STEM Problem Detail

Problem Statement

Prompt

Give me the solution for problem 7-32

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## Problem 7-32 Solution

We need to determine the internal normal force, shear force, and moment at points B and C on the curved rod.

**Given Information:**

- Curved rod with radius $$r = 2$$ ft
- Point B is at $$30°$$ from the vertical
- Point C is at $$45°$$ from the horizontal
- Applied load: $$P = 500$$ lb downward at point B

### Step 1: Support Reactions at A

The rod is fixed at point A.

Vertical force equilibrium:
\[\sum F_y = 0: \quad R_{Ay} - 500 = 0\]
\[R_{Ay} = 500 \text{ lb (upward)}\]

Horizontal force equilibrium:
\[\sum F_x = 0: \quad R_{Ax} = 0\]

Position of point B relative to A:

- Horizontal distance from A to B: $$x_{AB} = 2 - 2\sin(30°) = 1$$ ft
- Vertical distance from A to B: $$y_{AB} = 2\cos(30°) = 1.732$$ ft

Moment about A:
\[M_A = 500 \times 1 = 500 \text{ lb·ft}\]

### Step 2: Internal Forces at Point B

At point B, the tangent to the curve makes $$30°$$ with horizontal.

Normal direction: makes $$60°$$ with horizontal
Tangential direction: makes $$30°$$ with horizontal

Normal force $$N_B$$ (compression):
\[N_B = 500\cos(60°) = 500(0.5) = 250 \text{ lb}\]

Shear force $$V_B$$ (tangential):
\[V_B = 500\sin(60°) = 500(0.866) = 433 \text{ lb}\]

Moment $$M_B$$:
\[M_B = 500 \times 1 = 500 \text{ lb·ft}\]

### Step 3: Internal Forces at Point C

Point C is at $$45°$$ from horizontal. The tangent makes $$45°$$ with horizontal.

Position of C relative to A:

- Horizontal distance: $$2 - 2\cos(45°) = 0.586$$ ft
- Vertical distance: $$2\sin(45°) = 1.414$$ ft

Normal force $$N_C$$:
\[N_C = 500\cos(45°) = 500(0.707) = 354 \text{ lb}\]

Shear force $$V_C$$:
\[V_C = 500\sin(45°) = 500(0.707) = 354 \text{ lb}\]

Moment $$M_C$$:
\[M_C = 500 \times 0.586 = 293 \text{ lb·ft}\]

### Final Results:

At point B:
\[\boxed{N_B = 250 \text{ lb}, \quad V_B = 433 \text{ lb}, \quad M_B = 500 \text{ lb·ft}}\]

At point C:
\[\boxed{N_C = 354 \text{ lb}, \quad V_C = 354 \text{ lb}, \quad M_C = 293 \text{ lb·ft}}\]

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Problem #84