The cross-section of the rough-cut timber beam shown is loaded with (W) over six feet of the span. Determine the maximum value of (W #/ft.) if the allowable bending stress is Fb = 1,600 psi and the allowable shear stress is Fv = 85 psi.
The cross-section of the rough-cut timber beam shown is loaded with (W) over six feet of the span. Determine the maximum value of (W #/ft.) if the allowable bending stress is Fb = 1,600 psi and the allowable shear stress is Fv = 85 psi.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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The cross-section of the rough-cut timber beam shown is loaded with (W) over six feet of the span. Determine the maximum value of (W #/ft.) if the allowable bending stress is
Fb = 1,600 psi and the allowable shear stress is Fv = 85 psi.
![### Beam Analysis
#### Beam Cross Section
- **Dimensions:**
- Height: 12 inches
- Width: 6 inches
- **Details:**
- \( V_{\text{max}} = 3\omega \)
- \( M_{\text{max}} = 6\omega + \frac{1}{2}(3)(3\omega) = 10.5\omega \)
- Area (\( A \)): \( 72 \, \text{in}^2 \)
- Moment of Inertia (\( I_x \)):
\[
I_x = \frac{(6)(12^3)}{12} = 864 \, \text{in}^4
\]
- Distance to the outermost fiber (\( c \)): 6 inches
- Section Modulus (\( S_x \)):
\[
S_x = \frac{I_x}{c} = \frac{864 \, \text{in}^4}{6 \, \text{in}} = 144 \, \text{in}^3
\]
- Bending stress (\( F_b \)): \( 1,600 \, \text{psi} \)
- Shear stress (\( F_v \)): \( 85 \, \text{psi} \)
#### Diagram
- **Configuration:**
- Simply supported beam with a span (\( L \)) of 10 feet.
- Uniform distributed load (\( \omega \)) throughout 6 feet midspan, flanked by sections of 2 feet on either side.
- **Reactions:**
- Support reactions at both ends are \( 3\omega \).
- **Shear Force Diagram (V):**
- Shear force value starts from \( +3\omega \) on the left-hand side, dropping to 0 at midspan due to the distributed load, and then transitions to \( -3\omega \) on the right-hand side.
- **Bending Moment Diagram (M):**
- The maximum bending moment (\( M_{\text{max}} \)) is \( 10.5\omega \), located at the center of the distributed load.
This analysis evaluates the distribution of shear forces and bending moments in the beam under a uniform load, aiding in determining stress distributions and necessary dimensions](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ed8d57e-7060-482d-9b55-3648d7da0142%2Fdbc4ab5c-946d-4cd7-86db-d8c813cfaefb%2F901vzvr_processed.png&w=3840&q=75)
Transcribed Image Text:### Beam Analysis
#### Beam Cross Section
- **Dimensions:**
- Height: 12 inches
- Width: 6 inches
- **Details:**
- \( V_{\text{max}} = 3\omega \)
- \( M_{\text{max}} = 6\omega + \frac{1}{2}(3)(3\omega) = 10.5\omega \)
- Area (\( A \)): \( 72 \, \text{in}^2 \)
- Moment of Inertia (\( I_x \)):
\[
I_x = \frac{(6)(12^3)}{12} = 864 \, \text{in}^4
\]
- Distance to the outermost fiber (\( c \)): 6 inches
- Section Modulus (\( S_x \)):
\[
S_x = \frac{I_x}{c} = \frac{864 \, \text{in}^4}{6 \, \text{in}} = 144 \, \text{in}^3
\]
- Bending stress (\( F_b \)): \( 1,600 \, \text{psi} \)
- Shear stress (\( F_v \)): \( 85 \, \text{psi} \)
#### Diagram
- **Configuration:**
- Simply supported beam with a span (\( L \)) of 10 feet.
- Uniform distributed load (\( \omega \)) throughout 6 feet midspan, flanked by sections of 2 feet on either side.
- **Reactions:**
- Support reactions at both ends are \( 3\omega \).
- **Shear Force Diagram (V):**
- Shear force value starts from \( +3\omega \) on the left-hand side, dropping to 0 at midspan due to the distributed load, and then transitions to \( -3\omega \) on the right-hand side.
- **Bending Moment Diagram (M):**
- The maximum bending moment (\( M_{\text{max}} \)) is \( 10.5\omega \), located at the center of the distributed load.
This analysis evaluates the distribution of shear forces and bending moments in the beam under a uniform load, aiding in determining stress distributions and necessary dimensions
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