Question

1、internal force in bending:
example knowing conditions are p, a, l, as shown in the figure. determine the internal forces on the section at the distance x to the end a.
solution: ①determine external forces

Explanation:

Step1: Analyze Equilibrium Conditions

For the beam, apply static equilibrium equations. Sum of vertical forces: $\sum F_y = 0 \implies Y_A + R_B - P = 0$. Sum of moments about $A$: $\sum M_A = 0 \implies R_B \cdot l - P \cdot a = 0 \implies R_B = \frac{P \cdot a}{l}$. Then $Y_A = P - \frac{P \cdot a}{l} = \frac{P(l - a)}{l}$. Horizontal force: $X_A = 0$ (no horizontal loads).

Step2: Cut the Beam at Distance \( x \)

Consider the left segment (from \( A \) to the section at \( x \)). For vertical shear force \( V \): sum of vertical forces on left segment: $V = Y_A$ (if \( x \leq a \)) or $V = Y_A - P$ (if \( x > a \)). For bending moment \( M \): moment about the section: $M = Y_A \cdot x$ (if \( x \leq a \)) or $M = Y_A \cdot x - P \cdot (x - a)$ (if \( x > a \)).

Answer:

External forces: \( X_A = 0 \), \( Y_A = \frac{P(l - a)}{l} \), \( R_B = \frac{P a}{l} \). Internal forces (shear \( V \) and bending moment \( M \)) depend on \( x \):

• If \( x \leq a \): \( V = \frac{P(l - a)}{l} \), \( M = \frac{P(l - a)}{l}x \)
• If \( x > a \): \( V = \frac{P(l - a)}{l}-P = -\frac{P a}{l} \), \( M = \frac{P(l - a)}{l}x - P(x - a) = \frac{P a}{l}(l - x) \)