Question

example 4. a simple beam ab subjected to a concentrated load p at point c. solution: (1)the reactions at the supports $r_a = \frac{b}{l}p$, $r_b = \frac{a}{l}p$

Explanation:

Step1: Identify the problem type

This is a problem about beam mechanics, specifically finding support reactions for a simply supported beam with a concentrated load. It falls under the subfield of Engineering (within Natural Science). The key principles here are static equilibrium (sum of forces and moments equal to zero).

Step2: Apply static equilibrium for vertical forces

The sum of vertical forces should be zero: \( R_A + R_B - P = 0 \), so \( R_A + R_B = P \).

Step3: Apply moment equilibrium about point A

Taking moments about \( A \), the moment due to \( R_A \) is zero (distance from \( A \) is zero), the moment due to \( P \) is \( -P \times a \) (clockwise), and the moment due to \( R_B \) is \( +R_B \times l \) (counter - clockwise). So, \( \sum M_A=0\Rightarrow R_B\times l - P\times a = 0\Rightarrow R_B=\frac{a}{l}P \).

Step4: Solve for \( R_A \)

Using the force equilibrium equation \( R_A+R_B = P \) and substituting \( R_B=\frac{a}{l}P \), we get \( R_A=P - R_B=P-\frac{a}{l}P=\frac{l - a}{l}P \). But since \( l=a + b \), \( l - a=b \), so \( R_A=\frac{b}{l}P \).

Answer:

The reactions at the supports are \( R_A=\boldsymbol{\frac{b}{l}P} \) and \( R_B=\boldsymbol{\frac{a}{l}P} \)