Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9780073398242
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 7.2, Problem 7.51P
To determine

Draw the shear force and bending moment diagrams.

Find the maximum absolute values of shear force and bending moment.

Expert Solution & Answer
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Answer to Problem 7.51P

The maximum absolute shear force is |Vmax|=90lb_.

The maximum absolute bending moment is |Mmax|=1,400lb-in._

Explanation of Solution

Assumption:

Apply the sign convention for calculating the equations of equilibrium as below.

  • For the horizontal forces equilibrium condition, take the force acting towards right side as positive (+) and the force acting towards left side as negative ().
  • For the vertical forces equilibrium condition, take the upward force as positive (+) and downward force as negative ().
  • For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Apply the following sign convention for calculating the bending moment at any section x-x while approaching from the left hand side.

  • Take clockwise moment as positive and anticlockwise moment as negative

Apply the following sign convention for calculating the shear force at any section x-x while approaching from the left hand side.

  • Take downward force as negative and upward force as positive.

Calculation:

Show the free-body diagram of the beam as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  1

Refer to Figure 1:

Find the value of angle θ.

tanθ=oppositesideadjacentside=1420θ=34.99°

Find the force in the cable CD by taking moment about point H.

MH=050(26)+FCDcos34.99°(4)FCDsin34.99°(20)+100(10)50(6)=01,300+3.277FCD11.469FCD+1,000300=02,0008.192FCD=0

FCD=244.141lb

Find the vertical reaction at point H by resolving the vertical component of forces.

Fy=050100+FCDsin34.99°+Hy50=0200+244.141sin34.99°+Hy=0Hy=60lb()

Find the horizontal reaction at point H by resolving the horizontal component of forces.

Fx=0HxFCDcos34.99°=0Hx244.141cos34.99°=0Hx=200lb()

Show the calculated reaction values of the beam as in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  2

Consider a section x from left end of the beam.

Section AE (0x<6in.):

Show the free body diagram of the section as in Figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  3

Find the shear force at the section by resolving the vertical component of forces.

Fy=050V=0V=50lb (1)

Find the bending moment at the section by taking moment about the section.

Mx=0M+50(x)=0M=50x (2)

When x=0;

Refer to Equation (1);

V=50lb

Substitute 0 for x in Equation (2).

M=50(0)=0

When x=6in.;

Refer to Equation (1);

V=50lb

Substitute 6 in. for x in Equation (2).

M=50(6)=300lb-in.

Section EF (6in.<x<16in.):

Show the free body diagram of the section as in Figure 4.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  4

Find the shear force at the section by resolving the vertical component of forces.

Fy=050+140V=0V=90lb (3)

Find the bending moment at the section by taking moment about the section.

Mx=0M+50(x)140(x6)800=0M=140(x6)50x+800 (4)

When x=6in.;

Refer to Equation (3);

V=90lb

Substitute 6 in. for x in Equation (4).

M=140(66)50(6)+800=0300+800=500lbin.

When x=16in.;

Refer to Equation (3);

V=90lb

Substitute 16 in. for x in Equation (4).

M=140(166)50(16)+800=1,400800+800=1,400lb-in.

Section FG (16in.<x<26in.):

Show the free body diagram of the section as in Figure 5.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  5

Find the shear force at the section by resolving the vertical component of forces.

Fy=050+140100V=0V=10lb (5)

Find the bending moment at the section by taking moment about the section.

Mx=050(x)140(x6)800+100(x16)+M=0M=140(x6)+80050x100(x16) (6)

When x=16in.;

Refer to Equation (5).

V=10lb

Substitute 16 in. for x in Equation (6).

M=140(166)+80050(16)100(1616)=1,400+8008000=1,400lb-in.

When x=26in.;

Refer to Equation (5).

V=10lb

Substitute 26 in. for x in Equation (6).

M=140(266)+80050(26)100(2616)=2,800+8001,3001,000=1,300lb-in.

Section GB (26in.<x32in.):

Show the free body diagram of the section as in Figure 6.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  6

Find the shear force at the section by resolving the vertical component of forces.

Fy=0V50=0V=50lb (7)

Find the bending moment at the section by taking moment about the section.

Mx=0M50(32x)=0M=50(32x) (8)

When x=26in.;

Refer to Equation (7).

V=50lb

Substitute 26 in. for x in Equation (8).

M=50(3226)=300lb-in.

When x=32in.;

Refer to Equation (7).

V=50lb

Substitute 32 in. for x in Equation (8).

M=50(3232)=0

Tabulate the calculated shear force values as in Table 1.

Distance, x (in.)Shear force, V (lb)
0–50
6 (Just left)–50
6(Just right)90
16 (Just left)90
16 (Just right)–10
26 (Just left)–10
26 (Just right)50
3250

Plot the shear force diagram as in Figure 7.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  7

Tabulate the calculated bending moment values as in Table 2.

Distance, x (in.)Bending moment, M (lb-in.)
00
6 (Just left)–300
6 (Just right)500
161,400
26 (just left)1,300
26 (Just right)–300
320

Plot the bending moment diagram as in Figure 8.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.2, Problem 7.51P , additional homework tip  8

Refer to the Figure (7);

The maximum absolute shear force is |Vmax|=90lb_.

Refer to the Figure (8);

The maximum absolute bending moment is |Mmax|=1,400lb-in._

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Chapter 7 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 7.1 - A semicircular rod is loaded as shown. 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