NUMERICAL ON 3 HINGED PARABOLIC ARCH TO DETERMINE MAXIMUM BM

A three hinged parabolic arch ACB of span 20m and rise 4m carries a uniformly distributed load of 20KN/m run on the left half of the span. Find the Maximum bending moment off the arch.

Step 1:

Applying vertical equilibrium condition for the arch

V _a + V _b = 20(10)

V _a + V _b = 200KN……… (1)

Taking moments about support A

V _b (20) = 20 (10) (5)

V _b = 50KN

Substitute the value of V _b in eq (1)

Therefore, V _{a =} 150KN

Taking moments about crown C

H x 4 - V _b x 10 = 0

H = 125 KN

Step 2:

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

Taking moment about the section x-x

M _x-x = 150x- 20(x²/2) - H(y)

W k t

y = 4hx (L-x)/L²     (Since it is a parabolic Arch)

y = 4(4) (x) (20-x)/20²

y = 0.8x – 0.04x² ……. (2)

Therefore, Substitute the value of y in M _x-x Equation

M _x-x = 150x - 20(x²/2) – 125(0.8x – 0.04x²)

M _x-x = 50x – 5x² ………. (3)

In order to obtain Maximum moment, the value of x is essential. 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

50 – 10x = 0

x = 5m

Substitute the value of x in the eq 3

M _x-x = 50(5) – 5(5)²

M _x-x = 125 KN -m

Therefore Maximum Bending moment is 125 KN –m