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
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 18P
To determine

The reaction force equation at point A.

The reaction force equation at point C.

The shear force equation for section AB.

The shear force equation for section BC.

The bending moment equation for section AB.

The bending moment equation for section BC.

The deflection equations for section AB.

The deflection equations for section BC.

Expert Solution & Answer
Check Mark

Answer to Problem 18P

The reaction force equation at point A is R1=wa2l(2la).

The reaction force equation at point C is R2=wa22l.

The shear force equation for section AB is VAB=w2l[2l(ax)a2].

The shear force equation for section BC is VBC=wa22l.

The bending moment equation for section AB is MAB=wx2l(2ala2lx).

The bending moment equation for section BC is MBC=wa22l(lx).

The deflection equations for section AB is yAB=wx24EIl[2ax2(2la)lx3a2(2la)2].

The deflection equations for section BC is yBC=yAB+w24EI(xa)4.

Explanation of Solution

Write the balanced force equation in vertical direction.

    R1+R2=wa                                                                      (I)

Here, the reaction at point A is R1, the load on the beam is w, the length of the section BC is a and the reaction at point B is R2.

Take the net moment about point C.

    R1lwa(la2)=0R1l=wa(2la)2R1=wa2l(2la)

Thus, the reaction force at point A is wa2l(2la).

Substitute wa2l(2la) for R1 in Equation (I).

    wa2l(2la)+R2=wawawa22l+R2=waR2=wa22l

Thus, the reaction force at point C is wa22l.

Take a section at a distance x from left end as shown in below figure.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 4, Problem 18P , additional homework tip  1

Figure (1)

Write the shear force equation for part AB.

    VAB=R1wx                                                                          (II)

Here, the shear force for the section AB is VAB.

Substitute wa2l(2la) for R1 in Equation (II).

    VAB=wa2l(2la)wx=w2l[(2laa2)2lx]=w2l[2l(ax)a2]

Thus, the shear force equation for region AB is w2l[2l(ax)a2].

Take a section at a distance x from left end as shown in below figure.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 4, Problem 18P , additional homework tip  2

Figure (2)

Write the shear force equation for part BC.

    VBC=R1wa                                                                (III)

Here, the shear force for the section BC is VBC.

Substitute wa2l(2la) for R1 in Equation (III).

    VBC=wa2l(2la)wa=wawa22lwa=wa22l

Thus, the shear force equation for region BC is wa22l.

Write the moment equation for section AB using figure (1).

    MAB=R1xwxx2                                                           (IV)

Here, the moment for the section AB is MAB.

Substitute wa2l(2la) for R1 in Equation (IV).

    MAB=wa2l(2la)xwxx2=wx2l(2ala2lx)

Thus, the bending moment equation for region AB is wx2l(2ala2lx).

Write the moment equation for section BC using figure (2).

    MBC=R1xwa(xa2)                                                                  (V)

Here, the moment for the section BC is MBC.

Substitute wa2l(2la) for R1 in Equation (V).

    MBC=wa2l(2la)xwa(xa2)=w2l(2alxa2x2alx+a2l)=w2l(a2x+a2l)=wa22l(lx)

Thus, the bending moment equation for region BC is wa22l(lx).

Write the bending moment equation for section AB using figure (1).

    EId2yABdx2=MAB                                                                          (VI)

Here, Young’s modulus of the beam is E and the moment of inertia of the beam is I.

Substitute wx2l(2ala2lx) for MAB in Equation (VI).

    EId2yABdx2=wx2l(2ala2lx)

Integrate Equation (VI).

    EIdyABdx=w2l[(2ala2)x22lx33]+C1                                     (VII)

Here, the integration constant is C1.

Integrate Equation (VII).

    EIyAB=w2l[(2ala2)x36lx412]+C1x+C2                                (VIII)

Here, the second integration constant is C2.

Substitute 0 for x and 0 for yAB in Equation (VIII).

    EI0=w2l[(2ala2)0l0]+C10+C20=0+0+C2C2=0

Substitute 0 for C2 in the Equation (VIII).

    EIyAB=w2l[(2ala2)x36lx412]+C1x+0EIyAB=w2l[(2ala2)x36lx412]+C1x                            (IX)

Here, the deflection for the section AB is yAB.

Substitute wa224l(a24al4l2) for C1 and 0 for C2 in Equation (VIII).

    EIyAB=w2l[(2ala2)x36lx412]+wa224l(a24al4l2)x+0=w24l[4alx3+2a2x3+lx4+a4x4a3lx4a2l2x]yAB=wx24EIl[4alx2+2a2x2+lx3+a44a3l4a2l2]=wx24EIl[2ax2(2la)lx3a2(2la)2]

Thus, the beam deflection equation for region AB is yAB=wx24EIl[2ax2(2la)lx3a2(2la)2].

Write the bending moment equation for section BC using figure (2).

    EId2yBCdx2=MBC                                                                            (X)

Here, the moment for the section BC is MBC.

Substitute wa22l(lx) for MBC in Equation (X)

    EId2yBCdx2=wa22l(lx)

Integrate the Equation (X).

    EIdyBCdx=wa22l(lxx22)+C3                                                     (XI)

Here, the first integration constant is C3.

Integrate the Equation (XI).

    EIyBC=wa22l(lx22x36)+C3x+C4                                             (XII)

Here, the second integration constant is C4.

Substitute l for x and 0 for yBC in Equation (XII).

    EI0=wa22l(ll22l36)+C3l+C4C3l+C4=wa2l26C4=wa2l26C3l

Substitute wa2l26C3l for C4 in the Equation (XII).

    EIyBC=wa22l(lx22x36)+C3x+(wa2l26C3l)                          (XIII)

At x=a, yAB=yBC or EIyAB=EIyBC.

Equate the right hand side of equation (XI) and (XIII) and substitute a for x.

    w2l[(2ala2)x36lx412]+C1x={wa22l(lx22x36)+C3x+(wa2l26C3l)}w2l[(2ala2)a36la412]+C1a={wa22l(la22a36)+C3a+(wa2l26C3l)}C1a={wa46wa512lwa424wa44+wa512lwa2l26+C3(al)}C1a=wa48wa2l26+C3(al)  (XIV)

At x=a,

    dyABdx=dyBCdxEIdyABdx=EIdyBCdx                                                                        (XIV)

Substitute w2l[(2ala2)x22lx33]+C1 for EIdyABdx and wa22l(lxx22)+C3 for EIdyBCdx and a for x in Equation (XIV)

    w2l[(2ala2)a22la33]+C1=wa22l(laa22)+C3w2l[a3la42a3l3]+C1=w2l[a3la42]+C3w2l2a3l3+wa44l+C1=w2la3l+wa44l+C3C1+wa36=C3               (XV)

Substitute C1+wa36 for C3 in Equation (XV).

    C1a=wa48wa2l26+(C1+wa36)(al)C1a=wa48wa2l26+C1aC1l+wa46wa3l6C1l=wa424wa3l6wa2l26C1=wa224l(a24al4l2)                                   (XVI)

Substitute wa224l(a24al4l2) for C1 in Equation (XV).

    wa224l(a24al4l2)+wa36=C3wa424lwa36wa2l6+wa36=C3wa224l(a24l)=C3

Substitute wa224l(a24l) for C3 and 0 for C4 in Equation (XII).

    EIyBC=wa22l(lx22x36)+C3x+(wa2l26C3l)=wa2x24+wa2x312l+wa224l(a24l)(xl)wa2l26=w24[6a2x2+2a2x3l+a2l(a2xa2l4lx+4l2)4a2l2]=w24l[2a2x3+a4xa4l4la2x+4l2a2]+w24[6a2x24a2l2]

Solve the equation further,

    yBC=w24EIl[2a2x3+a4xa4l4la2x+4l2a2]+w24EI[6a2x24a2l2]=wx24EIl[2ax2(2la)lx3a2(2la)2]+w24EI(xa)4 (XVII)

Substitute wx24EIl[2ax2(2la)lx3a2(2la)2] for yAB in (XVII).

    yBC=yAB+w24EI(xa)4

Thus, the beam deflection equation for region BC is yBC=yAB+w24EI(xa)4.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 4 Solutions

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

Ch. 4 - A simply supported beam loaded by two forces is...Ch. 4 - Using superposition, find the deflection of the...Ch. 4 - A rectangular steel bar supports the two...Ch. 4 - An aluminum tube with outside diameter of 2 in and...Ch. 4 - The cantilever shown in the figure consists of two...Ch. 4 - Using superposition for the bar shown, determine...Ch. 4 - A simply supported beam has a concentrated moment...Ch. 4 - Prob. 18PCh. 4 - Using the results of Prob. 418, use superposition...Ch. 4 - Prob. 20PCh. 4 - Consider the uniformly loaded simply supported...Ch. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - 429 to 434 For the steel countershaft specified in...Ch. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - For the steel countershaft specified in the table,...Ch. 4 - For the steel countershaft specified in the table,...Ch. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - The cantilevered handle in the figure is made from...Ch. 4 - Prob. 42PCh. 4 - The cantilevered handle in Prob. 384, p. 154, is...Ch. 4 - A flat-bed trailer is to be designed with a...Ch. 4 - The designer of a shaft usually has a slope...Ch. 4 - Prob. 46PCh. 4 - If the diameter of the steel beam shown is 1.25...Ch. 4 - For the beam of Prob. 4-47, plot the magnitude of...Ch. 4 - Prob. 49PCh. 4 - 4-50 and 4-51 The figure shows a rectangular...Ch. 4 - and 451 the ground at one end and supported by a...Ch. 4 - The figure illustrates a stepped torsion-bar...Ch. 4 - Consider the simply supported beam 5 with a center...Ch. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Solve Prob. 410 using singularity functions. Use...Ch. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Solve Prob. 413 using singularity functions. Since...Ch. 4 - Prob. 61PCh. 4 - Solve Prob. 419 using singularity functions to...Ch. 4 - Using singularity functions, write the deflection...Ch. 4 - Determine the deflection equation for the...Ch. 4 - Use Castiglianos theorem to verify the maximum...Ch. 4 - Use Castiglianos theorem to verify the maximum...Ch. 4 - Solve Prob. 415 using Castiglianos theorem.Ch. 4 - Solve Prob. 452 using Castiglianos theoremCh. 4 - Determine the deflection at midspan for the beam...Ch. 4 - Using Castiglianos theorem, determine the...Ch. 4 - Solve Prob. 441 using Castiglianos theorem. Since...Ch. 4 - Solve Prob. 442 using Castiglianos theorem.Ch. 4 - The cantilevered handle in Prob. 384 is made from...Ch. 4 - Solve Prob. 450 using Castiglianos theorem.Ch. 4 - Solve Prob. 451 using Castiglianos theorem.Ch. 4 - The steel curved bar shown has a rectangular cross...Ch. 4 - Repeat Prob. 476 to find the vertical deflection...Ch. 4 - For the curved steel beam shown. F = 6.7 kips....Ch. 4 - A steel piston ring has a mean diameter of 70 mm....Ch. 4 - For the steel wire form shown, use Castiglianos...Ch. 4 - 4-81 and 4-82 The part shown is formed from a...Ch. 4 - 4-81 and 4-82 The part shown is formed from a...Ch. 4 - Repeat Prob. 481 for the vertical deflection at...Ch. 4 - Repeat Prob. 482 for the vertical deflection at...Ch. 4 - A hook is formed from a 2-mm-diameter steel wire...Ch. 4 - The figure shows a rectangular member OB, made...Ch. 4 - Prob. 87PCh. 4 - For the wire form shown, determine the deflection...Ch. 4 - Prob. 89PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92PCh. 4 - Solve Prob. 492 using Castiglianos method and...Ch. 4 - An aluminum step bar is loaded as shown. (a)...Ch. 4 - The steel shaft shown in the figure is subjected...Ch. 4 - Repeat Prob. 495 with the diameters of section OA...Ch. 4 - The figure shows a 12- by 1-in rectangular steel...Ch. 4 - For the beam shown, determine the support...Ch. 4 - Solve Prob. 498 using Castiglianos theorem and...Ch. 4 - Consider beam 13 in Table A9, but with flexible...Ch. 4 - Prob. 101PCh. 4 - The steel beam ABCD shown is simply supported at C...Ch. 4 - Prob. 103PCh. 4 - A round tubular column has outside and inside...Ch. 4 - For the conditions of Prob. 4104, show that...Ch. 4 - Link 2, shown in the figure, is 25 mm wide, has...Ch. 4 - Link 3, shown schematically in the figure, acts as...Ch. 4 - The hydraulic cylinder shown in the figure has a...Ch. 4 - The figure shows a schematic drawing of a...Ch. 4 - If drawn, a figure for this problem would resemble...Ch. 4 - Design link CD of the hand-operated toggle press...Ch. 4 - Find the maximum values of the spring force and...Ch. 4 - As shown in the figure, the weight W1 strikes W2...Ch. 4 - Part a of the figure shows a weight W mounted...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Understanding Torsion; Author: The Efficient Engineer;https://www.youtube.com/watch?v=1YTKedLQOa0;License: Standard YouTube License, CC-BY