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A beam is subjected to the uniform load WcW_cWc​ per unit length and MA,VAM_A, V_A MA​,VA​ & VBV_BVB​ are the reactions as sh
Question

A beam is subjected to the uniform load WcW_c per unit length and MA,VAM_A, V_A  & VBV_B are the reactions as shown in the figure. The expression for shear force V(x) and Bending moment M(x) is given by 

A.

V(x)=VA+Wcx,M(x)=MA+VAxWcx22V(x) = V_A + W_c x, M(x) = M_A + V_A x- \frac{W_c x^2}{2} ​​

B.

V(x)=VAWcx,M(x)=MAVAxWcx22V(x) = V_A - W_c x, M(x) = M_A - V_A x -\frac{W_c x^2}{2} ​​

C.

V(x)=VA+Wcx,M(x)=MAVAx+Wcx22V(x) = V_A + W_c x, M(x) = M_A - V_A x +\frac{W_c x^2}{2} ​​

D.

V(x)=VAWcx,M(x)=MA+VAxWcx22V(x) = V_A - W_c x, M(x) = M_A + V_A x -\frac{W_c x^2}{2} ​​

Correct option is A

Let:x=distance from the left end (point A)VA=shear force at point AMA=moment at point AShear Force Expression V(x):For a UDL, shear force linearly decreases along the length:V(x)=VA+wcxBending Moment Expression M(x):The bending moment is obtained by integrating the shear force:M(x)=MA+0xV(x) dx=MA+0x(VAwcx) dx=>M(x)=MA+VAxwcx22\text{Let:} \\\quad x = \text{distance from the left end (point A)} \\\quad V_A = \text{shear force at point A} \\\quad M_A = \text{moment at point A}{ \quad \textbf{Shear Force Expression } V(x):} \\\text{For a UDL, shear force linearly decreases along the length:} \\\boxed{V(x) = V_A + w_c x}\\{ \quad \textbf{Bending Moment Expression } M(x):} \\\text{The bending moment is obtained by integrating the shear force:} \\M(x) = M_A + \int_0^x V(x) \, dx = M_A + \int_0^x (V_A - w_c x) \, dx \\\Rightarrow M(x) = M_A + V_A x - \frac{w_c x^2}{2}​​

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