7k/f 12' Draw Shear & Moment Diagrams. Find Vmax, V=0, Mmax, M=0
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![# Analyzing a Cantilever Beam with a Triangular Distributed Load
### Problem Statement
Consider a cantilever beam fixed at point A and subjected to a triangular distributed load with a maximum intensity of 7 kips per foot (7 k/f) over a length of 12 feet. The objectives are to:
1. Draw the Shear and Moment Diagrams.
2. Find the maximum values of shear force (Vmax) and bending moment (Mmax).
### Given Data
- Maximum load intensity, \( w = 7 \frac{\text{kips}}{\text{foot}} \)
- Length of the beam, \( L = 12 \text{ feet} \)
### Loading Details
The load is a triangular distributed load, which linearly decreases from 7 kips per foot at point A to 0 kips per foot at the free end.
### Steps to Analyze
1. **Support Reactions:**
- As the beam is fixed at A, it provides reactions in the vertical and horizontal directions, and a moment.
2. **Shear Diagram (V):**
- For a triangular distributed load, the shear force will vary parabolically.
- The maximum shear force will occur at the fixed end (point A).
3. **Moment Diagram (M):**
- The moment diagram will be cubic due to the triangular load.
- Maximum moment occurs at the fixed point.
### Shear Force Calculation
Using the property of a triangular load, the equivalent point load \( P \) acts at \( \frac{1}{3} \) the distance from the larger end (which is at the fixed support).
\[
P = \frac{1}{2} \cdot w_{max} \cdot L = \frac{1}{2} \cdot 7 \cdot 12 = 42 \text{ kips}
\]
The resultant point load acts at:
\[
x = \frac{1}{3} \cdot 12 = 4 \text{ feet}
\]
So, the reaction at the fixed end A is:
\[
V_A = P = 42 \text{ kips}
\]
### Moment Calculation
The maximum moment at the fixed end:
\[
M_A = P \cdot x = 42 \cdot 4 = 168 \text{ kip-feet}
\]
### Shear and Moment Di](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50eb2229-1d0c-4dc7-966b-88adcd3ef11b%2Ff5aee5f8-b01d-4f94-86f1-601b11d25c83%2Fe9k0co_processed.jpeg&w=3840&q=75)
Transcribed Image Text:# Analyzing a Cantilever Beam with a Triangular Distributed Load
### Problem Statement
Consider a cantilever beam fixed at point A and subjected to a triangular distributed load with a maximum intensity of 7 kips per foot (7 k/f) over a length of 12 feet. The objectives are to:
1. Draw the Shear and Moment Diagrams.
2. Find the maximum values of shear force (Vmax) and bending moment (Mmax).
### Given Data
- Maximum load intensity, \( w = 7 \frac{\text{kips}}{\text{foot}} \)
- Length of the beam, \( L = 12 \text{ feet} \)
### Loading Details
The load is a triangular distributed load, which linearly decreases from 7 kips per foot at point A to 0 kips per foot at the free end.
### Steps to Analyze
1. **Support Reactions:**
- As the beam is fixed at A, it provides reactions in the vertical and horizontal directions, and a moment.
2. **Shear Diagram (V):**
- For a triangular distributed load, the shear force will vary parabolically.
- The maximum shear force will occur at the fixed end (point A).
3. **Moment Diagram (M):**
- The moment diagram will be cubic due to the triangular load.
- Maximum moment occurs at the fixed point.
### Shear Force Calculation
Using the property of a triangular load, the equivalent point load \( P \) acts at \( \frac{1}{3} \) the distance from the larger end (which is at the fixed support).
\[
P = \frac{1}{2} \cdot w_{max} \cdot L = \frac{1}{2} \cdot 7 \cdot 12 = 42 \text{ kips}
\]
The resultant point load acts at:
\[
x = \frac{1}{3} \cdot 12 = 4 \text{ feet}
\]
So, the reaction at the fixed end A is:
\[
V_A = P = 42 \text{ kips}
\]
### Moment Calculation
The maximum moment at the fixed end:
\[
M_A = P \cdot x = 42 \cdot 4 = 168 \text{ kip-feet}
\]
### Shear and Moment Di
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