Solution to Problem 403 | Shear and Moment Diagrams
A D V E R T I S E M E N T
Problem 403
Beam loaded as shown in Fig. P-403. See the instruction.

Solution 403
From the load diagram:
[math]\Sigma M_B = 0[/math]
[math]5R_D + 1(30) = 3(50)[/math]
[math]R_D = 24 \, \text{kN}[/math]
[math]\Sigma M_D = 0[/math]
[math]5R_B = 2(50) + 6(30)[/math]
[math]R_B = 56 \, \text{kN}[/math]
Segment AB:
[math]V_{AB} = –30 \, \text{kN}[/math]
[math]M_{AB} = –30x \, \text{kN}\cdot\text{m}[/math]
Segment BC:
[math]V_{BC} = –30 + 56[/math]
[math]V_{BC} = 26 \, \text{kN}[/math]
[math]M_{BC} = –30x + 56(x – 1)[/math]
[math]M_{BC} = 26x – 56 \, \text{kN}\cdot\text{m}[/math]
Segment CD:
[math]V_{CD} = –30 + 56 – 50[/math]
[math]V_{CD} = –24 \, \text{kN}[/math]
[math]M_{CD} = –30x + 56(x – 1) – 50(x – 4)[/math]
[math]M_{CD} = –30x + 56x – 56 – 50x + 200[/math]
[math]M_{CD} = –24x + 144 \, \text{kN}\cdot\text{m}[/math]

To draw the Shear Diagram:
- In segment AB, the shear is uniformly distributed over the segment at a magnitude of –30 kN.
- In segment BC, the shear is uniformly distributed at a magnitude of 26 kN.
- In segment CD, the shear is uniformly distributed at a magnitude of –24 kN.
To draw the Moment Diagram:
- The equation MAB = –30x is linear, at x = 0, MAB = 0 and at x = 1 m, MAB = –30 kN·m.
- MBC = 26x – 56 is also linear. At x = 1 m, MBC = –30 kN·m; at x = 4 m, MBC = 48 kN·m. When MBC = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B.
- MCD = –24x + 144 is again linear. At x = 4 m, MCD = 48 kN·m; at x = 6 m, MCD = 0.
- Printer-friendly version
- Add new comment
- 49013 reads
Recent comments
1 hour 4 min ago
1 week 5 days ago
2 weeks 4 days ago
2 weeks 5 days ago
2 weeks 5 days ago
4 weeks 2 days ago
4 weeks 4 days ago
5 weeks 1 day ago
6 weeks 1 day ago
15 weeks 2 days ago