The maximum bending moment due to an isolated load in a three-hinged parabolic arch (span l, rise h) having one of its hinges at the crown, occurs on either side of the crown at a distance
A
L/(2√3)
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B
L/4
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C
h/4
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D
L/(3√2)
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Solution
The correct option is AL/(2√3) Given :
Let us assume as isolated load W acting at distance x from end A.
Since there is no horizontal external force so horizontal reactions at end A and B will beHA=HB=H.
Taking moment about (A), we have
wx−VB×2=0
⇒VB=WxL
∵TakingbendingmomentatC:
Wehave(BM)C=0
⇒VB×L2−H×h=0
⇒H=VBL2h=Wx2h
Now,(BM)D=VB(L−x)−H×y
Since,y=4hxL2(L−x)(Equationofparabolicarch)
(BM)D=WxL(L−x)−H×4hxL2(L−x)
⇒(BM)D=Wx(L−x)L−Wx2h×4hxL2(L−x)
⇒(BM)D=Wx(L−x)L−2Wx2L2(L−x)
∴Forabsolutemaximumbendingmoment d(BM)Ddx=0
⇒d(PWx(L−x)2)dx−d(2Wx2L2(L−x))dx=0
⇒WL(L−2x)−2WL2(2Lx−3x2)=0
⇒(L−2x)−2L(2Lx−3x2)=0
⇒6x2−6xL+L2=0
⇒x=(6±2√312)L
x = 0.78867L, 0.2113l
So, distance of maximum moments for given load condition either side of crown.