SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
10th Edition
ISBN: 9781259489563
Author: BUDYNAS
Publisher: INTER MCG
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Chapter 4, Problem 27P
To determine

The deflection of the shaft at point A.

The slope of the shaft at point A.

Expert Solution & Answer
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Answer to Problem 27P

The deflection of the shaft at point A is 0.142in.

The slope of the shaft at point A is 0.00115rad.

Explanation of Solution

Write the expression for the moment of inertia of the shaft.

    I=πd464                                                                 (I)

Here, the diameter of the shaft is d and the moment of inertia is I.

Draw the free body diagram of the beam.

SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I, Chapter 4, Problem 27P , additional homework tip  1

Figure (1)

The free body diagram of the beam in the direction of y-axis is shown figure (1).

Write the deflection equation along y-axis for beam 6 and beam 10 using Table A-9.

    yA=F1yb1x6EIl(x2+b12l2)+F2ya2x6EIl(l2x2)                                               (II)

Here, the force component at point A along y-axis is F1y, the location of point A from the point C is b1, the distance of point A from left end is x, the total length of the beam between point O point C is l , Young modulus of the material is E , moment of inertia of the beam is I, the force component at point B along y-axis is F2y and the location of point B from the point C is a2.

Write the force component at point A along y-axis.

    F1y=F1cos20°                                                                                 (III)

Here, the force component at A along y-axis is F1y and the force at point A is F1.

Write the force component at point B along y-axis.

    F2y=F2sin20°                                                                                  (IV)

Here, the force component at B along y-axis is F2y and the force at point B is F2.

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 27P , additional homework tip  2

Figure (2)

Write the deflection equation along z-axis for beam 6 and beam 10 using Table A-9.

    zA=F1zb1x6EIl(x2+b12l2)+F2za2x6EIl(l2x2)                                           (IV)

Here, the force component at point A along z-axis is F1z and the force component at point B along z-axis is F2z.

Write the force component at point A along z-axis.

    F1z=F1sin20°                                                                                   (V)

Here, the force component at A along z-axis is F1z and the force at point A is F1.

Write the force component at point B along z-axis.

    F2z=F2cos20°                                                                                 (VI)

Here, the force component at B along z-axis is F2z and the force at point B is F2.

Write the expression for net displacement at point A.

    δ=yA2+zA2                                                                                       (VII)

Here, the net deflection at point A is δ.

Write the slope equation at along z-axis at point A.

    (θA)z=(dydx)=ddx{F1yb1x6EIl(x2+b12l2)+F2ya2x6EIl(l2x2)}=F1yb16EIl(3x2+b12l2)+F2ya26EIl(l23x2)                            (VIII)

Here, the slope along z-axis at point is (θA)z.

Write the slope equation at along y-axis at point A.

    (θA)y=(dzdx)=ddx{F1zb1x6EIl(x2+b12l2)+F2za2x6EIl(l2x2)}=F1zb16EIl(3x2+b12l2)F2za26EIl(l23x2)                           (IX)

Here, the slope along z-axis at point is (θA)y.

Write the expression for the net slope at point A.

    ΘA=(θA)z2+(θA)y2                                                                                (X)

Here, the net slope is ΘA.

Conclusion:

Substitute 1.25in for d in the equation (I).

    I=π×(1.25in)464=7.669964in4=0.1198in4

Substitute 300lbf for F1 in the equation (III).

    F1y=(300lbf)cos20°=(300lbf)(0.9396)=281.91lbf

Substitute 750lbf for F2 in the equation (IV).

    F2y=(750lbf)sin20°=(750lbf)(0.3420)=256.52lbf

Substitute 300lbf for F1 in the equation (V).

    F1z=(300lbf)sin20°=(300lbf)(0.3420)=102.61lbf

Substitute 750lbf for F2 in the equation (VI).

    F2z=(750lbf)cos20°=(750lbf)(0.9396)=704.77lbf

Substitute 281.91lbf for F1y, 16in for x, 14in for b1 , 30in for l, 30×106psi for E, 0.1198in4 for I and 256.52lbf for F2y in the equation (II).

    yA={(281.91lbf)(14in)(16in)6(30×106psi)(0.1198in4)(30in)((16in)2+(14in)2(30in)2)+(256.52lbf)(9in)(16in)6(30×106psi)(0.1198in4)(30in)((30in)2(16in)2)}={(9.7613×105)(448)+(5.709×105)(644)}(lbf/psiin2)(1psi1lbf/in2)=[(0.0437)+(0.0367)]in=0.0805in

Substitute 102.61lbf for F1z, 16in for x, 14in for b1 , 30in for l, 30×106psi for E, 0.1198in4 for I and 704.77lbf for F2z in the equation (IV).

    zA={(102.61lbf)(14in)(16in)6(30×106psi)(0.1198in4)(30in)((16in)2+(14in)2(30in)2)+(704.77lbf)(9in)(16in)6(30×106psi)(0.1198in4)(30in)((30in)2(16in)2)}={(3.5529×105)(448)+(1.5687×104)(644)}(lbf/psiin2)(1psi1lbf/in2)=[(0.0159)+(0.101)]in=0.1169in

Substitute 0.0805in for yA and 0.1169in for zA in the equation (VII).

    δ=(0.0805in)2+(0.1169in)2=0.00648in2+0.0136in2=0.142in

Thus, the net displacement of the shaft at point A is 0.142in.

Substitute 281.91lbf for F1y , 16in for x , 14in for b1 , 30in for l , 30×106psi for E , 0.1198in4 for I and 256.52lbf for F2y in the equation (VIII).

    (θA)z={(281.91lbf)(14in)6(30×106psi)(0.1198in4)(30in)(3(16in)2+(14in)2(30in)2)+(256.52lbf)(9in)6(30×106psi)(0.1198in4)(30in)((30in)23(16in)2)}={(6.1008×106)(64)+(3.5687×106)(132)}(lbf/psiin2)(1psi1lbf/in2)=[(0.000390)+(0.000471)]in=8.06×105rad

Substitute 102.61lbf for F1z , 16in for x , 14in for b1 , 30in for l , 30×106psi for E , 0.1198in4 for I and 704.77lbf for F2z in the equation (IX).

    (θA)y={(102.61lbf)(14in)6(30×106psi)(0.1198in4)(30in)(3(16in)2+(14in)2(30in)2)+(704.77lbf)(9in)6(30×106psi)(0.1198in4)(30in)((30in)23(16in)2)}={(2.2205×106)(64)+(9.8048×106)(132)}(lbf/psiin2)(1psi1lbf/in2)=[(0.000142)+(0.00129)]in=0.00115rad

Substitute 8.06×105rad for (θA)z and 0.00115rad for (θA)y in the equation (X).

    ΘA=(8.06×105rad)2+(0.00115rad)2=6.49×109rad2+1.3225×106rad2=0.00115rad

Thus, the net slope of the shaft at point A is 0.00115rad.

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Chapter 4 Solutions

SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I

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