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AB is a simply supported beam of span 10 m. A uniformly distributed moving load of length
5 m occupies the beam in such a way that its. one end is at support A and the other end in
the span of the beam. The support reaction at A is
ANS 75 kN upward?
?
5 m occupies the beam in such a way that its. one end is at support A and the other end in
the span of the beam. The support reaction at A is
ANS 75 kN upward?
?
Most Upvoted Answer
AB is a simply supported beam of span 10 m. A uniformly distributed mo...
Calculation of support reaction at A:
Given data:
- Span of the beam (L) = 10 m
- Length of the uniformly distributed load (l) = 5 m
Analysis:
To calculate the support reaction at A, we need to consider the moment equilibrium and vertical force equilibrium at support A.
Moment equilibrium:
Taking moments about support A, we can analyze the clockwise and anticlockwise moments acting on the beam.
- Clockwise moments:
- Reaction at A (Ra) = 0 (since it is acting vertically)
- Uniformly distributed load (w) = (w * l * l/2) (acting at the midpoint of the load)
- Anticlockwise moments:
- Reaction at A (Ra) = Ra * L/2 (acting at the midpoint of the beam)
Setting the clockwise moments equal to the anticlockwise moments, we have:
0 + (w * l * l/2) = Ra * L/2
Simplifying the equation, we get:
(w * l * l/2) = Ra * L/2
Vertical force equilibrium:
Taking vertical forces acting on the beam into consideration, we can analyze the upward and downward forces.
- Upward forces:
- Reaction at A (Ra) = Ra
- Downward forces:
- Uniformly distributed load (w) = w * l
Setting the upward forces equal to the downward forces, we have:
Ra = w * l
Solution:
Substituting the value of Ra from the vertical force equilibrium equation into the moment equilibrium equation, we have:
(w * l * l/2) = (w * l) * L/2
Simplifying the equation, we get:
l/2 = L/2
5/2 = 10/2
Ra = 75 kN (upward)
Therefore, the support reaction at A is 75 kN upward.
Conclusion:
The support reaction at A for the simply supported beam with a uniformly distributed moving load occupying a length of 5 m is 75 kN upward.
Given data:
- Span of the beam (L) = 10 m
- Length of the uniformly distributed load (l) = 5 m
Analysis:
To calculate the support reaction at A, we need to consider the moment equilibrium and vertical force equilibrium at support A.
Moment equilibrium:
Taking moments about support A, we can analyze the clockwise and anticlockwise moments acting on the beam.
- Clockwise moments:
- Reaction at A (Ra) = 0 (since it is acting vertically)
- Uniformly distributed load (w) = (w * l * l/2) (acting at the midpoint of the load)
- Anticlockwise moments:
- Reaction at A (Ra) = Ra * L/2 (acting at the midpoint of the beam)
Setting the clockwise moments equal to the anticlockwise moments, we have:
0 + (w * l * l/2) = Ra * L/2
Simplifying the equation, we get:
(w * l * l/2) = Ra * L/2
Vertical force equilibrium:
Taking vertical forces acting on the beam into consideration, we can analyze the upward and downward forces.
- Upward forces:
- Reaction at A (Ra) = Ra
- Downward forces:
- Uniformly distributed load (w) = w * l
Setting the upward forces equal to the downward forces, we have:
Ra = w * l
Solution:
Substituting the value of Ra from the vertical force equilibrium equation into the moment equilibrium equation, we have:
(w * l * l/2) = (w * l) * L/2
Simplifying the equation, we get:
l/2 = L/2
5/2 = 10/2
Ra = 75 kN (upward)
Therefore, the support reaction at A is 75 kN upward.
Conclusion:
The support reaction at A for the simply supported beam with a uniformly distributed moving load occupying a length of 5 m is 75 kN upward.
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AB is a simply supported beam of span 10 m. A uniformly distributed moving load of length5 m occupies the beam in such a way that its. one end is at support A and the other end inthe span of the beam. The support reaction at A isANS 75 kN upward??
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