Question

figure 1
120 kn
70 kn
30kn
a
b
2 m
3 m
2 m
1 m
8 m

Explanation:

To solve for the reactions at supports \( A \) (pin support) and \( B \) (roller support) of the simply - supported beam, we use the principles of static equilibrium. The two main equilibrium equations for a beam in static equilibrium are:

1. Sum of vertical forces (\(\sum F_y = 0\))

Let \( R_A \) be the vertical reaction at \( A \) and \( R_B \) be the vertical reaction at \( B \). The downward forces are \( 120\space kN \), \( 70\space kN \) and \( 30\space kN \). So,
\[
R_A+R_B - 120 - 70 - 30=0
\]
\[
R_A + R_B=220\space kN\quad(1)
\]

2. Sum of moments about a point (\(\sum M = 0\)). We take moments about point \( A \)

The moment due to a force \( F \) at a distance \( d \) from \( A \) is \( M = F\times d \) (clock - wise moments are positive and anti - clockwise moments are negative, or vice - versa. Here we take clock - wise moments as positive for simplicity).

• The moment due to \( 120\space kN \) about \( A \): The distance of \( 120\space kN \) from \( A \) is \( 2\space m \), so \( M_{120}=120\times2 \)
• The moment due to \( 70\space kN \) about \( A \): The distance of \( 70\space kN \) from \( A \) is \( 2 + 3=5\space m \), so \( M_{70}=70\times5 \)
• The moment due to \( 30\space kN \) about \( A \): The distance of \( 30\space kN \) from \( A \) is \( 2+3 + 2=7\space m \), so \( M_{30}=30\times7 \)
• The moment due to \( R_B \) about \( A \): The distance of \( R_B \) from \( A \) is \( 8\space m \), so \( M_{R_B}=R_B\times8 \)
• The moment due to \( R_A \) about \( A \) is \( 0 \) (since the distance from \( A \) to \( R_A \) is \( 0 \))

Using \(\sum M_A = 0\) (taking clock - wise moments as positive):
\[
120\times2+70\times5 + 30\times7-R_B\times8=0
\]
First, calculate the left - hand side terms:
\[
120\times2 = 240
\]
\[
70\times5=350
\]
\[
30\times7 = 210
\]
Sum of these terms: \( 240 + 350+210=800 \)
So,
\[
800-8R_B = 0
\]
\[
8R_B=800
\]
\[
R_B=\frac{800}{8}=100\space kN
\]

3. Substitute \( R_B \) into equation (1) to find \( R_A \)

From equation (1): \( R_A+R_B = 220 \)
Substitute \( R_B = 100\space kN \)
\[
R_A+100 = 220
\]
\[
R_A=220 - 100=120\space kN
\]

Final Answers

The reaction at support \( A \) (\( R_A \)) is \(\boldsymbol{120\space kN}\) (upward) and the reaction at support \( B \) (\( R_B \)) is \(\boldsymbol{100\space kN}\) (upward)

Answer:

To solve for the reactions at supports \( A \) (pin support) and \( B \) (roller support) of the simply - supported beam, we use the principles of static equilibrium. The two main equilibrium equations for a beam in static equilibrium are:

1. Sum of vertical forces (\(\sum F_y = 0\))

Let \( R_A \) be the vertical reaction at \( A \) and \( R_B \) be the vertical reaction at \( B \). The downward forces are \( 120\space kN \), \( 70\space kN \) and \( 30\space kN \). So,
\[
R_A+R_B - 120 - 70 - 30=0
\]
\[
R_A + R_B=220\space kN\quad(1)
\]

2. Sum of moments about a point (\(\sum M = 0\)). We take moments about point \( A \)

The moment due to a force \( F \) at a distance \( d \) from \( A \) is \( M = F\times d \) (clock - wise moments are positive and anti - clockwise moments are negative, or vice - versa. Here we take clock - wise moments as positive for simplicity).

• The moment due to \( 120\space kN \) about \( A \): The distance of \( 120\space kN \) from \( A \) is \( 2\space m \), so \( M_{120}=120\times2 \)
• The moment due to \( 70\space kN \) about \( A \): The distance of \( 70\space kN \) from \( A \) is \( 2 + 3=5\space m \), so \( M_{70}=70\times5 \)
• The moment due to \( 30\space kN \) about \( A \): The distance of \( 30\space kN \) from \( A \) is \( 2+3 + 2=7\space m \), so \( M_{30}=30\times7 \)
• The moment due to \( R_B \) about \( A \): The distance of \( R_B \) from \( A \) is \( 8\space m \), so \( M_{R_B}=R_B\times8 \)
• The moment due to \( R_A \) about \( A \) is \( 0 \) (since the distance from \( A \) to \( R_A \) is \( 0 \))

Using \(\sum M_A = 0\) (taking clock - wise moments as positive):
\[
120\times2+70\times5 + 30\times7-R_B\times8=0
\]
First, calculate the left - hand side terms:
\[
120\times2 = 240
\]
\[
70\times5=350
\]
\[
30\times7 = 210
\]
Sum of these terms: \( 240 + 350+210=800 \)
So,
\[
800-8R_B = 0
\]
\[
8R_B=800
\]
\[
R_B=\frac{800}{8}=100\space kN
\]

3. Substitute \( R_B \) into equation (1) to find \( R_A \)

From equation (1): \( R_A+R_B = 220 \)
Substitute \( R_B = 100\space kN \)
\[
R_A+100 = 220
\]
\[
R_A=220 - 100=120\space kN
\]

Final Answers

The reaction at support \( A \) (\( R_A \)) is \(\boldsymbol{120\space kN}\) (upward) and the reaction at support \( B \) (\( R_B \)) is \(\boldsymbol{100\space kN}\) (upward)