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

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

Consider the left segment (from \( A \) to \( x \)). For \( x \leq a \):

• Shear force \( V(x) \): Sum of vertical forces on left segment: \( V(x) = Y_A = \frac{P(l - a)}{l} \).
• Bending moment \( M(x) \): Moment about the cut section: \( M(x) = Y_A \cdot x = \frac{P(l - a)x}{l} \).

For \( x > a \):

• Shear force \( V(x) \): Sum of vertical forces on left segment: \( V(x) = Y_A - P = \frac{P(l - a)}{l} - P = -\frac{P a}{l} \).
• Bending moment \( M(x) \): Moment about the cut section: \( M(x) = Y_A \cdot x - P \cdot (x - a) = \frac{P(l - a)x}{l} - P(x - a) = \frac{P a (l - x)}{l} \).

Answer:

For \( x \leq a \): Shear force \( V(x) = \frac{P(l - a)}{l} \), Bending moment \( M(x) = \frac{P(l - a)x}{l} \);
For \( x > a \): Shear force \( V(x) = -\frac{P a}{l} \), Bending moment \( M(x) = \frac{P a (l - x)}{l} \);
Horizontal force \( X_A = 0 \).