Question

1. a simple beam of 8 meters in length is loaded with concentrated loads as shown. if the weight of the beam is neglected, plot the shear and moment diagram of the figure. figure 1

120 kn 70 kn 30kn
a ▲ ● b
2 m 3 m 2 m 1 m
8 m

Explanation:

Step 1: Find Reactions at Supports

Let \( R_A \) be the reaction at \( A \) (pin support) and \( R_B \) be the reaction at \( B \) (roller support). Sum of vertical forces: \( R_A + R_B = 120 + 70 + 30 = 220\space \text{kN} \). Sum of moments about \( A \): \( R_B \times 8 = 120 \times 2 + 70 \times (2 + 3) + 30 \times (2 + 3 + 2) \). Calculate right - hand side: \( 120\times2=240 \), \( 70\times5 = 350 \), \( 30\times7=210 \). Sum: \( 240 + 350+210=800 \). Then \( R_B=\frac{800}{8}=100\space \text{kN} \). Then \( R_A=220 - 100 = 120\space \text{kN} \).

Step 2: Shear Force Calculation (SF)

- **From \( A \) to 2m (before 120kN load):** \( SF = R_A=120\space \text{kN} \) (constant).
- **At 2m (after 120kN load):** \( SF = 120 - 120 = 0\space \text{kN} \).
- **From 2m to 2 + 3=5m (before 70kN load):** \( SF = 0\space \text{kN} \) (constant).
- **At 5m (after 70kN load):** \( SF = 0 - 70=- 70\space \text{kN} \).
- **From 5m to 5 + 2 = 7m (before 30kN load):** \( SF=-70\space \text{kN} \) (constant).
- **At 7m (after 30kN load):** \( SF=-70 - 30=-100\space \text{kN} \).
- **From 7m to 8m (to support \( B \)):** \( SF=-100\space \text{kN} \), and at \( B \), \( SF=- 100+R_B=-100 + 100 = 0\space \text{kN} \).

Step 3: Bending Moment Calculation (BM)

- **At \( A \) (\( x = 0 \)):** \( BM = 0\space \text{kNm} \).
- **At 2m (under 120kN load):** \( BM=R_A\times2=120\times2 = 240\space \text{kNm} \).
- **At 5m (under 70kN load):** \( BM=R_A\times5-120\times(5 - 2)=120\times5-120\times3=600 - 360 = 240\space \text{kNm} \) (or \( BM = R_B\times(8 - 5)-30\times(8 - 7)=100\times3-30\times1 = 300 - 30 = 270\space \text{kNm} \)? Wait, no, let's recalculate. Wait, from \( A \): \( BM=120\times5-120\times3 - 70\times0\) (wait, the distance from 120kN load to 5m is \( 5 - 2=3m \), and the 70kN load is at 5m. So \( BM = 120\times5-120\times3=240\space \text{kNm} \). Wait, maybe my first approach was wrong. Let's use the formula \( BM=\sum(\text{load}\times\text{distance from section to load})\) from the left. At \( x = 5m \): \( BM = 120\times5-120\times(5 - 2)=120\times5 - 120\times3=600 - 360 = 240\space \text{kNm} \). At \( x = 7m \): \( BM=120\times7-120\times(7 - 2)-70\times(7 - 5)=120\times7-120\times5-70\times2=840 - 600 - 140 = 100\space \text{kNm} \). At \( B \) (\( x = 8m \)): \( BM = 0\space \text{kNm} \) (since it's a roller support, moment at \( B \) is zero. Let's check with right - hand side: \( BM=30\times1+70\times(1 + 2)+120\times(1 + 2+3)-R_B\times8 \)? No, better to use left - hand side: \( BM=120\times8-120\times6-70\times3 - 30\times1=960-720 - 210 - 30=0 \), which matches.

Step 4: Plotting Shear Force Diagram (SFD)

- The SFD starts at \( R_A = 120\space \text{kN} \) at \( x = 0 \). It remains constant until \( x = 2m \), where it drops by 120kN to 0. It remains 0 until \( x = 5m \), where it drops by 70kN to - 70kN. It remains - 70kN until \( x = 7m \), where it drops by 30kN to - 100kN. Then it increases by \( R_B = 100\space \text{kN} \) to 0 at \( x = 8m \).

Step 5: Plotting Bending Moment Diagram (BMD)

- The BMD starts at 0 at \( x = 0 \). It increases linearly to \( x = 2m \) (since SF is constant) to 240kNm. Then, since SF is 0 from \( x = 2m \) to \( x = 5m \), the BM remains constant at 240kNm. From \( x = 5m \) to \( x = 7m \), the SF is - 70kN (constant), so the BM decreases linearly. The slope of BM is equal to SF. So the change in BM from \( x = 5m \) to \( x = 7m \) (a distance of 2m) is \( SF\times\Delta x=-70\times2=-140\space \text{kNm} \). So at \( x = 7m \), \( BM = 240-140 = 100\space \text{kNm} \). From \( x = 7m \) to \( x = 8m \), the SF is - 100kN, so the change in BM is \( - 100\times1=-100\space \text{kNm} \), so it goes from 100kNm to 0 at \( x = 8m \).

Answer:

Shear Force Diagram (SFD) Characteristics:

• From \( A(0m) \) to \( 2m \): \( SF = 120\space \text{kN} \) (constant).
• From \( 2m \) to \( 5m \): \( SF = 0\space \text{kN} \) (constant).
• From \( 5m \) to \( 7m \): \( SF=-70\space \text{kN} \) (constant).
• From \( 7m \) to \( B(8m) \): \( SF=-100\space \text{kN} \) (constant, and at \( B \) it becomes 0 due to \( R_B \)).

Bending Moment Diagram (BMD) Characteristics:

• From \( A(0m) \) to \( 2m \): Linear increase from \( 0 \) to \( 240\space \text{kNm} \).
• From \( 2m \) to \( 5m \): Constant at \( 240\space \text{kNm} \).
• From \( 5m \) to \( 7m \): Linear decrease from \( 240\space \text{kNm} \) to \( 100\space \text{kNm} \).
• From \( 7m \) to \( B(8m) \): Linear decrease from \( 100\space \text{kNm} \) to \( 0 \).

(To plot the diagrams, use the above values at key points (\( x = 0,2,5,7,8 \)) and connect them with appropriate lines based on the nature of shear force (constant or changing) and bending moment (linear or constant) in each segment.)