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 \( M = F\times d \). The reaction \( R_B \) creates a counter - clockwise moment (positive) about \( A \) and the loads create clockwise moments (negative) about \( A \).

• The distance of \( 120\space kN \) from \( A \) is \( 2\space m \).
• The distance of \( 70\space kN \) from \( A \) is \( 2 + 3=5\space m \).
• The distance of \( 30\space kN \) from \( A \) is \( 2+3 + 2 = 7\space m \).
• The distance of \( R_B \) from \( A \) is \( 8\space m \).

Using the moment equilibrium equation \(\sum M_A=0\):
\[
R_B\times8-120\times2 - 70\times5-30\times7 = 0
\]
First, calculate the products:

• \( 120\times2 = 240\)
• \( 70\times5=350\)
• \( 30\times7 = 210\)

Then,
\[
8R_B-240 - 350-210=0
\]
\[
8R_B-(240 + 350+210)=0
\]
\[
8R_B - 800=0
\]
\[
8R_B=800
\]
\[
R_B=\frac{800}{8}=100\space kN
\]

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

From \( R_A+R_B = 220 \), substitute \( R_B = 100 \):
\[
R_A+100=220
\]
\[
R_A=220 - 100=120\space kN
\]

Final Answers

The reaction at \( A \) (\( R_A \)) is \(\boldsymbol{120\space kN}\) (upward) and the reaction at \( B \) (\( R_B \)) 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 \( M = F\times d \). The reaction \( R_B \) creates a counter - clockwise moment (positive) about \( A \) and the loads create clockwise moments (negative) about \( A \).

• The distance of \( 120\space kN \) from \( A \) is \( 2\space m \).
• The distance of \( 70\space kN \) from \( A \) is \( 2 + 3=5\space m \).
• The distance of \( 30\space kN \) from \( A \) is \( 2+3 + 2 = 7\space m \).
• The distance of \( R_B \) from \( A \) is \( 8\space m \).

Using the moment equilibrium equation \(\sum M_A=0\):
\[
R_B\times8-120\times2 - 70\times5-30\times7 = 0
\]
First, calculate the products:

• \( 120\times2 = 240\)
• \( 70\times5=350\)
• \( 30\times7 = 210\)

Then,
\[
8R_B-240 - 350-210=0
\]
\[
8R_B-(240 + 350+210)=0
\]
\[
8R_B - 800=0
\]
\[
8R_B=800
\]
\[
R_B=\frac{800}{8}=100\space kN
\]

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

From \( R_A+R_B = 220 \), substitute \( R_B = 100 \):
\[
R_A+100=220
\]
\[
R_A=220 - 100=120\space kN
\]

Final Answers

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