Shear Bending Graphs 07D68F

1. **Problem Statement:** We are given the shear force and bending moment expressions for the segment $0 \leq x < a$ of a beam: $$V(x) = R_A = \frac{P(L - a)}{L}$$ $$M(x) = R_A \cdot x = \frac{P(L - a)}{L} \cdot x$$ We need to understand and sketch the graphs of $V(x)$ and $M(x)$ over this interval. 2. **Understanding the Shear Force $V(x)$:** - The shear force $V(x)$ is constant over the interval $0 \leq x < a$ because it equals $R_A$, which is a constant reaction force. - The formula $R_A = \frac{P(L - a)}{L}$ shows that $R_A$ depends on the load $P$, the beam length $L$, and the position $a$. 3. **Understanding the Bending Moment $M(x)$:** - The bending moment $M(x)$ varies linearly with $x$ since $M(x) = R_A \cdot x$. - At $x=0$, $M(0) = 0$. - At $x=a$, $M(a) = R_A \cdot a = \frac{P(L - a)}{L} \cdot a$. 4. **Sketching the Graphs:** - **Shear Force $V(x)$:** A horizontal line at height $R_A = \frac{P(L - a)}{L}$ from $x=0$ to $x=a$. - **Bending Moment $M(x)$:** A straight line starting at $(0,0)$ and increasing linearly to $(a, \frac{P(L - a)}{L} \cdot a)$. 5. **Summary:** - $V(x)$ is constant and positive (assuming $P$, $L$, and $a$ positive). - $M(x)$ is a linear function increasing from zero to $\frac{P(L - a)}{L} a$ over $0 \leq x < a$. **Final answers:** $$V(x) = \frac{P(L - a)}{L}$$ $$M(x) = \frac{P(L - a)}{L} x$$