A two hinged parabolic arch of span 20 meter and rise 4 meter uniformly distributed load of 40 kN/m on right half span find the reaction at the supports and draw BMD.

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written 7.9 years ago by

teamques10 ★ 69k

modified 2.9 years ago by

pedsangini276 • 4.8k

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Mumbai University > Civil Engineering > Sem 5 > Structural Analysis – II

Marks: 8 Marks

Year: May 2015

structural analysis

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1 Answer

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written 7.9 years ago by

teamques10 ★ 69k

ΣMA = 0

401015 – VB*20 = 0

VB = 300 KN

ΣFy = 0

VA - 40*10 + 300 = 0

VA = 100 KN

H = ∫ { ( My dx / EI ) / ( ∫ y² dx / EI ) }

Numerator = ∫ My dx / EI

= ∫0 – 10 [ 100X * ( 4*4X * (20 – X ) / 20² ) ] dx / EI + ∫0-10 (300X – 40X² / 2) * (4 * 4X * ( 20 – X ) / 20 ²) dx / EI

= ∫0-10 ( 100 X ) (0.8X – 0.04X²) dx / EI + ∫0-10 ( 300 X – 20 X²)

( 0.8X – 0.04 X² ) dx / EI

= 16666.67 + 26000

=42666.67

Denominator = ∫ y² dx

= ∫ 0-20 ( 4 * 4X * ( 20 – X ) / 20² )² dx / EI

= ∫0-20 ( 0.8X - 0.04 X² )

= 170.67

H = Numerator / Denominator = 42.66 * 10³ / 170.67

H = 250 KN

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