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A 3 hinged arch of span $40m$ and rise 8m carries concentrated loads of 200 KN and 150 KN at a distances of 8 m & 16 m from the left end and an UDL of 50 KN/m on the right half of the span

Subject : Structural Analysis 1

Topic : Hinge Arches

Difficulty : Medium

1 Answer
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Support $rx^n$ :

$\sum M_A = 0 (\circlearrowright +ve)$

$200 \times 8 + 150 \times 16 + (50 \times 20) \times 30 - V_B \times 40 = 0$

$\boxed{V_B = 850 KN}$

$\sum F_Y = 0 (\uparrow + ve)$

$V_A - 200 - 150 - (50 \times 20) + 850 = 0$

$\boxed{V_A = 500 KN}$

To find horizontal thrust:

Taking $BM_c$ = 0

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consider part (CA)

$500 \times 20 - 150 \times 4 - 200 \times 12 - H \times 8 = 0$

$\boxed{H = 875 KN}$

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