since ∆L ∝ L, doubling L would also double ∆L:
∆L2 = 0.500 mm ∆L2 = 4.00 mm
13-39. A cylinder 4 cm in diameter is filled with oil. What total force must be exerted on the oil to produce a 0.8 percent decrease in volume? Compare the forces necessary if the oil is replaced by water? By mercury?
[ ∆V/V = -0.008 ]
−P −F π D 2 π (0.04 m) 2 B = = -3 2 A= = = 1.26 x 10 m ; ∆V / V A ∆V 4 4 V
(
B= For oil:
−F ; (1.26 x 10-3 )(−0.008)
F = (1.005 x 10-5 m2)B
F = (1.005 x 10-5 m2)(1.7 x 109 Pa);
183
Foil = 17,090 N
)
Physics, 6th Edition
Chapter 13. Elasticity 13-39. (Cont.) For water: For mercury:
F = (1.005 x 10-5 m2)(2.1 x 109 Pa); F = (1.005 x 10-5 m2)(27 x 109 Pa);
Fw = 21,100 N Fm = 271,400N
*13-40. A 15 kg ball is connected to the end of a steel wire 6 m long and 1.0 mm in diameter. The other end of the wire is connected to a high ceiling, forming a pendulum. If we ignore the small change in length, what is the maximum speed that the ball may have as it passes through its lowest point without exceeding the elastic limit? How much will the length of the wire increase under the limiting stress? What effect will this change have on the maximum velocity? [ D = 1 mm = 0.001 m ] 6m
π D 2 π (0.001 m) 2 A= = = 7.85 x 10-7 m 2 4 4
T v
The maximum speed is that for which the tension causes the stress to exceed the elastic limit for steel (2.48 x 108 Pa). F = 2.48 x 108 Pa; A T − mg =
T = F = (2.48 x 108 Pa)(7.85 x 10-7 m 2 ) ; Tmax = 195 N
mv 2 R (T − mg ) ; mg = (15 kg)(9.8 m/s2) = 147 N ; mv 2 = TR − mgR; v 2 = R m v=
Y=
mg
(6 m)(195 N - 147 N) ; 15 kg
vmax = 4.37 m/s
FL FL (195 N)(6 m) ; ∆L = = ; A ∆L AY (7.85 x 10-7 m 2 )(207 x 109 Pa)
∆L = 7.19 mm
The stretch ∆L of the wire under this load will increase the radius R of the path from 6.000 m to 6.007 m. The larger radius provides a smaller centripetal force so that (T = Fc + mg) is reduced. That would then permit a slightly greater maximum velocity up to the point where the tension again reaches the maximum allowed.
184
Physics, 6th Edition
Chapter 13. Elasticity
*13-41. A cylinder 10 in. in diameter is filled to a height of 6 in. with glycerin. A piston of the same diameter pushes downward on the liquid with a force of 800 lb. The compressibility of glycerin is 1.50 x 10-6 in.2/lb. What is the stress on the glycerin? How far down does the [ R = 5 in.; A = πR2 = 78.5 in.2 ]
piston move?
stress = P =
F 800 lb = ; A 78.5 in.2
P = 10.2 lb/in.2
Vo = Aho = (78.5 in.2)(6 in.) = 471.2 in.3;
k =−
∆V PV0
∆V = -kPVo = -(1.5 x 10-6 in.2/lb)(10.2 lb/in.2)(471.2 in.3); ∆V = 0.00721 in.3 ∆V = A ∆h
∆h =
∆V 0.00721 in.3 = ; A 78.5 in.2
∆ h = 9.18 x 10-5 in.
*13-42. The twisting of a cylindrical shaft (Fig. 13-10) through an angle θ is an example of a shearing strain. An analysis of the situation shows that the angle of twist in radians is
θ=
2τ πSR 4
where τ is the applied torque, l is the length of cylinder, R is the radius of cylinder, and S is the shear modulus. If a torque of 100 lb ft is applied to the end of a cylindrical steel shaft 10 ft long and 2 in. in diameter, what will be the angle of twist in radians? Be careful of the consistent units: τ = 100 lb ft = 1200 lb in.; l = 10 ft = 120 in.
θ=
2τ l 2(1200 lb in.)(120 in.) = ; 4 π SR π (12 x 106 lb/in.2 )(1 in.) 4
185
θ = 0.00764 rad
Physics, 6th Edition
Chapter 13. Elasticity
*13-43. An aluminum shaft 1 cm in diameter and 16 cm tall is subjected to a torsional shearing stress as explained in the previous problem. What applied torque will cause a twist of 10 as defined in Fig. 13-10. [ θ = 10 = 0.01745 rad; R = (D/2) = 0.005 m ]
θ=
2τ l ; π SR 4
τ=
πτ SR 4 π (0.01745 rad)(23.7 x 109 Pa)(0.005 m) 4 = 2l 2(0.16 m)
τ = 2.54 N m
*13-44. Two sheets of aluminum on an aircraft wing are to be held together by aluminum rivets of cross-sectional area 0.25 in.2. The shearing stress on each rivet must not exceed one-tenth of the elastic limit for aluminum. How many rivets are needed if each rivet supports the same fraction of the total shearing force of 25,000 lb? The maximum stress allowed for each rivet is F 1 = (19, 000 lb/in.2 ) = 1900 lb/in.2 10 A This means a shearing force of:
F = (1900 lb/in.2)(0.25 in.2) = 475 lb/rivet
Now we can find the number of rivets N as follows: N=
25, 000 lb = 52.7 rivets; 475 lb/rivet
186
N = 53 rivets
