SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
10th Edition
ISBN: 9781259489563
Author: BUDYNAS
Publisher: INTER MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 32P
To determine

The slope of the shaft at each bearing.

Expert Solution & Answer
Check Mark

Answer to Problem 32P

The slope of the shaft at bearing point O is 0.00750rad and the slope of the shaft at bearing point C is 0.00558rad.

Explanation of Solution

The figure below shows the free body diagram of pulley A.

SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I, Chapter 4, Problem 32P , additional homework tip  1

Figure (1)

The figure below shows the free body diagram of pulley B.

SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I, Chapter 4, Problem 32P , additional homework tip  2

Figure (2)

The given assumption is that the belt tension on the loose side is 15% of the tension on the tight side.

Write the expression for the tensions on the loose side in terms of tension on the tight side.

T2=0.15T1 (I)

Here, the tension on the tight side is T1 and the tension on the loose side is T2.

Write the equation to balance the tension on the counter shaft.

T=0(TA1TA2)dA2(T2T1)dB2=0 (II)

Here, the tension on the tight side of pulley A is TA1, the tension on the loose side of pulley A is TA2, diameter of shaft A is dA and the diameter of shaft B is dB.

Substitute 0.15T1 for T2 in Equation (II).

(TA1TA2)dA2+(0.15T1T1)dB2=0T1=10.85((TA1TA2)dAdB) (III)

Write the expression for net applied load at point A.

FA=(TA1+TA2) (IV)

Here, the tension force exerted by the pulley at each side at point A are TA1 and TA2 respectively.

Write the expression for net applied load at point B.

FB=T1+T2 (V)

Here, the tension force exerted by the pulley at each side at point B are T1 and T2 respectively.

Write the equation for moment of inertia of the shaft.

I=πd464 (VI)

Here, the diameter of the shaft is d.

The free body diagram of the beam in the direction of y-axis is shown below.

SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I, Chapter 4, Problem 32P , additional homework tip  3

Figure (3)

Write the equation for force component along y-axis.

FAy=FAsin45°

Here, the net force at point A is FA.

Write the equation for force component along z-axis.

FAz=FAcos45°

Write the deflection equation for portion OA using Table A-9 for beam 6.

yOA=FAyb1x6EIl(x2+b12l2)

Here, Young’s modulus of the shaft material is E, , the location of applied point load at point A from right end bearing point C is b1 and the distance of any given section from left end is x.

Substitute FAsin45° for FAy in the above Equation.

yOA=(FAsin45°)b1x6EIl(x2+b12l2)

Write the expression for net slope of the shaft at point O along z-axis.

(θO)z=(dyOAdx)x=0

Substitute (FAsin45°)b1x6EIl(x2+b12l2) for yOA.

(θO)z={ddx(FAsin45°)b1x6EIl(x2+b12l2)}x=0={16EIl[(FAsin45°)b1(3x2+b12l2)]}x=0=16EIl[(FAsin45°)b1(3(0)2+b12l2)]=(FAsin45°)b16EIl(b12l2) (VII)

The free body diagram of the beam in the direction of z-axis is shown below.

SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I, Chapter 4, Problem 32P , additional homework tip  4

Figure (4)

Write the deflection equation along z-axis using Table A-9 for beam 6.

zOA=FAzb1x6EIl(x2+b12l2)+FBb2x6EIl(x2+b22l2)

Here, the force component along z-axis is FAz , the net force at point B is FB , the location of applied point load at point B from right end bearing point C is b2 and the distance of any given section from left end is x.

Write the expression for net slope of the shaft at point O along y-axis.

(θO)y=(dzOBdx)x=0

Substitute FAzb1x6EIl(x2+b12l2)+FBb2x6EIl(x2+b22l2) for zOB.

(θO)y={ddxFAzb1x6EIl(x2+b12l2)+FBb2x6EIl(x2+b22l2)}x=0={16EIl[FAzb1(3x2+b12l2)+FBb2(3x2+b22l2)]}x=0=16EIl[FAzb1(3(0)2+b12l2)+FBb2(3(0)2+b22l2)]=FAzb16EIl(b12l2)FBb26EIl(b22l2)

Substitute FAcos45° for FAz in the above Equation.

(θO)y=(FAcos45°)b16EIl(b12l2)FBb26EIl(b22l2) (VIII)

Write the expression for net slope at point O.

ΘO=(θO)z2+(θO)y2 (IX)

Write the deflection equation for portion BC using Table A-9 for beam 6.

yBC=FAya1(lx)6EIl(x2+a122lx)

Here, the location of applied point load at point A from left end bearing point O is a1 and the distance of any given section from left end is x.

Write the expression for net slope of the shaft at point C along z-axis.

(θC)z=(dyBCdx)x=l

Substitute FAya1(lx)6EIl(x2+a122lx) for yBC.

(θC)z={ddxFAya1(lx)6EIl(x2+a122lx)}x=l={16EIl[FAya1(6lx2l23x2a12)]}x=l=16EIl[FAya1(6ll2l23l2a12)]=FAya16EIl(l2a12)

Substitute FAsin45° for FAy in the above Equation.

(θC)z=(FAsin45°)a16EIl(l2a12) (X)

Write the deflection equation for portion BC using Table A-9 for beam 6.

zBC=FAza1(lx)6EIl(x2+a122lx)+FBa2(lx)6EIl(x2+a222lx)

Here, the location of applied point load at point B from left end bearing point O is a2 and the distance of any given section from left end is x.

Write the expression for net slope of the shaft at point C along y-axis.

(θC)y=(dzBCdx)x=l

Substitute the value of zBC.

(θC)y={ddxFAza1(lx)6EIl(x2+a122lx)+FBa2(lx)6EIl(x2+a222lx)}x=l={16EIl[FAza1(6lx2l23x2a12)+FBa2(6lx2l23x2a22)]}x=l=16EIl[FAza1(6ll2l23l2a12)+FBa2(6ll2l23l2a22)]=FAza16EIl(l2a12)FBa26EIl(l2a22)

Substitute FAcos45° for FAz in the above Equation.

(θC)y=(FAcos45°)a16EIl(l2a12)FBa26EIl(l2a22) (XI)

Write the expression for net slope at point C.

ΘC=(θC)z2+(θC)y2 . (XII)

Conclusion:

Substitute 300N for TA1, 45N for TA2, 250mm for dA and 300mm for dB  in Equation (III).

T1=10.85((300N45N)×250mm300mm)=10.85(212.5N)=250N

Substitute 250N for T1 in Equation (I)

T2=0.15(250N)=37.5N

Substitute 300N for TA1 and 45N for TA2 in Equation (IV).

FA=(300N+45N)=345N

Substitute 250N for T1 and 37.5N for T2 in Equation (V).

FB=250N+37.5N=287.5N

Substitute 20mm for d in Equation (VI).

I=π(20mm)464=7854mm4

Substitute 345N for FA , 550mm for b1 , 850mm for l , 207×103MPa for E and 7854mm4 for I in Equation (VII).

(θO)z=((345N)sin45°)(550mm)6(207×103MPa)(7854mm4)(850mm)((550mm)2(850mm)2)=[(1.618×1081/mm2)(42000mm2)]=0.00680rad

Thus, the slope of the shaft at bearing point O along z-axis is 0.00680rad.

Substitute 345N for FA , 550mm for b1 , 850mm for l , 207×103MPa for E and 7854mm4 for I and 150mm for b2 in Equation (VIII).

(θO)y=[((345N)cos45°)(550mm)((550mm)2(850mm)2)6(207×103MPa)(7854mm4)(850mm)(287.5N)(150mm)6(207×103MPa)(7854mm4)(850mm)((150mm)2(850mm)2)]=[(0.00680rad)(3.6426×103rad)]0.00316rad

Thus, the slope of the shaft at bearing point O along y-axis is 0.00316rad.

Substitute 0.00680rad for (θO)z and 0.00316rad for (θO)y in Equation (IX).

ΘO=(0.00680rad)2+(0.00316rad)2=5.622×105rad20.00750rad

Thus, the net slope of the shaft at bearing point O is 0.00750rad.

Substitute 345N for FA , 300mm for a1 , 850mm for l , 207×103MPa for E and 7854mm4 for I in Equation (X).

(θC)z=((345N)sin45°)(300mm)6(207×103MPa)(7854mm4)(850mm)((850mm)2(300mm)2)=[(8.83×1091/mm2)(632500mm2)]=0.00558rad

Thus, the slope of the shaft at bearing point C along z-axis is 0.00558rad.

Substitute 345N for FA, 550mm for b1, 850mm for l, 207×103MPa for E, 7854mm4 for I and 150mm for b2 in Equation (XI).

(θC)y=[(345cos45°)(300)6(207×103)(7854)(850)(85023002)(287.5)(700)6(207×103)(7854)(850)(85027002)]=6.04×105rad

Thus, the slope of the shaft at bearing point C along y-axis is 6.04×105rad.

Substitute 0.00558rad for (θC)z and 6.04×105rad for (θC)y in Equation (XII).

ΘC=(0.00558rad)2+(6.04×105rad)2=3.11×105rad2=0.00558rad

Thus, the net slope of the shaft at bearing point C is 0.00558rad.

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 MECH.ENGINEERING DESIGN-EBK>I

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
EVERYTHING on Axial Loading Normal Stress in 10 MINUTES - Mechanics of Materials; Author: Less Boring Lectures;https://www.youtube.com/watch?v=jQ-fNqZWrNg;License: Standard YouTube License, CC-BY