By looking the example, could you, please, solve this problem clearly? ### Example 6.4: Beam Shear and Moment Diagrams#### Objective:Draw the shear and moment diagrams for the beam shown in Figure 6-7a.#### Solution:**Support Reactions:**- The reactions at the supports are shown on the free-body diagram of the beam (Figure 6-7d).**Shear and Moment Functions:**- There is a discontinuity of distributed load and also a concentrated load at the beam’s center. Therefore, two regions must be considered in order to describe the shear and moment functions for the entire beam.1. **For 0 ≤ x₁ ≤ 5 m (Figures 6-7b):** - \( \Sigma F_y: \quad 0_+ = 0_+ - 5.75 \, \text{kN} + V = 0 \) *\[ V = 5.75 \, \text{kN} \]* - \( \Sigma M: \quad -80 \, \text{kN} \cdot \text{m} + M = (5.75x) \, \text{kN} \cdot \text{m} \) *\[ M = (5.75x + 80) \, \text{kN} \cdot \text{m} \]* 2. **For 5 m < x₂ ≤ 10 m (Figures 6-7c):** - \( \Sigma F_y: \quad 0_+ = 5.75 \, \text{kN} - 15 \, \text{kN} - 5 \, \text{kN}/\text{m} \cdot (x₂ - 5) + V = 0 \) *\[ V = (15.75 - 5x) \, \text{kN} \]* - \( \Sigma M: \quad -80 \, \text{kN} \cdot \text{m} - 5.75x + 15 \cdot x₂ + 5 \, \text{kN}/\text{m} \cdot (x₂ - 5) + M = 0 \) *\[ M = (-2.5x² + 15.75x + 92.5) \,

Given: The simply supported beam with the loads is shown below:Calculation: The free body diagram of the beam is shown below: Determine the reactions as follows: There will no horizontal reaction at point A as there is no horizontal load. Take moment about point A as follows: ∑MA=0Cy8+50=105+4552Cy=6.25 kN Take the summation of force in y- direction as follows: ∑Fy=0Cy+Ay=10+45Ay=23.75 kN Take sections x-x in each of segments AB and BC at distance x from the left support. Sign convention for the bending moment and shear forces is shown below: For segment AB, Shear forces are: SFx=+23.75-4xSo, at A (x=0)SFA=+23.75-40=+23.75 kNAt B x=5 mSFB=+23.75-45=+3.75 kN Bending moments are: Mx=+23.75x-4xx2=23.75x-2x2So, at A (x=0)MA=+23.750-202=+0 kN-mAt B x=5 mMB=+23.755-252=+68.75 kN-m For segment BC, Shear forces are: SFx=+23.75-45-10=-6.25So, at B (x=5 m)SFB=-6.25 kNAt C x=8 mSFC=-6.25 kN Bending moments are: Mx=+23.75x-45x-2.5-10x-5=23.75x-20x-2.5-10x-5So, at B (x=5)MB=23.755-205-2.5-105-5=+68.75 kN-mAt C x=8 mMC=23.758-208-2.5-108-5=+50 kN-m Draw the all shear force and bending moment calculated above as follows:

Answered: Draw the shear and moment diagrams for…

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Draw the shear and moment diagrams for the beam shown below. 10 kN 4 kN/m 50 kN · m 5 m 3т

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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Question

By looking the example, could you, please, solve this problem clearly?

### Example 6.4: Beam Shear and Moment Diagrams

#### Objective:
Draw the shear and moment diagrams for the beam shown in Figure 6-7a.

#### Solution:

**Support Reactions:**
- The reactions at the supports are shown on the free-body diagram of the beam (Figure 6-7d).

**Shear and Moment Functions:**
- There is a discontinuity of distributed load and also a concentrated load at the beam’s center. Therefore, two regions must be considered in order to describe the shear and moment functions for the entire beam.

1. **For 0 ≤ x₁ ≤ 5 m (Figures 6-7b):**

- \( \Sigma F_y: \quad 0_+ = 0_+ - 5.75 \, \text{kN} + V = 0 \)
*\[ V = 5.75 \, \text{kN} \]*

- \( \Sigma M: \quad -80 \, \text{kN} \cdot \text{m} + M = (5.75x) \, \text{kN} \cdot \text{m} \)
*\[ M = (5.75x + 80) \, \text{kN} \cdot \text{m} \]*

2. **For 5 m < x₂ ≤ 10 m (Figures 6-7c):**

- \( \Sigma F_y: \quad 0_+ = 5.75 \, \text{kN} - 15 \, \text{kN} - 5 \, \text{kN}/\text{m} \cdot (x₂ - 5) + V = 0 \)
*\[ V = (15.75 - 5x) \, \text{kN} \]*

- \( \Sigma M: \quad -80 \, \text{kN} \cdot \text{m} - 5.75x + 15 \cdot x₂ + 5 \, \text{kN}/\text{m} \cdot (x₂ - 5) + M = 0 \)
*\[ M = (-2.5x² + 15.75x + 92.5) \,

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