Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9781259639272
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 7.3, Problem 7.92P

(a)

To determine

Find the value of P and Q.

(a)

Expert Solution
Check Mark

Answer to Problem 7.92P

The value of the unknown forces is P=0.540lb();Q=1.860lb()_.

Explanation of Solution

Given information:

The bending moment at point D is MD=+5.96kip-ft.

The bending moment at point E is ME=+6.84kip-ft.

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 section DE as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.3, Problem 7.92P , additional homework tip  1

Find the shear force at point E by taking moment about point D.

MD=05.96kip-ft×1,000lb1kip250×4×42+VE(4)+6.84kip-ft×1,000lb1kip=05,9602,000+4VE+6,840=0VE=280lb()

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

Fy=0VD250×4+VE=0VD1,000+280=0VD=720lb()

Show the free-body diagram of the section AD as in Figure 2.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.3, Problem 7.92P , additional homework tip  2

Find the value of force P by taking moment about point A.

MA=05.96kips-ft×1,000lb1kip720(4)P(2)250×4×42=05,9602,8802P2,000=0P=540lb×1kip1,000lb

P=0.540lb()

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

Fx=0Ax=0

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

Fy=0Ay250×4P720=0Ay1,000540720=0Ay=2,260lb()

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

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.3, Problem 7.92P , additional homework tip  3

Find the value of force Q by taking moment about point B.

MB=06.84kips-ft×1,000lb1kip+280(4)+Q(2)+250×4×42=06,840+1,120+2Q+2,000=0Q=1,860lb×1kip1,000lb

Q=1.860lb()

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

Fy=0By250×4Q280=0By1,0001,860280=0By=3,140lb()

Show the free-body diagram of the beam as in Figure 4.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.3, Problem 7.92P , additional homework tip  4

Therefore, the value of the unknown forces is P=0.540lb();Q=1.860lb()_.

(b)

To determine

Draw the shear force and bending moment diagrams.

(b)

Expert Solution
Check Mark

Explanation of Solution

Show the calculation of shear force as follows;

Shear force at x=0;

V|x=0(Justleft)=0V|x=0(Justright)=2,260lb

Shear force at x=2ft;

V|x=2ft(Justleft)=2,260250×2=1,760lbV|x=2ft(Justright)=1,760540=1,220lb

Shear force at x=10ft;

V|x=10ft(Justleft)=2,260250×10540=780lbV|x=10ft(Justright)=7801,860=2,640lb

Shear force at x=12ft;

V|x=12ft(Justleft)=2,260250×125401,860=3,140lbV|x=12ft(Justright)=3,140+3,140=0

Tabulate the calculated shear force values as in Table 1.

Distance, x (ft)Shear force, V (lb)
0 (Just left)0
0 (Just right)2,260
2 (Just left)1,760
2 (Just right)1,220
10 (Just left)–780
10 (Just right)–2,640
12 (Just left)–3,140
12 (Just right)0

Plot the shear force diagram as in Figure 5.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.3, Problem 7.92P , additional homework tip  5

Location of maximum bending moment;

The maximum bending moment occurs where the shear force changes it’s sign.

V=0

Refer to the shear force diagram the shear force changes in the region CF.

Show the similar triangle of the shear force diagram for the region CF as in Figure 6.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.3, Problem 7.92P , additional homework tip  6

Use the similar triangle method.

x8x=1,220780780x=9,7601,220x2,000x=9,760x=4.88ft

Therefore, the maximum bending moment occurs at a distance 6.88 ft from the left end A.

Show the calculation of bending moment as follows;

Bending moment at x=0;

M|x=0=0

Bending moment at x=2ft;

M|x=2ft=2,260×2250×2×22=4,520500=4,020lb-ft

Bending moment at x=6.88ft;

M|x=6.88ft=2,260×6.88250×6.88×6.882540×4.88=15,548.85,916.82,635.5=6,996.5lb-ft

Bending moment at x=10ft;

M|x=10ft=2,260×10250×10×102540×8=22,60012,5004,320=5,780lb-ft

Bending moment at x=12ft;

M|x=12ft=2,260×12250×12×122540×101,860×2=27,12018,0005,4003,720=0

Tabulate the calculated bending moment values as in Table 2.

Distance, x (ft)Bending moment, M (lb-ft)
00
24,020
6.886,996.5
105,780
120

Plot the bending moment diagram as in Figure 7.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 7.3, Problem 7.92P , additional homework tip  7

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Chapter 7 Solutions

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

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