Question

A uniform beam (EI=constant ) PQ in the form of a quarter-circle of radius R is fixed at end P and free at the end Q, where a load W is applied as shown. The vertical downward displacement, δq, at the loaded point Q is given by: δq=β(wR3EI) Find the value of β (correct to 4-decimal places). 0.7854

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Solution

The correct option is A 0.7854δB=β(WR3EI) We need to find β Mxx=−WRsinθ ∂Mxx∂w=−Rsinθ ∂U∂w=∫Mxx(∂Mxx/∂w)dxEI =∫π/20(wR2sin2θ)RdθEI ⇒∂U∂w=wR32EI∫π/202sin2θdθ =wR32EI∫π/20(1−cos2θ)dθ ⇒∂U∂w=δB=π4(wR3EI) Thus β=π4=0.7854

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