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

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  1. \(\rm V(x)=V_{A}+W_{e}x, M(x)=M_{A}+V_{A} x- \frac{W_{e} x^{2}}{2} \)
  2. \(\rm V(x)=V_{A}-W_{e} x, M(x)=M_{A}-V_{A} x- \frac{W_{e} x^{2}}{2} \)
  3. \(\rm V(x)=V_{A}+W_{e} x, M(x)=M_{A}-V_{A} x- \frac{W_{e} x^{2}}{2} \)
  4. \(\rm V(x)=V_{A}-W_{e} x,~ M(x)=M_{A}+V_{A} x- \frac{W_{e} x^{2}}{2}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm V(x)=V_{A}-W_{e} x,~ M(x)=M_{A}+V_{A} x- \frac{W_{e} x^{2}}{2}\)
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Detailed Solution

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Concept:

The beam is subjected to a uniformly distributed load (UDL) of intensity \(W_e\) N/m over its length.

Let \(x\) be the distance from the fixed support A.

Shear Force Calculation:

Consider a section at distance \(x\) from A.

Upward force due to support A = \(V_A\)

Downward load from 0 to x = \(W_e \cdot x\)

So, shear force at distance x is:

\(V(x) = V_A - W_e \cdot x\)

Bending Moment Calculation:

Bending moment due to:

  • \(M_A\) — fixed moment at A (positive)
  • \(V_A \cdot x\) — moment due to vertical force at A
  • \(\frac{W_e x^2}{2}\) — moment of the distributed load acting at center (x/2)

So, bending moment at distance x is:

\(M(x) = M_A + V_A x - \frac{W_e x^2}{2}\)

Final Expressions:

\(V(x) = V_A - W_e x\)

\(M(x) = M_A + V_A x - \frac{W_e x^2}{2}\)

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