Draw the shear and moment diagrams for the beam shown below. 10 kN 4 kN/m 50 kN · m 5 т 3 т
Draw the shear and moment diagrams for the beam shown below. 10 kN 4 kN/m 50 kN · m 5 т 3 т
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
![**Example 6.4**
**Objective:**
Draw the shear and moment diagrams for the beam shown in Fig. 6–7a.
**Solution:**
**Support Reactions:**
The reactions at the supports are shown on the free-body diagram of the beam, Fig. 6–7d.
**Shear and Moment Functions:**
Since there is a discontinuity of distributed load and also a concentrated load at the beam’s center, two regions of \( x \) must be considered in order to describe the shear and moment functions for the entire beam.
- **For \( 0 \leq x_1 \leq 5 \, \text{m} \), Fig. 6–7b:**
\[
\begin{align*}
+\sum F_y = 0 &; \quad 5.75 \, \text{kN} - V = 0 \\
V &= 5.75 \, \text{kN} \quad \text{(1)}
\end{align*}
\]
\[
\begin{align*}
\sum M = 0 &; \quad -80 \, \text{kN} \cdot \text{m} - 5.75 \, \text{kN} \cdot x_1 + M = 0 \\
M &= (5.75 \, x_1 + 80) \, \text{kN} \cdot \text{m} \quad \text{(2)}
\end{align*}
\]
- **For \( 5 \, \text{m} < x_2 \leq 10 \, \text{m} \), Fig. 6–7c:**
\[
\begin{align*}
+\sum F_y = 0 &; \quad 5.75 \, \text{kN} - 15 \, \text{kN} - 5 \, \text{kN/m} \cdot (x_2 - 5) - V = 0 \\
V &= (15.75 - 5x_2) \, \text{kN} \quad \text{(3)}
\end{align*}
\]
\[
\begin{align*}
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4f46a7d-33e6-4d1b-8c9b-a4f6142b1fc8%2Fa0524c0e-7d85-4865-ab3d-39e5d069a5d3%2Fmjigbpv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example 6.4**
**Objective:**
Draw the shear and moment diagrams for the beam shown in Fig. 6–7a.
**Solution:**
**Support Reactions:**
The reactions at the supports are shown on the free-body diagram of the beam, Fig. 6–7d.
**Shear and Moment Functions:**
Since there is a discontinuity of distributed load and also a concentrated load at the beam’s center, two regions of \( x \) must be considered in order to describe the shear and moment functions for the entire beam.
- **For \( 0 \leq x_1 \leq 5 \, \text{m} \), Fig. 6–7b:**
\[
\begin{align*}
+\sum F_y = 0 &; \quad 5.75 \, \text{kN} - V = 0 \\
V &= 5.75 \, \text{kN} \quad \text{(1)}
\end{align*}
\]
\[
\begin{align*}
\sum M = 0 &; \quad -80 \, \text{kN} \cdot \text{m} - 5.75 \, \text{kN} \cdot x_1 + M = 0 \\
M &= (5.75 \, x_1 + 80) \, \text{kN} \cdot \text{m} \quad \text{(2)}
\end{align*}
\]
- **For \( 5 \, \text{m} < x_2 \leq 10 \, \text{m} \), Fig. 6–7c:**
\[
\begin{align*}
+\sum F_y = 0 &; \quad 5.75 \, \text{kN} - 15 \, \text{kN} - 5 \, \text{kN/m} \cdot (x_2 - 5) - V = 0 \\
V &= (15.75 - 5x_2) \, \text{kN} \quad \text{(3)}
\end{align*}
\]
\[
\begin{align*}
\
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