Question
Download Solution PDFA 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
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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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