Find the reactions at supports and draw BMD for a two hinged parabolic arch loaded as shown in figure.

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written 7.8 years ago by

teamques10 ★ 68k

ΣMA = 0

VB * 40 – 40 * 20 * 30 – 20 * 10 = 0

VB = 605 KN

ΣV = 0

VA +VB = 40 * 20 + 20

mVA + VB = 820

VA = 215 KN

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

Numerator = ∫ My dx / EI

= ∫10 – 20 (215x – 20x * y / 2 ) dx / EI + ∫0-10 (605x * y) dx / EI
+ ∫10-20 ( 605x – 40 (x – 10 )² * y / 2 ) dx / EI

= 143 * 10³ + 82 * 10³ + 406 * 10³

= 631 * 10³

Y = 4hx ( L – x ) / L² = 4 * 5 * x ( 40 – x ) / 40²

= 0.5x – 0.0125x²

Denominator = 2 ∫ y² dx = 2 ∫ 0-20 ( 0.5x – 0.0125x² )² = 533.33

H = Numerator / Denominator = 631 * 10³ / 533.33 = 1183.13

Mx at LHS = VA * x – H * y – 20 * ( x-10 )

= 215x – 1183.13 * ( 0.5x – 0.0125x² ) – 20x + 200

= 215 – 591.56 + 14.78 * 2x + 29.57x - 20 + 200

29.57x = 196.56

X = 6.65 m

Mx at RHS = VB * x – H * y – 40x² / 2

= 605x – 1183.13 * (0.5x – 0.0125x²) – 40 * x² / 2

= 605 – 591.56 + 29.57x – 20 * 2x

10.43x = 13.44

X = 1.28 m

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