NUMERICAL ON 3 HINGED PARABOLIC ARCH WITH SUPPORTS AT DIFFERENT LEVELS

NUMERICAL ON 3 HINGED PARABOLIC ARCH WITH SUPPORTS AT DIFFERENT LEVELS

A three hinged parabolic arch ACB is hinged at supports A and B with rise of 3m and 6.75 m respectively from crown. Span of the arch is 22.5 m .The arch carries a uniformly distributed load of 30KN/m from A to C. Determine the maximum positive and negative bending moments.

Step 1:

Determining the values of L₁ and L₂

By the property of parabola

(L₁)²/ (h₁) = (L₂)²/ (h₂)

L₁ = (L₁ + L₂) (√ (h₁) / (√ (h₁) + √ (h₂))

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L₁ + L₂ = L

Therefore,

L₁ = (L) (√ (h₁) / (√ (h₁) + √ (h₂))

L₁ = (22.5) (√ (3) / (√ (3) + √ (6.75))

L₁ = 9m

L₂ = L- L₁

L₂ = 13.5m

Step 2:

Applying vertical equilibrium condition for the arch

V _a + V _b = 30(9) = 270KN….. (1)

Taking moment from left support about C

V _a (L₁) –H (h₁) -30(9) (4.5) = 0

V _a (9) –H (3) -30(9) (4.5) = 0….. (2)

Solving Eq 2

V _a = 0.33H + 135

Taking moment from right support about C

V _b(13.5) - H (h₂) = 0

V _b(13.5) - H (6.75) = 0….. (3)

Solving Eq 3

V _b = 0.5H

Substitute values of V _a and V _b in Eq 1

V _a + V _b = 270KN

0.33H + 135 +0.5H =270….. (4)

By solving the above equation (4) , H = 162 KN

Therefore,

V _a = 189KN

V _b = 81KN

Step 3: Determination of Maximum positive and Negative BM

Consider the section X-X at the Horizontal distance x from support A and vertical distance y from the arch rib

M _x-x = V _a (x) - 30(x) ²/2 –H (y)

y = 4hx (2 L₁-x)/ 2L₁²     (In case of arch with supports at different levele L is twice the span of individual side)

y = 4(3) x (18 - x)/ 18²   

M _x-x = 189 (x) - 30(x) ²/2 –162 (4(3) x (18 - x)/ 18²))

Simplifying the above equation

M _x-x = 81x – 9x²

Therefore differentiating the M _x-x wrt x and equating to zero to determine the value of x

(d M _x-x) / (d x) = 0

81 – 18x = 0

x = 4.5m

M _x-x = 81(4.5) – 9(4.5)²

M _x-x = 182.25 KN-m

Consider the section X-X at the Horizontal distance x from support B and vertical distance y from the arch rib

M _x-x = V _b (x) –H (y)

y = 4hx (2 L₂-x)/ 2L₂²    

y = 4(3) x (27 - x)/ 27²   

M _x-x = 81 (x) –162 (4(3) x (27 - x)/ 27² )

M _x-x = -81x + 6x²

Therefore differentiating the M _x-x wrt x and equating to zero to determine the value of x

(d M _x-x) / (d x) = 0

-81 + 12x = 0

x = 6.75m

M _x-x = -81(6.75) + 6(6.75)²

M _x-x = 273.38KN-m

Therefore, Max positive moment = 182.25 KN-m

Max negative moment = 273.38KN-m