Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
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Combined Loading
Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…
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Textbook Question
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Chapter 3, Problem 48P

3–48 and 3–49 The beam shown is loaded in the xy and xz planes.

  1. (a) Find the y- and z-components of the reactions at the supports.
  2. (b) Plot the shear-force and bending-moment diagrams for the xy and xz planes. Label the diagrams properly and provide the values at key points.
  3. (c) Determine the net shear-force and bending-moment at the key points of part (b).
  4. (d) Determine the maximum tensile bending stress. For Prob. 3–48, use the cross section given in Prob. 3–34, part (a). For Prob. 3–49, use the cross section given in Prob. 3–39, part (b).

Chapter 3, Problem 48P, 348 and 349 The beam shown is loaded in the xy and xz planes. (a) Find the y- and z-components of

(a)

Expert Solution
Check Mark
To determine

The y-components of the reactions at the supports of the beam.

The z-components of the reactions at the supports of the beam.

Answer to Problem 48P

The y-components of the reactions at the supports O and of the beam are 0.649 kN and 1.949 kN respectively.

The z-components of the reactions at the supports O and C of the beam are 1.625 kN and 1.375 kN respectively.

Explanation of Solution

The following figure shows the forces acting on the beam along x-z plane.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 48P , additional homework tip 1

Figure-(1)

Write the equilibrium equation of force applied on the beam in z-direction.

    ∑Fz=0R1z−(1.5 kN)−(2(sin30°) kN/m)(1.5 m)+R2z=0R1z+R2z−3 kN=0R1z+R2z=3 kN                                     (I)

Write the bending moment at the support O in x-z plane.

    ∑MO=0[(1.5 kN)(0.5 m)+(2(sin30°) kN/m)(1.5 m)(0.5 m+1 m+1.5 m2)−R2z(3 m)]=0(0.75 kN⋅m)+(1.5 kN)(2.25 m)−R2z(3 m)=0 (II)

The following figure shows the forces acting on the beam along x-y plane.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 48P , additional homework tip 2

Figure-(2)

Write the equilibrium equation of force applied on the beam in y-direction.

    ∑Fy=0R1y−(2cos(30°) kN/m)(1.5 m)+R2y=0R1y+R2y−2.598 kN=0R1y+R2y=2.598 kN                                                 (III)

Write the bending moment at the support O in x-y plane.

    ∑MO=0−(2cos(30°) kN/m)(1.5 m)(0.5 m+1 m+1.5 m2)+R2y(3 m)=0−(2.598 kN)(2.25 m)+R2y(3 m)=0          (IV)

Conclusion:

From Equation (II), R2z=1.375 kN.

Substitute 1.375 kN for R2z in Equation (I).

    R1z+1.375 kN=3 kNR1z=(3 kN)−(1.375 kN)R1z=1.625 kN

Thus, the z-components of the reactions at the supports O and C of the beam are 1.625 kN and 1.375 kN respectively.

Simplify Equation (IV).

    −(2.598 kN)(2.25 m)+R2y(3 m)=0R2y=5.8455 kN⋅m3R2y≈1.949 kN

Substitute 1.949 kN for R2y in Equation (III).

    R1y+1.949 kN=2.598 kNR1y=(2.598 kN)−(1.949 kN)R1y=0.649 kN

Thus, the y-components of the reactions at the supports O and of the beam are 0.649 kN and 1.949 kN respectively.

(b)

Expert Solution
Check Mark
To determine

The shear force and bending moment diagram of the beam.

Explanation of Solution

Write the expression for the moment at point A.

    MAz=−R2z(2.5 m)+(2sin30° kN/m)(1.5 m)(1.5 m2+1 m)        (V)

Write the expression for the moment at point B.

    MBz=−R2z(1.5 m)+(2sin30° kN/m)(1.5 m)(1.5 m2)                 (VI)

Write the equation for the shear force.

    Vz=−2sin30°x+R2z                                                                     (VII)

Write the expression for the maximum bending moment.

    Mzmax=−R2z(x)+(2sin30° kN/m)(x)(x2)                                 (VIII)

Write the expression for the bending moment at B.

    MBy=R2y(1.5 m)−(2cos30° kN/m)(1.5 m)(1.5 m2)                    (IX)

Write the equation for the shear force.

    Vy=2cos30°x−R2y                                                                          (X)

Write the expression for the maximum bending moment.

    Mymax=R2y(x)−(2cos30° kN/m)(x)(x2)                                     (XI)

Conclusion:

Substitute 1.375 kN for R2z in Equation (V).

    MAz=−(1.375 kN)(2.5 m)+(2sin30° kN/m)(1.5 m)(1.5 m2+1 m)=−(1.375 kN)(2.5 m)+(1 kN/m)(1.5 m)(1.75 m)=−0.8125 kN⋅m

Substitute 1.375 kN for R2z in Equation (VI).

    MBz=−(1.375 kN)(1.5 m)+(2sin30° kN/m)(1.5 m)(1.5 m2)=−(1.375 kN)(1.5 m)+(1 kN/m)(1.5 m)(0.75 m)=−0.9375 kN⋅m

The maximum bending moment occur at Vz=0 kN.

Substitute 0 kN for Vz, and 1.375 kN for R2z in Equation (VII).

    0 kN=−2sin30°x+1.375 kN0 kN=x kN/m+1.375 kNx=1.375 m

Substitute 1.375 m for x, and 1.375 kN for R2z in Equation (VIII).

    Mzmax=−(1.375 kN)(1.375 m)+(2sin30° kN/m)(1.375 m)(1.375 m2)=−(1.375 kN)(1.375 m)+(1 kN/m)(1.375 m)(1.375 m2)=−0.9453 kN⋅m

The following figure shows the shear force and bending moment diagram of the beam along x-z plane.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 48P , additional homework tip 3

Figure-(3)

Substitute 1.949 kN for R2y in Equation (IX).

    MBy=(1.949 kN)(1.5 m)−(2cos30° kN/m)(1.5 m)(1.5 m2)=(1.949 kN)(1.5 m)−(1.732 kN/m)(1.5 m)(0.75 m)=0.975 kN⋅m

The maximum bending moment occur at Vy=0 kN.

Substitute 0 kN for Vy, and 1.949 kN for R2y in Equation (X).

    0 kN=2cos30°x−1.949 kN0 kN=1.732x kN/m−1.949 kNx=1.125 m

Substitute 1.125 m for x, and 1.949 kN for R2y in Equation (XI).

    Mymax=1.949 kN(1.125 m)−(2cos30° kN/m)(1.125 m)(1.125 m2)=1.949 kN(1.125 m)−(1.732 kN/m)(1.125 m)(0.5625 m)=1.096 kN⋅m

The following figure shows the shear force and bending moment diagram of the beam along x-y plane.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 48P , additional homework tip 4

Figure-(4)

(c)

Expert Solution
Check Mark
To determine

The net bending moment at x=3 m.

The net shear force at x=3 m.

The net bending moment at x=1.5 m.

The net shear force at x=1.5 m.

The net bending moment at x=0.5 m.

The net shear force at x=0.5 m.

The net bending moment at x=0 m.

The net shear force at x=0 m.

Answer to Problem 48P

The net bending moment at x=3 m is 0 kN⋅m.

The net shear force at x=3 m is 2.385 kN.

The net bending moment at x=1.5 m is 1.352 kN⋅m.

The net shear force at x=1.5 m is 0.661 kN.

The net bending moment at x=0.5 m is 0.8749 kN⋅m.

The net shear force at x=0.5 m is 1.749 kN.

The net bending moment at x=0 m is 0 kN⋅m.

The net shear force at x=0 m is 1.749 kN.

Explanation of Solution

Write the expression for the net shear force on the beam.

    V=Vz2+Vy2                                                                        (XII)

Write the expression for the net bending moment on the beam.

    M=Mz2+My2                                                                  (XIII)

Conclusion:

The following table shows the magnitudes of bending moment and shear force along x-y and x-z plane from the shear force and bending moment diagram.

x(m)Vz(kN)Vy(kN)My(kN⋅m)Mz(kN⋅m)
0−1.6250.649100
0.5−1.6250.64910.81250.3246
1.5−0.1250.64910.93750.9737
31.375−1.94900

At x=0 m:

Substitute −1.625 kN for Vz, and 0.6491 kN for Vy in Equation (XII).

    V0=(−1.625 kN)2+(0.6491 kN)2=2.640625+0.42133 kN=1.749 kN

Thus, the net shear force at x=0 m is 1.749 kN.

Substitute 0 kN⋅m for Mz, and 0 kN⋅m for My in Equation (XIII).

    M0=(0 kN⋅m)2+(0 kN⋅m)2=0 kN⋅m

Thus, the net bending moment at x=0 m is 0 kN⋅m.

At x=0.5 m:

Substitute −1.625 kN for Vz, and 0.6491 kN for Vy in Equation (XII).

    V0.5=(−1.625 kN)2+(0.6491 kN)2=2.640625+0.42133 kN=1.749 kN

Thus, the net shear force at x=0.5 m is 1.749 kN.

Substitute 0.8125 kN⋅m for Mz, and 0.3246 kN⋅m for My in Equation (XIII).

    M0.5=(0.8125 kN⋅m)2+(0.3246 kN⋅m)2=(0.66015+0.10536) kN2⋅m2=0.8749 kN⋅m

Thus, the net bending moment at x=0.5 m is 0.8749 kN⋅m.

At x=1.5 m:

Substitute −0.125 kN for Vz, and 0.6491 kN for Vy in Equation (XII).

    V1.5=(−0.125 kN)2+(0.6491 kN)2=0.015625+0.42133 kN=0.661 kN

Thus, the net shear force at x=1.5 m is 0.661 kN.

Substitute 0.9737 kN⋅m for Mz, and 0.9375 kN⋅m for My in Equation (XIII).

    M1.5=(0.9737 kN⋅m)2+(0.9375 kN⋅m)2=1.82699 kN⋅m=1.352 kN⋅m

Thus, the net bending moment at x=1.5 m is 1.352 kN⋅m.

At x=3 m:

Substitute 1.375 kN for Vz, and −1.949 kN for Vy in Equation (XII).

    V3=(1.375 kN)2+(−1.949 kN)2=1.890625+3.798601 kN=2.385 kN

Thus, the net shear force at x=3 m is 2.385 kN.

Substitute 0 kN⋅m for Mz, and 0 kN⋅m for My in Equation (XIII).

    M3=(0 kN⋅m)2+(0 kN⋅m)2=0 kN⋅m

Thus, the net bending moment at x=3 m is 0 kN⋅m.

(d)

Expert Solution
Check Mark
To determine

The maximum tensile bending stress on the beam.

Answer to Problem 48P

The maximum tensile bending stress on the beam is 73.52 MPa.

Explanation of Solution

The following diagram shows the cross-section of the beam.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 48P , additional homework tip 5

Figure-(5)

Write the expression for the second moment of area of the section about z-axis.

  Iz=Ia−2Ib                                                                       (XIV)

Here, the second moment area of the area a is Ia and the second moment of area of the area b is Ib.

Write the moment of inertia of the rectangular section.

    I=BH312                                                                               (XV)

Here total height of the section is H, and the width is H.

Write the expression for the moment of inertia about y-axis.

    Iy=2(hB312)+h2b2312                                                             (XVI)

Here, the height of the flange is h, height of the web is h2, and the width is b2.

Write the expression for maximum tensile bending stress.

    σmax=−MzyAIz+MyzAIy                                                       (XVII)

Conclusion:

Substitute 40 mm for B and 75 mm for H in Equation (XV).

    Ia=(40 mm)(75 mm)312=1406250 mm4

Substitute 34 mm for B and 25 mm for H in Equation (XV).

    Ib=(34 mm)(25 mm)312=44270.83 mm4

Substitute 1406250 mm4 for Ia and 44270.83 mm4 for Ib in Equation (XIV).

    Iz=1406250 mm4−2×44270.83 mm4=(1317708.34 mm4)(10−12 m41 mm4)≈1.32×10−6 m4

Substitute 40 mm for B, 25 mm for h, 25 mm for h2, and 6 mm for b2 in Equation (XVI).

    Iy=2((25 mm)(40 mm)312)+(25 mm)(6 mm)312=2(133333.33 mm4)+450 mm4=(267116.66 mm4)(10−12 m41 mm4)≈2.67×10−7 m4

Substitute 1.075 kN⋅m for Mz, −0.9408 kN⋅m for My, −37.5 mm for yA, −20 mm for zA, 1.32×10−6 m4 for Iz, and 2.67×10−7 m4 for Iy in Equation (XVII).

    σmax=−(1.075 kN⋅m)(−37.5 mm)1.32×10−6 m4+(−0.9408 kN⋅m)(−20 mm)2.67×10−7 m4=[−(1.075 kN⋅m)(−37.5 mm)(1 m1000 mm)1.32×10−6 m4+(−0.9408 kN⋅m)(−20 mm)(1 m1000 mm)2.67×10−7 m4]=[3.0539×103 kN/m2+7.0471×104kN/m2](1 MPa103 kN/m2)=73.52 MPa

Thus, the maximum tensile bending stress on the beam is 73.52 MPa.

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Chapter 3 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 3 - 31 to 34 Sketch a free-body diagram of each...Ch. 3 - 31 to 34 Sketch a free-body diagram of each...Ch. 3 - Sketch a free-body diagram of each element in the...Ch. 3 - 3-1 to 3-4 Sketch a free-body diagram of each...Ch. 3 - 35 to 38 For the beam shown, find the reactions at...Ch. 3 - 35 to 38 For the beam shown, find the reactions at...Ch. 3 - 35 to 38 For the beam shown, find the reactions at...Ch. 3 - For the beam shown, find the reactions at the...Ch. 3 - For the beam shown, find the reactions at the...Ch. 3 - Repeat Prob. 36 using singularity functions...
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