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A parabolic three-pinned arch has a span of 20 m and central rise 4 m. It is loaded
with a uniformly distributed load of 20 kN/m for a length of 8 m from the left end
support. Find the horizontal thrust at supports.
(1) 48kN
(2) 8kKN (3) 80kN (4) 98kN?
with a uniformly distributed load of 20 kN/m for a length of 8 m from the left end
support. Find the horizontal thrust at supports.
(1) 48kN
(2) 8kKN (3) 80kN (4) 98kN?
Most Upvoted Answer
A parabolic three-pinned arch has a span of 20 m and central rise 4 m....
**Given Data:**
- Span of the arch (L) = 20 m
- Central rise (h) = 4 m
- Uniformly distributed load (w) = 20 kN/m
- Length of the loaded portion (l) = 8 m
**Assumptions:**
- The arch is symmetrical.
- The weight of the arch itself is neglected.
- The arch is hinged at the supports.
- The arch is subjected to a uniformly distributed load over the loaded portion.
**Calculating the Vertical Reaction at the Supports:**
The vertical reaction at each support can be calculated using the equation: ΣVertical Forces = 0
Let R1 and R2 be the vertical reactions at the left and right supports, respectively.
- R1 + R2 = Total vertical load on the arch
The total vertical load on the arch can be calculated by integrating the distributed load over the loaded portion. Given that the distributed load is 20 kN/m for a length of 8 m, the total vertical load can be calculated as:
Total vertical load = 20 kN/m * 8 m = 160 kN
Therefore, we have:
- R1 + R2 = 160 kN
Since the arch is symmetrical, the vertical reactions at both supports will be equal. Thus, we can rewrite the equation as:
- 2R = 160 kN
Solving for R, we get:
- R = 80 kN
Therefore, the vertical reaction at each support is 80 kN.
**Calculating the Horizontal Thrust at Supports:**
The horizontal thrust at the supports can be calculated using the equation: ΣHorizontal Forces = 0
Let H1 and H2 be the horizontal thrusts at the left and right supports, respectively.
- H1 + H2 = 0
Since the arch is hinged at the supports, there are no horizontal forces acting on the arch. Therefore, the horizontal thrust at each support is zero.
**Answer:**
The horizontal thrust at the supports is zero. Therefore, none of the given options (1), (2), (3), or (4) is correct.
- Span of the arch (L) = 20 m
- Central rise (h) = 4 m
- Uniformly distributed load (w) = 20 kN/m
- Length of the loaded portion (l) = 8 m
**Assumptions:**
- The arch is symmetrical.
- The weight of the arch itself is neglected.
- The arch is hinged at the supports.
- The arch is subjected to a uniformly distributed load over the loaded portion.
**Calculating the Vertical Reaction at the Supports:**
The vertical reaction at each support can be calculated using the equation: ΣVertical Forces = 0
Let R1 and R2 be the vertical reactions at the left and right supports, respectively.
- R1 + R2 = Total vertical load on the arch
The total vertical load on the arch can be calculated by integrating the distributed load over the loaded portion. Given that the distributed load is 20 kN/m for a length of 8 m, the total vertical load can be calculated as:
Total vertical load = 20 kN/m * 8 m = 160 kN
Therefore, we have:
- R1 + R2 = 160 kN
Since the arch is symmetrical, the vertical reactions at both supports will be equal. Thus, we can rewrite the equation as:
- 2R = 160 kN
Solving for R, we get:
- R = 80 kN
Therefore, the vertical reaction at each support is 80 kN.
**Calculating the Horizontal Thrust at Supports:**
The horizontal thrust at the supports can be calculated using the equation: ΣHorizontal Forces = 0
Let H1 and H2 be the horizontal thrusts at the left and right supports, respectively.
- H1 + H2 = 0
Since the arch is hinged at the supports, there are no horizontal forces acting on the arch. Therefore, the horizontal thrust at each support is zero.
**Answer:**
The horizontal thrust at the supports is zero. Therefore, none of the given options (1), (2), (3), or (4) is correct.
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A parabolic three-pinned arch has a span of 20 m and central rise 4 m. It is loadedwith a uniformly distributed load of 20 kN/m for a length of 8 m from the left endsupport. Find the horizontal thrust at supports.(1) 48kN(2) 8kKN (3) 80kN (4) 98kN?
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