Applied Statics and Strength of Materials (6th Edition)
Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Chapter 13, Problem 13.1P

through 13.6 Calculate the reactions at points A and BFor the beams shown.

Chapter 13, Problem 13.1P, through 13.6 Calculate the reactions at points A and BFor the beams shown.

a.

Expert Solution
Check Mark
To determine

The reaction forces at points A and B

Answer to Problem 13.1P

  RA=72 kipupward

  RB=72 kipupward

Explanation of Solution

Given:

The beam diagram and loading are given as shown below:

  Applied Statics and Strength of Materials (6th Edition), Chapter 13, Problem 13.1P , additional homework tip  1

Concept Used:

Replacing the uniformly distribute load by a concentrated external load W acting through the midpoint of the beam length.

  Applied Statics and Strength of Materials (6th Edition), Chapter 13, Problem 13.1P , additional homework tip  2

Calculation:

  W=6 k/ft.×24 ft.=144 kip

Taking anticlockwise moment positive, the summation of all moment of forces at point A equals to zero

  RB×24W×12=0RB×24144×12=0RB=144×1224 kip=72 kip

For vertical equilibrium, the summation of forces in vertical direction must be zero

  FV=0RA+RBW=0RA=WRBRA=14472=72 kip

Conclusion:

Therefore, the reaction force at points A, RA=72 kip upward and the reaction force at points B, RB=72 kip upward.

b.

Expert Solution
Check Mark
To determine

The reaction forces at points A and B

Answer to Problem 13.1P

  RA=13.125 k

  RB=6.875 k

Explanation of Solution

Given:

The beam diagram and loading are given as shown below

  Applied Statics and Strength of Materials (6th Edition), Chapter 13, Problem 13.1P , additional homework tip  3

Concept Used:

The reaction forces at point A and B are shown below in the figure and assumed to act in the upward (positive) direction

  Applied Statics and Strength of Materials (6th Edition), Chapter 13, Problem 13.1P , additional homework tip  4

Calculation:

Taking anticlockwise moment positive, the summation of all moment of forces at point A equals to zero

  RB×1610×310×8=0RB×16=30+80RB=11016 k=6.875 k

For vertical equilibrium, the summation of forces in vertical direction must be zero

  FV=0RA+RB=10 k+10 kRA+6.875=10 k+10 kRA=(206.875) k=13.125 k

Conclusion:

Therefore, the forces are RA=13.125 k and RB=6.875 k .

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Chapter 13 Solutions

Applied Statics and Strength of Materials (6th Edition)

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