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A symmetrical parabolic arch of span 20 metres and rise 5 m is hinged at its springings and crowns and is subjected to uniformly distributed load of 100 kN/m throughout the span. The horizontal thrust at springings will be
- a)25 kN
- b)400 kN
- c)1000 kN
- d)15.625 kN
Correct answer is option 'C'. Can you explain this answer?
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A symmetrical parabolic arch of span 20 metres and rise 5 m is hinged ...
Given data:
Span of the arch (l) = 20 m
Rise of the arch (h) = 5 m
Uniformly distributed load (w) = 100 kN/m
To find: Horizontal thrust at springings
Solution:
1. The shape of the arch is symmetrical, which means that the horizontal thrust at the crown will be zero.
2. Let's consider a section of the arch at a distance x from one end. The load on this section will be wx, and the vertical reaction at this section will be y. Since the arch is in equilibrium, the horizontal component of the reaction (y') will balance the horizontal component of the load (wx'). Therefore, we can write:
y' = wx'/h
3. Integrating both sides of this equation from x = 0 to x = l/2, we get:
∫[0,l/2] y' dx = ∫[0,l/2] (w/h) x' dx
y(l/2) = (w/h) (l/2)^2
4. Similarly, we can write the equation for the other half of the arch:
y(l/2) = (w/h) (l/2)^2
5. The total horizontal thrust at the springings is the sum of the horizontal components of the reactions at the two ends. Since the horizontal component at one end is equal and opposite to the horizontal component at the other end, we can write:
Horizontal thrust = 2y' = 2(wx'/h)
6. Substituting the value of x' from the previous equation, we get:
Horizontal thrust = wl^2/8h
7. Substituting the given values, we get:
Horizontal thrust = 100 × 20^2/8 × 5
Horizontal thrust = 1000 kN
Therefore, the correct option is (c) 1000 kN.
Span of the arch (l) = 20 m
Rise of the arch (h) = 5 m
Uniformly distributed load (w) = 100 kN/m
To find: Horizontal thrust at springings
Solution:
1. The shape of the arch is symmetrical, which means that the horizontal thrust at the crown will be zero.
2. Let's consider a section of the arch at a distance x from one end. The load on this section will be wx, and the vertical reaction at this section will be y. Since the arch is in equilibrium, the horizontal component of the reaction (y') will balance the horizontal component of the load (wx'). Therefore, we can write:
y' = wx'/h
3. Integrating both sides of this equation from x = 0 to x = l/2, we get:
∫[0,l/2] y' dx = ∫[0,l/2] (w/h) x' dx
y(l/2) = (w/h) (l/2)^2
4. Similarly, we can write the equation for the other half of the arch:
y(l/2) = (w/h) (l/2)^2
5. The total horizontal thrust at the springings is the sum of the horizontal components of the reactions at the two ends. Since the horizontal component at one end is equal and opposite to the horizontal component at the other end, we can write:
Horizontal thrust = 2y' = 2(wx'/h)
6. Substituting the value of x' from the previous equation, we get:
Horizontal thrust = wl^2/8h
7. Substituting the given values, we get:
Horizontal thrust = 100 × 20^2/8 × 5
Horizontal thrust = 1000 kN
Therefore, the correct option is (c) 1000 kN.
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A symmetrical parabolic arch of span 20 metres and rise 5 m is hinged at its springings and crowns and is subjected to uniformly distributed load of 100 kN/m throughout the span. The horizontal thrust at springings will bea)25 kNb)400 kNc)1000 kNd)15.625 kNCorrect answer is option 'C'. Can you explain this answer?
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