VIPSolutionsOPTIONAL: Elastic Curve for Beam Deflections. Figure 3 is a front elevation view of a simply supported beam that carries a triangular load. W (N/m) . fy A — x B I Figure 3: Simply supported beam carrying a triangular load. The load decreases from W (N/m) at point A to zero at point B. Thus, the total beam loading is WL/2. [3a] (4 pts). Starting from first principles of engineering, show that the bending moment at point is: M(x) = [2] z(L—2)(2L—1). ©) [3b] (4 pts). Show that the clastic curve for beam deflection is given by (notice that in Figure 4, the y axis is pointing upwards): WL (1275 (214, (213 [= ve) = [35087] (313) ot [7] +20[7] -5[3)) ® [3¢] (2 pts). Show that the maximum beam curvature occurs at x = ln - & L.
Question:
OPTIONAL: Elastic Curve for Beam Deflections. Figure 3 is a front elevation view of a simply supported
beam that carries a triangular load.
W (N/m)
. fy
A — x B
I
Figure 3: Simply supported beam carrying a triangular load.
The load decreases from W (N/m) at point A to zero at point B. Thus, the total beam loading is WL/2.
[3a] (4 pts). Starting from first principles of engineering, show that the bending moment at point is:
M(x) = [2] z(L—2)(2L—1). ©)
[3b] (4 pts). Show that the clastic curve for beam deflection is given by (notice that in Figure 4, the y axis
is pointing upwards):
WL (1275 (214, (213 [=
ve) = [35087] (313) ot [7] +20[7] -5[3)) ®
[3¢] (2 pts). Show that the maximum beam curvature occurs at x = ln - & L.
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Asked by: Test User 100328
Created at: 2025-05-20 07:20:15
Asked by: Test User 100328
Created at: 2025-05-20 07:20:15
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This image provides an **optional section on the elastic curve for beam deflections** under a **triangular load**. Here’s a breakdown of the key points and equations:
---
### **Beam Setup:**
- **Simply supported beam** (supported at both ends A and B).
- **Triangular distributed load:** Maximum at A (W N/m), zero at B.
- **Span length:** \( L \)
#### **Total Load:**
- Area under the triangle: \( \frac{1}{2} \times W \times L = \frac{WL}{2} \)
---
### **Questions and Key Equations:**
#### **[3a] Bending Moment at Point \( x \):**
- **Goal:** Derive the bending moment \( M(x) \) using first principles.
- **Result:**
\[
M(x) = \left[\frac{W}{6L}\right] x (L - x)(2L - x)
\]
- **Interpretation:** This gives the internal bending moment at a distance \( x \) from the left end, due to the triangular load.
---
#### **[3b] Elastic Curve for Deflection:**
- **Goal:** Show the deflection curve equation for the beam.
- **Result:**
\[
y(x) = \left[\frac{WL^4}{360EI}\right] \left( 3\left[\frac{x}{L}\right]^5 - 15\left[\frac{x}{L}\right]^4 + 20\left[\frac{x}{L}\right]^3 - 8\left[\frac{x}{L}\right] \right)
\]
- Where \( E \) is the modulus of elasticity and \( I \) is the moment of inertia.
- **Interpretation:** This formula gives the vertical deflection \( y(x) \) along the beam.
---
#### **[3c] Maximum Beam Curvature:**
- **Goal:** Find where the maximum beam curvature occurs.
- **Result:**
\[
x = \left[1 - \frac{1}{\sqrt{3}}\right] L
\]
- **Interpretation:** The position along the beam where the curvature (second derivative of the elastic curve) is greatest.
---
### **Summary Table**
| Quantity | Formula |
|--------------------------|-----------------------------------------------------------------------------------------------------|
| Bending Moment \( M(x) \)| \( \left[\frac{W}{6L}\right] x (L - x)(2L - x) \) |
| Deflection \( y(x) \) | \( \left[\frac{WL^4}{360EI}\right] \left( 3\left[\frac{x}{L}\right]^5 - 15\left[\frac{x}{L}\right]^4 + 20\left[\frac{x}{L}\right]^3 - 8\left[\frac{x}{L}\right] \right) \) |
| Max Curvature Position | \( x = \left[1 - \frac{1}{\sqrt{3}}\right] L \) |
---
If you need the derivation steps for any of these results, let me know!
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