The beam consists of material with Problem 15.4 modulus of elasticity E = 14x 14x106 psi and is subjected 150,, 000 in− in−lb at its ends. (a) What is to couples M = 150 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 =
πr 4 π (2in (2in))4 = = 12. 12 .57in 57in4 4 4
(a) Using Equation (15-10) to determine the magnitude of the radius of curvature:
1 M 150, 150, 000 in−lb = = = 8. 8 .524×10 ρ EI (14×106 lb/ lb/in2 )(12. )(12.57 in4 )
−4
ANS:
−1
in
ρ = 1173. 1173.2 in = 97. 97.7 ft
Using Equation (15-12) to determine the applied moment:
σMAX =
ANS:
M yMAX (150, (150, 000 in−lb)(2in lb)(2in)) = I 12. 12.57 in4
σMAX = 23. 23 .9ksi
The materia materiall of the beam in ProbProbProblem Problem 15.5 lem 15.4 will safely support a tensile or compressive stress of 30, 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:
σMAX =
M yMAX → I
ANS:
M = 188, 188, 550 in−lb
M =
σMAX I (30, (30, 000 lb/ lb/in2 )(12. )(12.57 in4 ) = yMAX 2 in
The materia materiall of the beam in ProbProbProblem Problem 15.6 lem 15.4 will safely support a tensile or compressive stress of 30, hollow circular circular 30, 000 psi. If the beam has a hollow cross-sec cross-section, tion, with with 2-in. outer radius radius and and 1-in. 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:
I =
π 4 π (r − ri4 ) = (2 in)4 − (1 in)4 = 11. 11 .78 in4 4 o 4
Using Equation (15-12) to determine the applied moment:
σMAX =
M yMAX → I
ANS:
M = 176, 176, 700 in−lb
M =
σMAX I (30, (30, 000 lb/ lb/in2 )(11. )(11.78 in4 ) = yMAX 2 in
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 determinethe allowable moment:
σMAX =
M yMAX → I
M =
σMAX I (20×106 N/m2 )(1.85×10 = yMAX 0.08 m − 0.0475 m
−6
m4 )
−6
m4 )
M = 1138 N−m Using the maximum compressive stress in Equation (15-12) to determine the allowable moment:
σMAX =
M yMAX → I
M =
σMAX I (50×106 N/m2 )(1.85×10 = 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:
Free Body Diagram:
Summing moments about point B to determine the reaction at point A:
ΣM B =
1 (w0 )(L) 2
L −Ay (L) → Ay = w0 L/6 → By = w0 L/3 3
The bending moment is maximum where the shear force equals zero. Summing verticalforces on an arbitrarylength of theleft-hand portion of the beam:
ΣF y = 0 =
ANS:
w0 L 1 w0 x (x) − 6 2 L
x = 0.577L at the bottom of the cross-section
Summing moments on the free-body diagram at x = 0.577L:
M = −
w0 L 6
1 (0.577L)+ 2
w0 L
(0.577L) (0.577L)
M = w0 L2 (−0.064) Using Equation (15-12) to determine the bending stress:
(σT )MAX =
ANS:
My 0.064w0 L2 (h/2) = I (h(h)3 /12) 2
(σT )MAX = 0.384w0 L h3
0.577L 3
The beam consists of material that Problem 15.9 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:
Free Body Diagram:
The moment of inertia of the cross-section in case (a) is:
I a =
bh3 (0.0233 m)(0.060 m)3 = = 4.194×10 12 12
−7
m4
The moment of inertia for the cross-section in case (b) is:
(0.050 m)(0.060 m)3 (0.020 m) (0.040 m)4 I b = −2 12 12
= 6.87×10
−7
m4
Summing moments about point B to find the reaction at point A:
ΣM B = 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:
M MAX = (0.375F )(1.0 m) = 0.375F N−m For cross-section (a):
350×106 N/m2 =
ANS:
(0.375F )(0.03 m) 4.194×10 7 m4 −
F = 13 .05 kN
For cross-section (b):
350×106 N/m2 =
ANS:
(0.375F )(0.03 m) 6.87×10 7 m4 −
F = 21 .4 kN
Problem 15.10 If the beam in Problem 15.9 is sub jected 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:
Free Body Diagram:
The moment of inertia of the cross-section in case (a) is:
I a =
bh3 (0.0233 m)(0.060 m)3 = = 4.194×10 12 12
−7
m4
The moment of inertia for the cross-section in case (b) is:
(0.050 m)(0.060 m)3 (0.020 m) (0.040 m)4 I b = −2 12 12
= 6.87×10
Summing moments about point B to find the reaction at point A:
ΣM B = 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:
(σa )MAX =
ANS:
(1800 N−m)(0.03 m) 4.194×10 7 m4 −
(σa )MAX = 128.8 MPa
In case (b), the maximum bending stress is:
(σb )MAX =
ANS:
(1800 N−m)(0.03 m) 6.87×10 7 m4
(σb )MAX = 78.6 MPa
−
−7
m4
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 hasthe cross section (a); if it hasthe 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:
I a =
I b =
(4.47 in)(4.47 in)3 = 33 .27 in4 12
(6 in)(6 in)3 (4 in)(4 in)3 − = 86 .67 in4 12 12
Equilbiruim to find Ay + By
Free Body Diagram:
ΣM B = 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:
(σT )MAX =
ANS:
(195, 000 in−lb) (4.47 in/2) 33.27 in4
(σT )MAX = 13.1 ksi at the bottom of the cross-section
In case (b), the maximum tensile stress is:
(σT )MAX =
ANS:
(195, 000 in−lb)(3in) 86.67 in4
(σT )MAX = 6.75 ksi at the bottom of the cross-section
Problem 15.13 Thebeamin Problem 15.12 consists of material that will safely support a tensile or compressive stress of 30ksi. Based on this criterion, determine the largestdistributed load w0 (in lb/in) thebeamwillsafely 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).
ΣF y = 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 −w 0 2
85 in 2
85 in 4
= 903.1w0
To find the maximum allowable bending moment in case (a), the maximum allowable normal stress is used.
30, 000lb/in2 =
M (4.47 in/2) → M = 446, 577in−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
Solving equations [1] and [3] together:
ANS:
w0 = 960 lb/in2
[3]
[1]
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:
τ AVG =
ANS:
3V 3(8, 000 N) = 2A 2(0.04 m)(0.06 m)
τ 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 . Solution: (a) Using Equation (15-17) to determine the average stress at y = 0.01 m :
τ AVG
6V = bh3
ANS:
h 2
2
− y
2
6(8, 000 N) = (0.04 m)(0.06 m)3
0.06 m 2
2
2
− (0.01 m)
τ AVG = 4.44 MPa
(b) Using Equation (15-17) to determine the average stress at y = −0.02 m :
τ AVG
6V = bh3
ANS:
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
τ 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:
Free Body Diagram:
Summing moments about point B to determine Ay :
M B = 0 = [(6, 000 N/m)(8 m)] (4m)−Ay (8m) → Ay = 24, 000N ↑
Cut the FBD where x = 3 m and draw the FBD.
Summing vertical forces to determine the shear force V :
F y = 0 = 24, 000N−(6, 000N/m)(3m)−V → V = 6, 000N ↓
Using Equation (15-17) to determine the average stress at y = −0.05 m :
τ AVG
6V = bh3
ANS:
h 2
2
− y
τ AVG = 121 kPa
2
6(6, 000 N) = (0.25 m)(0.25 m)3
0.25 m 2
2
2
− (0.05 m)
Problem 15.28 Solve Problem 15.27 for the cross section at x = 80 in. Free Body Diagram:
Solution: Summing themomentsabout point B to determinethe reactionat point A:
M B = 0 = [(1500 lb/in)(60 in)] (30in)−Ay (120in) → Ay = 22, 500lb ↑
Draw the FBD at x = 80 in.
Summing the vertical forces to determine the shear force:
F y = 0 = 22, 500lb(−1, 500lb/in)(20in)+V → V = 7, 500lb ↑
(a) Using Equation (15-18) to determine the average shear stress at the neutral axis:
τ =
ANS:
3V 3(7, 500 lb) = 2A 2(1 in)(4 in)
τ AVG = 2, 812.5 lb/in2
(b) Using Equation (15-17) to determine the average stress at y = 1.5 in :
τ AVG
6V = bh3
ANS:
h 2
2
− y
2
6(7, 500 lb) = (1 in)(4 in)3
τ AVG = 1, 230.46 lb/in2
4 in 2
2
− (1.5 in)
2
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 themomentsabout point B to determinethe reactionat point A:
M B = 0 = [(1500 lb/in)(60 in)] (30in)−Ay (120in) → Ay = 22, 500lb ↑
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)
τ =
ANS:
τ 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:
τ AVG =
6v = 3 bh
=
h/2
6v bh3
6vb bh3
h/2
−
h/2
h/2
−
h/2
h/2
−
h2 = (y )2 2
h2 = ( y )2 2
h2 = (y )2 2
dA
bdy
dy
Doing the integration:
τ AVG
τ AVG
6v = 3 h
τ AVG
ANS:
6v = 3 h
h2 y (y )3 − 4 3
h3 h3 − 8 24
6v = 3 h
τ AVG = V
h/2
−
−h3
h/2
−h3
−
h3 −h3 − 4 12
4
=
6v h3
−
24
h3 6
By = 67, 500lb ↑
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 =
5.5 in = −2.75 in 2
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 =
(2 in)(8 in)3 (4 in)(2 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
Now calculating the average shear stress:
VQ (2, 400 lb)(−30.25 in3 ) = bI (2 in)(162.7 in 4 )
τ AVG =
ANS:
τ 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 =
(2 in)(8 in)3 (4 in)(2 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 in 2
Calculating Q:
Q = y A = (1.5 in)(8 in2 ) = 12 in3
Now calculating the average shear stress:
τ AVG =
ANS:
VQ (2, 400 lb)(12 in3 ) = bI (2 in)(162.7 in 4 )
τ AVG = 88.5 lb/in2
