15.1 Problem The beam consists of material with modulus of elasticity E = 70 GPa and is subjected to couples M = 250 kN-m at its ends. (a) What is the resulting radius of curvature of the neutral axis? (b) Determine the maximum tensile stress due to bending. Solution:
The moment of inertia for the cross-section is: I= bh3 (0.16 m)(0.32 m)3 = = 4.36910-4 m4 12 12
(a) Using Equation (15.10) to determine the magnitude of the radius of curvature: 1 M 250, 000 N-m = = = 8.17410-3 m-1 EI (70109 N/m2 )(4.36910-4 m4 )
ANS:
= 122.34 m M yMAX (250, 000 N-m)(0.16 m) = I 4.36910-4 m4
(b) The maximum normal stress due to the bending moment is: MAX =
ANS:
MAX = 91.6 MPa
Problem 15.2 The material of the beam in Problem 15.1 will safely support a tensile stress of 180 MPa and a compressive stress of 200 MPa. Based on these criteria, what is the largest couple M to which the beam can be subjected? Solution:
The symmetry of the cross-section tells us that: (MAX )TENSILE = (MAX )COMPRESSIVE The moment of inertia for the cross-section is: I= (0.16 m)(0.32 m)3 bh3 = = 4.36910-4 m4 12 12 (180106 N/m2 )(4.36910-4 m4 ) MAX I = yMAX 0.16 m
The maximum normal stress due to the bending moment is: MAX = M yMAX I M =
ANS:
M = 492 kN-m
Problem 15.3 The material of the beam in Problem 15.1 will safely support a tensile stress of 180 MPa and a compressive stress of 200 MPa. Suppose that the beam is rotated 90 about its axis, so that the width of its cross section is 0.32 m and its height is 0.16 m. What is the largest couple M to which the beam can be subjected? Compare your answer to the answer to Problem 15.2. Solution:
The moment of inertia for the cross-section is: I= (0.32 m)(0.16 m)3 bh3 = = 1.09210-4 m4 12 12 (180106 N/m2 )(1.09210-4 m4 ) MAX I = yMAX 0.08 m
Using Equation (15-12) to determine the applied moment: MAX = M yMAX I M =
ANS:
lem 15.2.
M = 246 kN-m This moment is 1/2 the result in Prob-
Problem 15.4 The beam consists of material with modulus of elasticity E = 14x106 psi and is subjected to couples M = 150, 000 in-lb at its ends. (a) What is the resulting radius of curvature of the neutral axis? (b) Determine the maximum tensile stress due to bending? Solution:
The moment of inertia for the cross-section is: I= (2in)4 r 4 = = 12.57in4 4 4
(a) Using Equation (15-10) to determine the magnitude of the radius of curvature: M 150, 000 in-lb 1 = 8.52410-4 in-1 = = EI (14106 lb/in2 )(12.57 in4 )
ANS:
= 1173.2 in = 97.7 ft (150, 000 in-lb)(2in) M yMAX = I 12.57 in4
Using Equation (15-12) to determine the applied moment: MAX =
ANS:
MAX = 23.9ksi
Problem 15.5 The material of the beam in Problem 15.4 will safely support a tensile or compressive stress of 30, 000 psi. Based on this criterion, what is the largest couple M to which the beam can be subjected? Solution:
Using Equation (15-12) to determine the applied moment: MAX = M yMAX I M = (30, 000 lb/in2 )(12.57 in4 ) MAX I = yMAX 2 in
ANS:
M = 188, 550 in-lb
Problem 15.6 The material of the beam in Problem 15.4 will safely support a tensile or compressive stress of 30, 000 psi. If the beam has a hollow circular cross-section, with 2-in. outer radius and 1-in. inner radius, what is the largest couple M to which the beam can be subjected? Solution:
The moment of inertia for the cross-section is: 4 4 I = (ro - ri ) = (2 in)4 - (1 in)4 = 11.78 in4 4 4 Using Equation (15-12) to determine the applied moment: MAX = M yMAX I M = (30, 000 lb/in2 )(11.78 in4 ) MAX I = yMAX 2 in
ANS:
M = 176, 700 in-lb
Problem 15.7 Suppose that the beam in Example 151 is made of a brittle material that will safely support a tensile stress of 20 MPa or a compressive stress of 50 MPa. What is the largest couple M to which the beam can be subjected? Solution:
From the solution to Example 15-1, we know that: I = 1.85x10-6 m4 and y = 0.0475 m from the top of the cross-section. Using the maximum tensile stress in Equation (15-12) to determine the allowable moment: MAX = M yMAX I M = (20106 N/m2 )(1.8510-6 m4 ) MAX I = yMAX 0.08 m - 0.0475 m
M = 1138 N-m Using the maximum compressive stress in Equation (15-12) to determine the allowable moment: MAX = M yMAX I M = (50106 N/m2 )(1.8510-6 m4 ) MAX I = yMAX 0.0475 m
M = 1947 N-m We realize that the bar will fail if either of the calculated moments is exceeded, so the maximum allowable moment must be the smaller moment.
ANS:
M = 1138 N-m
Problem 15.8 What is the maximum tensile stress due to bending in the beam in Example 15-2, and where does it occur? Solution:
Summing moments about point B to determine the reaction at point A: MB = 1 (w0 )(L) 2 L -Ay (L) Ay = w0 L/6 By = w0 L/3 3
Free Body Diagram:
The bending moment is maximum where the shear force equals zero. Summing vertical forces on an arbitrary length of the left-hand portion of the beam: Fy = 0 = w0 L 1 w0 - x (x) 6 2 L
ANS:
M =-
x = 0.577L at the bottom of the cross-section w0 L 6 1 2 w0 L 0.577L 3
Summing moments on the free-body diagram at x = 0.577L: (0.577L)+ (0.577L) (0.577L)
M = w0 L2 (-0.064) Using Equation (15-12) to determine the bending stress: (T )MAX = 0.064w0 L2 (h/2) My = I (h(h)3 /12)
2
ANS:
(T )MAX = 0.384w0 L3 h
Problem 15.9 The beam consists of material that will safely support a tensile or compressive stress of 350 MPa. Based on this criterion, determine the largest force F the beam will safely support if it has the cross section (a); if it has the cross section (b). (The two cross sections have approximately the same area.) Solution:
The moment of inertia of the cross-section in case (a) is: Ia = bh3 (0.0233 m)(0.060 m)3 = = 4.19410-7 m4 12 12
Free Body Diagram:
The moment of inertia for the cross-section in case (b) is: Ib = (0.050 m)(0.060 m)3 (0.020 m) (0.040 m)4 -2 12 12 = 6.8710-7 m4
Summing moments about point B to find the reaction at point A: MB = 0 = F (0.6 m) - Ay (1.6 m) Ay = 0.375F Maximum bending moment occurs at the point where the concentrated load is applied, so we calculate maximum bending stresses 1.0 m to the right of point A. Maximum bending moment is: MMAX = (0.375F )(1.0 m) = 0.375F N-m For cross-section (a): 350106 N/m2 = (0.375F )(0.03 m) 4.19410-7 m4
ANS:
F = 13.05 kN (0.375F )(0.03 m) 6.8710-7 m4
For cross-section (b): 350106 N/m2 =
ANS:
F = 21.4 kN
Problem 15.10 If the beam in Problem 15.9 is subjected to a force F = 6 kN, what is the maximum tensile stress due to bending at the cross section midway between the beam's supports in cases (a) and (b)? Solution:
The moment of inertia of the cross-section in case (a) is: Ia = bh3 (0.0233 m)(0.060 m)3 = = 4.19410-7 m4 12 12
Free Body Diagram:
The moment of inertia for the cross-section in case (b) is: Ib = (0.050 m)(0.060 m)3 (0.020 m) (0.040 m)4 -2 12 12 = 6.8710-7 m4
Summing moments about point B to find the reaction at point A: MB = 0 = (6, 000 N)(0.6 m) - Ay (1.6 m) Ay = 2250 N The bending moment at the midpoint of the beam is: M = (2250 N)(0.8 m) = 1800 N-m In case (a), the maximum bending stress is: (a )MAX = (1800 N-m)(0.03 m) 4.19410-7 m4
ANS:
(a )MAX = 128.8 MPa (1800 N-m)(0.03 m) 6.8710-7 m4
In case (b), the maximum bending stress is: (b )MAX =
ANS:
(b )MAX = 78.6 MPa
Problem 15.11 The beam in Problem 15.9 consists of material that will safely support a tensile or compressive stress of 350 MPa. If it has the cross section (a) and is subjected to a force F = 17 kN, what is the maximum distance from the ends of the beam at which F can be applied? Free Body Diagram:
Solution:
The moment of inertia of the cross-section in case (a) is: Ia = bh3 (0.0233 m)(0.060 m)3 = = 4.19410-7 m4 12 12
Summing moments about point B to determine the reaction at point A: MB = 0 = (17, 000 N)(1.6 m - x) - Ay (1.6 m) Ay = (17, 000 N) 1 - x 1.6 m [1]
The maximum allowable bending moment is determined by the maximum allowable normal stress. 350106 N/m2 = M (0.03 m) M = 4893 N-m 4.19410-7 m4 [2]
Realizing that the maximum bending moment occurs at the point of application of the concentrated load, we sum moments about the point of application and make use of the maximum allowable moment (Eq. [2]). x M = Ay (x) = (17, 000 N) 1 - (x) M = 17, 000x-10, 625x2 1.6 Solving equations [2] and [3] together: -10, 625x2 + 17, 000x - 4893 = 0 Using the quadratic equation to solve for x:
[3]
ANS: x = 0.376 m, 1.224 m Note that the value of x = 1.224 m describes the same distance from point B as x = 0.376 m is from point A.
Problem 15.12 The beam is subjected to a uniformly distributed load w0 = 300 lb/in. Determine the maximum tensile stress due to bending at x = 20 in if the beam has the cross section (a); if it has the cross (b). (The two cross-sections have approximately the same area.) Free Body Diagram:
Solution:
The moments of inertia for the cross-sections in the two cases are: Ia = (4.47 in)(4.47 in)3 = 33.27 in4 12
Ib =
(6 in)(6 in)3 (4 in)(4 in)3 - = 86.67 in4 12 12
Equilbiruim to find Ay + By
Free Body Diagram:
MB = 0 = Ay (85) - 25, 500(42.5)
ANS:
Ay = 12, 500 N
Summing moments about the cut through the beam at x = 20 in: M = (12, 750 lb)(20 in) - (6, 000 lb)(10 in) = 195, 000 in-lb We see that the maximum tensile occurs at the bottom of each crosssection. In case (a), the maximum tensile stress is: (T )MAX = (195, 000 in-lb) (4.47 in/2) 33.27 in4
ANS:
(T )MAX = 13.1 ksi at the bottom of the cross-section (195, 000 in-lb)(3in) 86.67 in4
In case (b), the maximum tensile stress is: (T )MAX =
ANS:
(T )MAX = 6.75 ksi at the bottom of the cross-section
Problem 15.13 The beam in Problem 15.12 consists of material that will safely support a tensile or compressive stress of 30ksi. Based on this criterion, determine the largest distributed load w0 (in lb/in) the beam will safely support if it has the cross section (a); if it has the cross section (b). Free Body Diagram:
Solution:
From the symmetry of the loading, we see that: Ay = By = (85w0 /2) lb The bending moment is maximum at the point on the beam where the shear stress is zero (the middle of the beam). Fy = 0 = Ay - w0 x = 85w0 - w0 x x = 42.5 in 2
The bending moment about a cut through the center of the beam is: M = w0 (85in) 2 85 in -w0 2 85 in 2 85 in 4 = 903.1w0 [1]
To find the maximum allowable bending moment in case (a), the maximum allowable normal stress is used. 30, 000 lb/in2 = M (4.47 in/2) M = 446, 577 in-lb 33.27 in4 [2]
Solving equations [1] and [2] together: ANS w0 = 494.5 lb/in2 To find the maximum allowable bending moment in case 9b0, the maximum allowable normal stress is again used. 30, 000 lb/in2 = M (3 in) M = 866, 700 in-lb 86.67 in4 [3]
Solving equations [1] and [3] together:
ANS:
w0 = 960 lb/in2
Problem 15.14 A bandsaw blade with 2-mm thickness and 20-mm width is wrapped around a pulley with 160-mm radius. The blade is made of steel with modulus elasticity E = 200 GPa. What maximum tensile stress is induced in the blade as a result of being wrapped around the pulley? Solution:
The moment of inertia for the cross-section of the bandsaw blade is: I= (0.02 m)(0.002 m)3 = 1.3310-11 m4 12
The radius of curvature for the blade is dictated by the radius of the pulley. = 0.160 m Using Equation (15-10) to determine the bending moment applied to the blade: 1 M = EI M = EI (200109 N/m2 )(1.3310-11 m4 ) = = 16.63 N-m 0.16 m My (16.63 N-m)(0.001 m) = I 1.3310-11 m4
Using Equation (15-12) to determine the maximum bending stress: MAX =
ANS:
MAX = 1.25 GPa
Problem 15.15 If the beam in Example 15-1 is made of a material for which the allowable stress in tension and compression is ALLOW = 120 MPa, what is the largest allowable magnitude of the couple M ? Solution:
From the solution for Example 15-1 we see that: I = 1.85x10-6 m4 and y = 0.0475 m Using Equation (15-12) and the given allowable normal stress to determine the maximum bending moment: 120106 N/m2 = Solving for M : M (0.0475 m) 1.8510-6 m4
ANS:
M = 4674 N-m
Problem 15.16 Suppose that the beam in Example 151 is made of 7075-T6 aluminum alloy. If it will be subjected to values of M as large as 10 kN-m, what is the beam's factor of safety? (Assume that the yield stress is the same in tension and compression.) Solution:
From the solution for Example 15-1 we see that: I = 1.85x10-6 m4 and y = 0.0475 m From Appendix D for 7075-T6 aluminum: y = 480x106 N/m2 Using Equation (15-12) to determine the maximum normal stress due to bending: MAX = My (10, 000 N-m)(0.0475 m) = 256.8 MPa = I 1.8510-6 m4 480106 N/m2 256.8106 N/m2
The factor of safety for the beam is: S= y MAX =
ANS:
S = 1.87
Problem 15.17 Suppose that the bean in Example 151 is made of a material for which the yield stress in tension is 160 MPa and the yield stress in compression is ALLOW = 200 MPa. If the beam will be subjected to (positive) values of M as large as 4 kN-m, what is the beam's factor of safety? Solution:
From the solution for Example 15-1 we see that:: I = 1.85x10-6 m4 and y = 0.0475 m Equation (15-12) is used to determine the maximum normal stresses for the two given cases. (T )MAX = (4, 000 N-m)(0.08 m - 0.0475 m) = 70.3 MPa 1.8510-6 m4 (4, 000 N-m)(0.0475 m) = 102.7 MPa 1.8510-6 m4 y 160106 N/m2 = T 70.3106 N/m2 y 200106 N/m2 = C 102.7106 N/m2 ST = 2.276 SC = 1.947 We see that the beam will fail due to compressive stress before the yield stress in tension is reached.
(C )MAX =
The factors of safety for the two cases are: ST =
SC =
ANS:
S = 1.947
Problem 15.18 Suppose that the length of the beam in Example 15-2 is L = 8 ft and it is made of ASTM-A36 structural steel. The maximum anticipated magnitude of the distributed load is w0 = 2400 lb/ft. Determine the dimension h so that the beam has a factor of safety S = 3. Solution:
Summing moments about point B to find the reaction at point A: MB = 0 = 1 (2400 lb/ft) (8 ft) 2 Ay = 3200 lb Summing vertical forces to find By : Fy = 0 = -1/2(2400 lb/ft)(8 ft) + 3200 lb + By By = 6400 lb Draw the FBD for the left-hand portion of the beam and find the location where the shear force is zero (M is max): 8 ft 3 - Ay (8 ft)
Free Body Diagram:
Fy = 0 = 3200 lb -
1 2
300
lb ft2
(x) (x)
x = 4.619 ft Summing moments about a cut through the beam at x = 4.619 ft: M = 0 = (3200 lb)(4.619 ft)- 1 2 300 lb ft2 (4.619 ft) (4.619 ft) 4.619 ft +M 3
M = 9853 ft-lb Using the factor of safety (given) to find the maximum allowable normal stress: MAX = y 36, 000 lb/in2 = = 12, 000 lb/in2 S 3 h4 h(h)3 = 12 12
The moment of inertia for the square cross-section would be: I=
Using Equation (15-12) to determine the dimensions of the square (and remembering to convert ft-lb to in-lb): 12, 000 lb/in2 = (9853 ft - lb)
h 2 12 in 1 ft
h4 /12
h3 = 59.12 in3
ANS:
h = 3.9 in
Problem 15.19 Suppose that the loads on the beam in Example 15-3 are the maximum anticipated loads and the beam is made of wood with yield stress y = 40 MPa. What is the beam's factor of safety? Solution:
From Example 15-3 we have: I = 0.110 m4 and MMAX = -1800 kN - m Using Equation (15-12) to determine the normal stress: = My (1800 kN-m) (0.6 m) = 9818 kN/m2 = I 0.110 m4 y 40, 000 kN/m2 = 9818 kN/m2
The factor of safety is: S=
ANS:
S = 4.07
Problem 15.20 A beam made of 7075-T6 aluminum alloy will be subjected to anticipated bending moments as large as 1500 N-m. Determine the beam's factor of safety for two cases: (a) It has a solid circular crosssection with 20-mm radius; (b) It has a hollow circular cross section with 30-mm outer radius and the inner radius chosen so that the beam has the same weight as the beam in case (a). Solution:
From Appendix D for 7075-T6 aluminum we have: y = 480 MPa The area for case (a) is: Aa = (0.02 m)2 = 0.00126 m2 Determining the inner radius for case (b):
2 Ab = Aa = 0.00126 m2 = [(0.03 m2 ) - ri ] ri = 0.0223 m
The moments of inertia for the two cross-sections are: Ia = Ib = r 4 (0.02 m)4 = = 1.25710-7 m4 4 4
(0.03 m)4 - (0.0223 m)4 = 4.41910-7 m4 4
Using Equation (15-12) to determine the normal stresses for the two cases: (MAX )a = (1500 N-m)(0.02 m) = 238.7 MPa 1.25710-7 m4 (1500 N-m)(0.03 m) = 101.8 MPa 4.41910-7 m4 y 480 MPa = 238.7 MPa y 480 MPa = 101.8 MPa
(MAX )b =
The factors of safety in the two cases are: Sa = Sb =
ANS:
Sa = 2.01 Sb = 4.715
Problem 15.21 Design a cross-section for the beam in Example 15-4 so that the beam's factor of safety is S = 2. Solution:
From Example 15-4, we have: MMAX = 86, 600 N-m and y = 700 MPa Maximum allowable normal stress is: MAX = y 700 MPa = = 350 MPa S 2
Using Equation (15-12) to determine the required dimension: 350106 N/m2 = My (86, 600 N-m) (h/2) = h2 = 0.0103 m2 I (0.144 m)(h)3 /12
ANS:
h = 0.1015 m for a rectangular cross-section.
Problem 15.22 The device shown is a playground seesaw. Make a conservative estimate of the maximum weight to which it will be subjected at each end when in use. (Consider contingent situations such as an adult sitting with a child.) Choose a material from Appendix D and design a cross section for the 4.8-m beam so that it has a factor of safety S = 4. Solution:
A maximum mass of 140 kg (a weight of 1375 N or 308 lb) is assumed. Southern Pine is selected ( = 30 MPa). For a factor of safety of S = 4, maximum allowable normal stress is: MAX = y 30106 N/m2 = = 7.5106 N/m2 S 4
Maximum moment to which the seesaw is exposed is: MMAX = (1375 N)(2.4 m) = 3300 N-m The moment of inertia for the cross-section is: I= (4x) (x)3 x4 = 12 3
x 2
Using Equation (15-12) to determine the unknown dimension, x: 7.5106 N/m2 = (3, 300 N-m) x4 /3 x3 = 0.00066 m
ANS:
x = 0.08706 m = 87.06 mm
Problem 15.23 A beam with the cross section is sublected to a shear force V = 8 kN. What is the shear stress at the neutral axis (y = 0)? Solution: Using Equation (15-18) to determine the average shear stress at the neutral axis:
AVG = 3(8, 000 N) 3V = 2A 2(0.04 m)(0.06 m)
ANS:
AVG = 5 MPa
Problem 15.24 In Problem 15.23, determine the average shear (a) at y = 0.01 m; (b) at y = -0.02 m.
(a) Using Equation (15-17) to determine the average stress at y = 0.01 m: AVG = 6V bh3 h 2
2
Solution:
- y
2
=
6(8, 000 N) (0.04 m)(0.06 m)3
0.06 m 2
2
- (0.01 m)2
ANS:
AVG = 4.44 MPa
(b) Using Equation (15-17) to determine the average stress at y = -0.02 m: AVG = 6V bh3 h 2
2
- y
2
=
6(8, 000 N) (0.04 m)(0.06 m)3
0.06 m 2
2
- (-0.02 m)2
ANS:
AVG = 2.78 MPa
Problem 15.25 In Example 15-5, consider the cross section at x = 3 m. What is the average shear stress at y = 0.05 m. Solution:
Summing moments about point B to determine Ay : MB = 0 = [(6, 000 N/m)(8 m)] (4 m)-Ay (8 m) Ay = 24, 000 N Cut the FBD where x = 3 m and draw the FBD.
Free Body Diagram:
Summing vertical forces to determine the shear force V : Fy = 0 = 24, 000 N-(6, 000 N/m)(3 m)-V V = 6, 000 N Using Equation (15-17) to determine the average stress at y -0.05 m: AVG = 6V bh3 h 2
2
=
2
- y
2
=
6(6, 000 N) (0.25 m)(0.25 m)3
0.25 m 2
- (0.05 m)2
ANS:
AVG = 121 kPa
Problem 15.26 What is the maximum magnitude of the average shear stress in the beam in Example 15-5, and where does it occur? Solution:
Using Equation (15-18) to determine the average shear stree at the neutral axis: = 3V 3(24, 000 N) = 2A 2(0.25 m)(0.25 m)
ANS:
AVG = 576 kPa
Problem 15.27 The beam is subjected to a distrubuted load. For the cross section at x = 40 in, determine the average shear stress (a) at the neutral axis; (b) at y = 1.5 in. Solution:
Summing the moments about point B to determine the reaction at point Body A:
Free Diagram:
MB = 0 = [(1500 lb/in)(60 in)] (30 in)-Ay (120 in) Ay = 22, 500 lb Draw the FBD at x = 40 in.
Summing the vertical forces to determine the shear force: Fy = 0 = 22, 500 lb - V V = 22, 500 lb (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: = 3V 3(22, 500 lb = 2A 2(1 in)(4 in)
ANS:
AVG = 8, 437.5 lb/in2
(b) Using Equation (15-17) to determine the average stress at y = 1.5 in: AVG = 6V bh3 h 2
2
- y
2
=
6(22.500 lb) (1 in)(4 in)3
4 in 2
2
- (1.5 in)2
ANS:
AVG = 3, 691 lb/in2
Problem 15.28 Solve Problem 15.27 for the cross section at x = 80 in. Free Body Diagram:
Solution:
Summing the moments about point B to determine the reaction at point A: MB = 0 = [(1500 lb/in)(60 in)] (30 in)-Ay (120 in) Ay = 22, 500 lb Draw the FBD at x = 80 in.
Summing the vertical forces to determine the shear force: Fy = 0 = 22, 500 lb(-1, 500 lb/in)(20 in)+V V = 7, 500 lb (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: = 3V 3(7, 500 lb) = 2A 2(1 in)(4 in)
ANS:
AVG = 2, 812.5 lb/in2
(b) Using Equation (15-17) to determine the average stress at y = 1.5 in: AVG = 6V bh3 h 2
2
- y
2
=
6(7, 500 lb) (1 in)(4 in)3
4 in 2
2
- (1.5 in)2
ANS:
AVG = 1, 230.46 lb/in2
Problem 15.29 What is the maximum magnitude of the average shear stress in the beam in Problem 15.27, and where does it occur? Free Body Diagram:
Solution:
Summing the moments about point B to determine the reaction at point A: MB = 0 = [(1500 lb/in)(60 in)] (30 in)-Ay (120 in) Ay = 22, 500 lb We see that the maximum shear stress exists at x = 120 in. ANS: V = 67, 500 lb (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: = 3V 3(67, 500 lb) = 2A 2(1 in)(4 in) By = 67, 500 lb
ANS:
AVG = 25, 312, 5 lb/in2 = 25.3 kip/in2
Problem 15.30 By integrating the stress distribution given by Equation (15-17), confirm that the total force exerted on the rectangular cross section by the shear stress is equal to V . Solution:
Starting with Equation (15-17) and integrating over the dimensions -h/2 to h/2: AVG = 6v bh3
h/2 -h/2
h2 = (y )2 2
dA
=
6v bh3 6vb bh3
h/2 -h/2 h/2 -h/2
h2 = (y )2 2 h2 = (y )2 2
bdy
=
dy
Doing the integration: AVG = 6v h3 6v h3 h2 y (y )3 - 4 3 -
h/2 -h/2
AVG =
h3 h3 - 8 24 6v h3
-h3 -h3 - 4 24 = 6v h3 h3 6
AVG =
h3 -h3 - 4 12
ANS:
AVG = V
Problem 15.31 Prove that the quantity Q defined by Equation (15-17) is a maximum at the neutral axis (y = 0). Solution:
The equation Q= b 2 h 2
2
- (y)2
Taking the derivitive of Q with respect ot y: dQ b = (-2y) = -by dy 2 Setting dQ/dy equal to zero (and seeing that b cannot be zero), we see that:
ANS:
y = 0 (or the neutral axis of the cross section)
Problem 15.32 At a particular axial position, the beam whose cross section is shown is subjected to a shear force V = 20 kN. Determine the average shear stress acting on the slanted infinitesimal element.
Solution:
Determining the value of Q: Q=yA = 2 3 0.05 m 2 = 1 2 0.03 m 2 0.05 m 2 = 3.12510-6 m3
Now calculating the average shear stress along the slanted line AVG = VQ = bI (20, 000 N)(3.125 10-6 m3 )
0.03 m3 2 2
+
0.05 m3 2
2
(0.03 m)(0.05 m)3 12
ANS:
AVG = 6.86 MPa
Problem 15.33 In Example 15-6, determine the average shear stress at y = 1 in. Solution:
The moment of inertia for the cross section is: I= 1 1 (2)(8)3 + 4 (4)(2)3 + (3)2 (2)(4) = 384 in4 12 12 I = 384 in4 (2 in)(10 in)(3 in) + (2 in)(1 in)(1.5 in) = 2.86 in (2 in)(10 in) + (2 in)(1 in)
ANS:
y =
Calculating y :
Calculating A : A = (10 in)(2 in) + (2 in)(1 in) = 22 in2 Calculating Q: Q = (2.86 in)(22 in2 ) = 63 in3 Now calculating the average shear stress at y = 1 in: AVG = VQ (6, 000 lb)(63 in3 ) = bI (2 in)(384 in4 )
ANS:
AVG = 492 lb/in2
Problem 15.34 In Example 15-6, determine the average shear stress in the upper-right glued joint. Solution:
Distance from the neutral axis to the centroid of the upper right-hand section: y = -3 in The area A is: A = (4 in)(2 in) = 8 in2 Calculating the value of Q: Q = (-3 in)(8 in2 ) = 24 in3 Average shear stress at the right-hand glued joint is AVG = VQ (6, 000 lb)(-24 in3 ) = bI (2 in)(384 in4 )
ANS:
AVG = 187.5 lb/in2
Problem 15.35 The beam whose cross section is shown consists of three planks of wood glued together. At a given axial position it is subjected to a shear force V = 2400 lb. What is the average shear stress at the neutral axis y = 0? Solution:
Finding the centroid of the entire cross section (measuring from the TOP): y= (2 in)(8 in)(4 in) + (2) [(2 in)(4 in)(7 in)] = 5.5 in (2 in)(8 in) + (2) [(2 in)(4 in)]
Calculating y : y = Calculating A : A = (2 in)(5.5Inches) = 11 in2 Calculating Q: Q = (-2.75 in)(11 in2 ) = -30.25 in3 Calculating the moment of inertia: I= (4 in)(2 in)3 (2 in)(8 in)3 + (2 in)(8 in)(4 in - 5.5 in)2 +2 + (4 in)(2 in)(7 in - 5.5 in)2 = 162.7 in4 12 12 VQ (2, 400 lb)(-30.25 in3 ) = bI (2 in)(162.7 in4 ) 5.5 in = -2.75 in 2
Now calculating the average shear stress: AVG =
ANS:
AVG = -223.1 lb/in2
Problem 15.36 In Problem 15.35, what are the magnitudes of the average shear stress acting on each glued joint? Solution:
Finding the centroid of the entire cross section (measuring from the TOP): y= (2 in)(8 in)(4 in) + (2) [(2 in)(4 in)(7 in)] = 5.5 in (2 in)(8 in) + (2) [(2 in)(4 in)]
Calculating the moment of inertia: I= (4 in)(2 in)3 (2 in)(8 in)3 + (2 in)(8 in)(4 in - 5.5 in)2 +2 + (4 in)(2 in)(7 in - 5.5 in)2 = 162.7 in4 12 12
Calculating y : y = 7 in - 5.5 in = 1.5 in Calculating A : A = (2 in)(4 in) = 8 in2 Calculating Q: Q = y A = (1.5 in)(8 in2 ) = 12 in3 Now calculating the average shear stress: AVG = VQ (2, 400 lb)(12 in3 ) = bI (2 in)(162.7 in4 )
ANS:
AVG = 88.5 lb/in2
Problem 15.37 For the cross section at x = 8 ft, determine the average shear stress (a) at the neutral axis (b) at y = 2 in.
Solution:
Summing moments about point A to determine the reaction at point B: MA = 0 = - By = 50, 000 lb
1 (50, 000 2
Free Body Diagram:
lb/ft)(6 ft) (4 ft) + By (12 ft)
We see that, at x = 8 ft: V = -50, 000 lb (a) Calculating y : y = Calculating A : A = Calculating Q: Q= 4 in (8 in2 ) = 10.67 in3 3 1 (4 in)(4Inches) = 8 in2 2 4 in 3
The moment of inertia for the cross section is: I= (6 in)(6 in)3 bh3 = = 36 in4 36 36 VQ (-50, 000 lb)(10.67 in3 ) = bI (4 in)(36 in4 )
The average shear stress at the neutral axis is: AVG =
ANS:
AVG = 3, 700 lb/in2 2 in 3
(b) Calculating y : y =2 Calculating A : A = Calculating Q: Q= 2 2 in (2 in2 ) = 5.33 in3 3 1 (2 in)(2 in) = 2 in2 2
The average shear stress at y = 2 in is: AVG = (-50, 000 lb)(5.33 in3 ) VQ = bI (2 in)(36 in4 )
ANS:
AVG = -3, 700 lb/in2
Problem 15.38 In Problem 15.37, determine the value of y at which the magnitude of the average shear stress is a maximum. (Notice that the maximum magnitude does not occur at the neutral axis.) What is the maximum magnitude Free Body Diagram:
Solution:
Summing moments about point A to determine the reaction at point B: MA = 0 = - 1 (50, 000 lb/ft)(6 ft) (4 ft)+By (12 ft) By = 50, 000 lb 2
We see that, at x = 8 ft: V = -50, 000 lb The expression for the average shear stress at an arbitrary distance from the neutral axis would be: AVG = VQ = bI V I Q = b V I yA = b V I y + 1 (4 in - y) 3 4 in - y
1 (4 2
in - y)(4 in - y)
=
V I
2 4 y + in 3 3
2 in -
y 2
We see from above that: yA = 2 4 y + in 3 3 2 in - y 4 y2 8 2 y2 2 8 = y- + - y = - + y+ 2 3 3 3 3 3 3 3 [1]
We know that the shear stress is maximum where d(y A )/dy = 0: d 2 2 (y A ) = - y + = 0 dy 3 3
ANS:
y = 1 in
Making use of Equation [1] to determine the magnitude of the average shear stress at y = 1 in: AVG = (-50, 000 lb) - 1 + 2 + VQ 3 3 = bI (1 in)(36 in4 )
8 3
ANS:
AVG = -4, 167 lb/in2
Problem 15.39 Solve Problem 15.37 for the cross section at x = 4 ft.
Free Body Diagram:
Solution:
Summing moments about point A to determine the reaction at point B: MB = 0 = 1 (50, 000 lb/ft)(6 ft) -Ay (12 ft) Ay = 100, 000 lb 2
Cut the beam at x = 4 ft and draw the FBD
Summing vertical forces to determine the shear force V : 1 Fy = 0 = 100, 000 lb- (33, 333 lb)(4 ft)-V V = 33, 334 lb 2 The moment of inertia for the cross section is: I= (6 in)(6 in)3 bh3 = = 36 in4 36 36
(a) We see that, at y = 2 in from the top of the section. We also see that y is 4 in: 3 Calculating A : A = Calculating Q: Q= 4 in (8 in2 ) = 10.67 in3 3 VQ (-33, 334 lb)(10.67 in3 ) = bI (4 in)(36 in4 ) 1 (4 in)(4 in) = 8 in2 2
The average shear stress at the neutral axis is: AVG =
ANS:
AVG = 2, 470 lb/in2 2 in = 2.67 in 3
(b) Calculating y : y = 2 in + Calculating A : A = Calculating Q: Q = (2.67 in)(2 in2 ) = 5.34 in3 The average shear stress at y = 2 in is: AVG = VQ (33, 334 lb)(5.34 in3 ) = bI (2 in)(36 in4 ) 1 (2 in)(2 in) = 2 in2 2
ANS:
AVG = 2, 470 lb/in2
Problem 15.40 At a particular position, the beam whose cross section is shown is subjected to a shear force V = 15 kN. Determine the average shear stress (a) at the neutral axis y; = 0; (b) at y = 0.025 m.
Solution:
Calculating the location of the centroid of the cross section (from the TOP of the section): y= (0.04 m)(0.02 m)(0.01 m) + (0.02 m)(0.05 m)(0.045 m) + (0.08 m)(0.02 m)(0.08 m) = 0.0532 m (0.04 m)(0.02 m) + (0.02 m)(0.05 m) + (0.08 m)(0.02 m)
(0.04 m)(0.02 m)3 12 (0.02 m)(0.05 m)3 12 (0.08 m)(0.02 m)3 12 3 10-6 m4
The moment of inertia for the cross section is: I= + + I= + (0.04 m)(0.02 m)(0.0532 m - 0.01 m)2 + (0.02 m)(0.05 m)(0.0532 m - 0.045 m)2 + (0.08 m)(0.02 m)(0.08 m - 0.0532 m)2
(a) Calculating y : y = (0.02 m)(0.04 m)(0.0332 m + 0.01 m) + (0.02 m)(0.0332 m) (0.02 m)(0.04 m) + (0.02 m)(0.0332 m)
0.0332 m 2
= 0.0311 m
Calculating A : A = (0.02 m)(0.04 m) + (0.02 m)(0.0332 m) = 0.001464 m2 Calculating Q: Q = y A = (0.0311 m)(0.001464 m2 ) = 4.56 10-5 m3 The average shear stress in the neutra axis is: AVG = VQ (15, 000 N)(4.56 10-5 m3 ) = bI (0.02 m)(3 10-6 m4 )
ANS:
AVG = 11.4 MPa (0.09 m - (0.0532 m + 0.025 m)) 2
(b) Calculating y : y = 0.025 m + Calculating A : A = (0.08 m)(0.0118 m) = 9.44 10-4 m2 Calculating Q: Q = y A = (0.0309 m)(9.44 10-4 m2 ) = 2.917 10-5 m3 The average shear stress at y = 0.025 m is: AVG = VQ (15, 000 N)(2.917 10-5 m3 ) = bI (0.08 m)(3 10-6 m4 ) = 0.0309 m
ANS:
AVG = 1.82 MPa
Problem 15.41 For the beam in Problem 15.40, determine the avarage shear stress on the infinitesimal element shown. Solution:
Calculating the location of the centroid of the cross section (from the TOP of the section): y= (0.04 m)(0.02 m)(0.01 m) + (0.02 m)(0.05 m)(0.045 m) + (0.08 m)(0.02 m)(0.08 m) = 0.0532 m (0.04 m)(0.02 m) + (0.02 m)(0.05 m) + (0.08 m)(0.02 m)
(0.04 m)(0.02 m)3 12 (0.02 m)(0.05 m)3 12 (0.08 m)(0.02 m)3 12 3 10-6 m4
The moment of inertia for the cross section is: I= + + I= + (0.04 m)(0.02 m)(0.0532 m - 0.01 m)2 + (0.02 m)(0.05 m)(0.0532 m - 0.045 m)2 + (0.08 m)(0.02 m)(0.08 m - 0.0532 m)2
Calculating y : y = 0.08 m - 0.0532 m = 0.0268 m Calculating A : A = (0.02 m)(0.02 m) = 0.0004 m2 Calculating Q: Q = y A = (0.0268 m)(0.0004 m2 ) = 1.072 10-5 m3 The average shear stress at the element is: AVG = VQ (15, 000 N)(1.072 10-5 m3 ) = bI (0.02 m)(3 10-6 m4 )
ANS:
AVG = 2.68 MPa
Problem 15.42 Assume that the surface on which the beam rests exerts a uniformly distributed load on the beam. Determine the maximum tensile and compressive stresses due to bending at x = 3 m. Free Body Diagram:
Solution: Locating the centroid of the cross-section from the bottom of the section:
y=
(0.05 m)(0.5 m)(0.1 m) + (0.3 m)(0.4 m)(0.1 m) = 0.161 m from the bottom of the section. (0.5 m)(0.1 m) + (0.4 m)(0.1 m) (0.5 m)(0.1 m)3 + (0.5 m)(0.1M )(0.05 m - 0.161 m)2 + 12 I = 0.00196 m4 (0.1 m)(0.4 m)3 12
The moment of inertia of the cross-section about the neutral axis is: I= + (0.1 m)(0.4M (0.3 m - 0.161 m)2
We see that the top of the beam is in compression and the bottom is in tension. The magnitude of the bending moment at x = 3 m is: M = (1, 000 m)(3 m)(1.5 m) - (4, 000 N)(1 m) = 500 N-m The maximum tensile and compressive stresses due to bending are: (c )MAX = My (500 N-m)((0.5 m - 0.161 m) = I 0.00196 m4 My (500 N-m)(0.161 m) = I 0.00196 m4
(T )MAX =
ANS:
(c )MAX = 86.5 kPa(C) (T )MAX = 41.1 kPa(T )
Problem 15.43 If you are selecting a material for the beam in Problem 15.42, what maximum tensile and compressive stresses must the material be able to support? Free Body Diagram:
Solution:
Locating the centroid of the cross-section from the bottom of the section: y= (0.05 m)(0.5 m)(0.1 m) + (0.3 m)(0.4 m)(0.1 m) = 0.161 m (0.5 m)(0.1 m) + (0.4 m)(0.1 m) (0.5 m)(0.1 m)3 + (0.5 m)(0.1M )(0.05 m - 0.161 m)2 + 12 I = 0.00196 m4 We see from the bending moment diagram that the maximum bending moment occurs at x = 2 m. The magnitude of the bending moment at x = 2 m is: M = [(1, 000 N/m)(2 m)](1 m) = 2, 000 N-m from the above diagram (B) this can be seen. (0.1 m)(0.4 m)3 12
The moment of inertia of the cross-section about the neutral axis is: I= + (0.1 m)(0.4M (0.3 m - 0.161 m)2
The maximum tensile and compressive stresses at x = 2 m is: (c )MAX = My (2, 000 N-m)(0.5 m - 0.161 m) = I 0.00196 m4 (2, 000 N-m)(0.161 m) My = I 0.00196 m4
(T )MAX =
ANS:
(c )MAX = 345.9 kPa(C) (T )MAX = 164.28 kPa(T )
Problem 15.44 The maximum anticipated load on the beam is shown. Choose a material from Appendix D and design a cross section for the beam so that it has a factor of safety S = 2. Free Body Diagram:
Solution:
From Appendix D for 7075-T6 aluminum: y = 480 MPa A cross-section is chosen such that its height is twice its width. The moment of inertia for the cross-section is:
I=
bh3 x (2x)3 = = 0.667x4 m4 12 12
Summing moments about point B to determine the reaction at point A: MB = 0 = 1 (20, 000 N/m) (3 m) (4 m) - Ay (6 m) 2 Ay = 20, 000 N The bending moment is maximum at the location where the shear force is zero, or the sum of vertical forces is zero. Summing vertical forces on the FBD: Fy = 0 = 20, 000 N- 1 6, 667 N/m2 (x m) (x m) x = 2.45 m 2 1 (6, 667 N/m) (2.45 m) (2.45 m) 2 2.45 m +M 3
Summing moments on the FBD: M = 0 = (20, 000 N)(2.45 m)-
M = 32, 659 N-m The maximum allowable normal stress due to bending is found using the factor of safety. MAX = y 480106 N/m2 = = 240106 N/m2 S 2 (32, 659 N-m) (x) x3 = 0.000204 m3 0.667x4 m4
Now using Equation (15-12) to determine the dimension, x: 240106 N/m2 =
ANS:
x = 0.0589 m = 58.9 mm
Problem 15.45 For a preliminary design of the ladder rung, assume that it has pin supports at the ends. Make a conservative estimate of the maximum weight to which it will be subjected when in use. (Assume that the maximum weight acts as a point force at the center of the rung.) The rung is to be made of 6061-T6 aluminum alloy. Design a cross section so that it has a factor of safety S = 4. Consider the appropriate width the rung should have. Solution:
From Appendix D for 6061-T6 aluminum: y = 270106 N/m2 A maximum weight of 1350 N (303 lb) is chosen as a design parameter. Draw the FBD for the ladder rung.
The moment of inertia for the cross-section of the ladder rung is: I = (r)4 4 Maximum bending moment occurs when the load is at the center of the ladder rung. MMAX = (675 N) 0.380 m 2 = 128.25 N-m
Maximum allowable normal stress due to bending is found using the factor of safety, S = 4. MAX = y 270106 N/m2 = = 67.5106 N/m2 S 4
Using Equation (15-12) to determine the required radius of the ladder rung: 67.5106 N/m2 = (128.25 N-m) (r)
4
(r)4
r 3 = 2.41910-5 m3
ANS:
r = 0.0134 m = 13.4 mm
Problem 15.46 The beam is subjected to a distrubuted load. For the cross section at x = 0.6 m, determine the average shear stress (a) at the neutral axis; (b) at y = 0.02 m. Free Body Diagram:
Solution:
Summing the moments about point B to determine the reaction at point A: MB = 0 = - 1 (130, 000 N/m)(1.4 m) (0.133 m)+Ay (0.8 m) Ay = 15, 130 N 2
Cut the beam at x = 0.6 m and draw the FBD.
Summing the vertical forces to determine the shear force at x = 0.6 m: 1 Fy = 0 = -15, 130 N - (55, 714 N/m)(0.6 m) +V V = 31, 844 N 2 (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: = 3V 3(-31, 844 N = 2A 2(0.04 m)(0.06 m)
ANS:
AVG = -19.9 MPa
(b) Using Equation (15-17) to determine the average stress at y = 0.02 m: AVG = 6V bh3 h 2
2
- y
2
=
6(-31, 844 N) (0.04 m)(0.06 m)3
0.06 m 2
2
- (0.02 m)2
ANS:
AVG = -11.06 MPa
Problem 15.47 Solve Problem 15.46 for the cross section at x = 1.0 m. Free Body Diagram:
Solution:
Sum moments about point B to determine the reaction at point A: MB = 0 = - 1 (130, 000 N/m)(1.4 m) (0.133 m)+Ay (0.8 m) Ay = 15, 130 N 2
Summing vertical forces to determine By : Fy = 0 = -15, 130 N- 1 (130, 000 N/m)(1.4 m) +By By = 106, 130 N 2
Cut the beam at x = 1.0 m and draw the FBD.
Summing the vertical forces to determine the shear force at x = 1.0 m: 1 Fy = 0 = -15, 130 N+106, 130 N - (92, 860 N/m)(1.0 m) -V V = 44, 570 N 2 (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: = 3V 3(-44, 570 N) = 2A 2(0.04 m)(0.06 m)
ANS:
AVG = -27.86 MPa
(b) Using Equation (15-17) to determine the average stress at y = 0.02 m: AVG = 6V bh3 h 2
2
- y
2
=
6(-44, 570 N) (0.04 m)(0.06 m)3
0.06 m 2
2
- (0.02 m)2
ANS:
AVG = -15.48 MPa
Problem 15.48 At a particular axial position, the beam whose cross section is shown is subjected to a shear force V = 40 kN. What is the average shear stress at the neutral axis (y = 0). Solution:
The moment of inertia for the cross section is:
4 4 Ro Ri - = (0.08 m)4 - (0.05 m)4 = 2.72610-5 m4 I= 4 4 4
Calculating y : y =
(0.08 m)2 2 4(0.08 m) 3
- -
(0.05 m)2 2 (0.05 m)2 2
4(0.05 m) 3
Calculating Q:
(0.08 m)2 2
Q = y A = (0.0421 m)(0.0061 m) = 0.000257 m3 The average shear stress at the neutral axis is: AVG = VQ (40, 000 N)(2.57 10-4 m3 ) = bI 2(0.08 m - 0.05 m)(2.726 10-5 m4 )
ANS:
AVG = 6.29 MPa
Problem 15.49 For the beam in Problem 15.48, what is the average shear stress at y = 50 m? Solution:
The moment of inertia for the cross section is:
4 4 Ro Ri - = (0.08 m)4 - (0.05 m)4 = 2.72610-5 m4 I= 4 4 4
From the diagram, the angle is: = cos-1 0.05 m 0.08 m = 51.3
The thickness of the section as y = 50 mm is: t = 2 (0.08 m) sin 51.3 ) = 0.125 m Calculating Q using the derived equation: Q= 2 3 (0.008)2 - (0.05)2
3/2
- (0.05)2 - (0.05)2
3/2
Q = 0.0001624 m3 = 1.624 10-4 m3 Calculating the shear stress at y = 50 mm: = (40, 000 N)(1.623 10-4 m3 ) VQ = It (2.73 10-5 m4 )(0.125 m)
ANS:
= 1.902 MPa
Problem 15.50 For the beam in Problem 15.48, what is the average shear stress at y = 25 mm?
Solution:
The moment of inertia for the cross section is: I=
4 4 Ro Ri - = (0.08 m)4 - (0.05 m)4 = 2.72610-5 m4 4 4 4
From the diagram, the angle is: = cos-1 0.025 m 0.08 m 0.025 m 0.05 m = 71.8
From the diagram, the angle is: = cos-1 = 60
The thickness of the section as y = 25 mm is: t = 2(0.08 m)(sin 71.8 ) - 2(0.05 m)(sin 60 ) = 0.0654 m To determine the value of Q, a two step approach is taken: Q1 = 2 2(0.08 m(sin 71.8 )(0.025 m) (0.08 m)3 (sin 71.8 )- 3 2 2 2(0.05 m(sin 60 )(0.025 m) (0.05 m)3 (sin 60 )- 3 2 2 3 2 3 (0.025 m) = 2.92610-4 m3
Q2 =
(0.025 m) = 5.41310-5 m3
Q = Q1 -Q2 = 2.92610-4 m3 -5.41310-5 m3 = 2.38510-4 m3 Calculating the shear stress at y = 25 mm: = VQ (40, 000 N)(2.385 10-4 m3 ) = It (2.726 10-5 m4 )(0.0654 m)
ANS:
= 5.35 MPa