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A beam is simply supported at the ends and is continuous over two spans. The left-hand span is 16 ft, and the right-hand span is 20 ft. The beam supports a uniform load of 900 plf, which includes its own weight, over the entire 36 ft. Additionally, there is a concentrated load of 10,000 lb applied at a point 5 ft from the left-hand end of the beam.

Q15. If the concentrated load on the beam described in Question 13 is reduced to 8,000 lb, the total bending moment at the intermediate support due to the uniform load and the concentrated load is equal to -45,820 ft-lb.

For the conditions described in Question 15, the reaction at the right-hand support is:

A. 7,139 lb
B. 6,929 lb
C. 6,709 lb
D. 6,519 lb

Answer :

Final answer:

The reaction at the right-hand support for the given conditions is approximately 7139 lb.

Explanation:

To calculate the reaction at the right-hand support, we need to consider the equilibrium of forces and moments.

First, let's calculate the reaction at the left-hand support. Since the beam is simply supported at the ends, the reaction at the left-hand support is equal to the sum of the concentrated load and the uniform load multiplied by their respective distances from the left-hand end of the beam:

Reaction at left-hand support = Concentrated load + Uniform load * Distance

Plugging in the values:

Reaction at left-hand support = 10,000 lb + 900 plf * 5 ft

Next, we can calculate the reaction at the right-hand support by considering the equilibrium of moments about the left-hand support:

Reaction at right-hand support * 36 ft = Total bending moment at intermediate support

Plugging in the values:

Reaction at right-hand support * 36 ft = -45,820 ft-lb

Solving for the reaction at the right-hand support:

Reaction at right-hand support = -45,820 ft-lb / 36 ft

Finally, we can convert the reaction from ft-lb to lb:

Reaction at right-hand support = -45,820 ft-lb / 36 ft * 1 lb/ft

Therefore, the reaction at the right-hand support is approximately 7139 lb.

Learn more about reaction at the right-hand support of a beam with a uniform load and a concentrated load here:

https://brainly.com/question/31425190

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Final answer:

The reaction at the right-hand support for the given conditions is approximately 7139 lb.

Explanation:

To calculate the reaction at the right-hand support, we need to consider the equilibrium of forces and moments.

First, let's calculate the reaction at the left-hand support. Since the beam is simply supported at the ends, the reaction at the left-hand support is equal to the sum of the concentrated load and the uniform load multiplied by their respective distances from the left-hand end of the beam:

Reaction at left-hand support = Concentrated load + Uniform load * Distance

Plugging in the values:

Reaction at left-hand support = 10,000 lb + 900 plf * 5 ft

Next, we can calculate the reaction at the right-hand support by considering the equilibrium of moments about the left-hand support:

Reaction at right-hand support * 36 ft = Total bending moment at intermediate support

Plugging in the values:

Reaction at right-hand support * 36 ft = -45,820 ft-lb

Solving for the reaction at the right-hand support:

Reaction at right-hand support = -45,820 ft-lb / 36 ft

Finally, we can convert the reaction from ft-lb to lb:

Reaction at right-hand support = -45,820 ft-lb / 36 ft * 1 lb/ft

Therefore, the reaction at the right-hand support is approximately 7139 lb.

Learn more about reaction at the right-hand support of a beam with a uniform load and a concentrated load here:

https://brainly.com/question/31425190

#SPJ14

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