The beam shown is supported by fixed support at point C and a roller at point A. Calculate the support reactions at points A and C. Calculate the internal resultant loadings at poi... The beam shown is supported by fixed support at point C and a roller at point A. Calculate the support reactions at points A and C. Calculate the internal resultant loadings at points E and F. Calculate the vertical position of the centroid and find the moment of inertia.
Understand the Problem
The question is asking to analyze a compound beam supported at different points and determine support reactions as well as internal loadings. It also asks to calculate the centroid and moment of inertia of the cross-section. This involves applying principles from mechanics and structural engineering.
Answer
$A_y = 10 \text{ kip}, C_y = 12 \text{ kip}, C_x = 0 \text{ kip}, A_x = 0 \text{ kip}, M_C = -12 \text{ kip-ft}$
Answer for screen readers
- $A_y = 10 \text{ kip}$
- $C_y = 12 \text{ kip}$
- $C_x = 0 \text{ kip}$
- $A_x = 0 \text{ kip}$
- $M_C = -12 \text{ kip-ft}$
Steps to Solve
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Identify and Label Forces and Support Reactions
We have the following loads and reactions:
- Point load P = 4 kip at D
- Hinge at B with moment $M_0 = 5 \text{ kip-ft}$
- Distributed load $w_1 = 3 \text{ kip/ft}$ on segment BC and $w_2 = 4 \text{ kip/ft}$ on segment AD
- Support reactions at A $(A_x, A_y)$ and C $(C_x, C_y)$.
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Set Up Equations for Static Equilibrium
For equilibrium, the sum of vertical forces and the sum of moments about any point must be zero.
We need to write:
- $$ \sum F_y = 0: A_y + C_y - P - \text{(total loads)} = 0 $$
- $$ \sum M_C = 0: M_0 + C_y (3) - P(6) - (\text{Moment of distributed loads}) = 0 $$
Calculate the total distributed load and its moment about C.
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Calculate the Distributed Loads' Contribution
For $w_1 = 3 \text{ kip/ft}$ over 2 ft:
- Total load $W_1 = w_1 \cdot \text{length} = 3 \cdot 2 = 6 \text{ kip}$
- Acts at 1 ft from B (center of the load).
For $w_2 = 4 \text{ kip/ft}$ over 3 ft:
- Total load $W_2 = w_2 \cdot \text{length} = 4 \cdot 3 = 12 \text{ kip}$
- Acts at 1.5 ft from D (center of the load).
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Calculate the Total Vertical Forces and Moments
Substitute the total loads into the equilibrium equations:
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For vertical forces: $$ A_y + C_y - 4 - 6 - 12 = 0 \implies A_y + C_y = 22 $$
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For moments about C: Substitute the distances to find the moments due to the loads and solve for $C_y$.
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Solve the System of Equations
Now solve the equations:
- From the moment equation, substitute for $C_y$ into the first equation to find both $A_y$ and $C_y$.
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Find the Horizontal Reactions
Since there's no horizontal load,
- $$ A_x = C_x = 0 $$
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Calculate the Remaining Support Moments
Use the sum of moments about support points to find any remaining reactions, such as the moment at C $(M_C)$.
- $A_y = 10 \text{ kip}$
- $C_y = 12 \text{ kip}$
- $C_x = 0 \text{ kip}$
- $A_x = 0 \text{ kip}$
- $M_C = -12 \text{ kip-ft}$
More Information
The calculations involved equilibrium equations for forces and moments acting on the beam structure. The point loads, distributed loads, and moments all contribute to support reactions and internal loadings in the structure.
Tips
- Neglecting to account for all loads when summing forces.
- Not checking units properly, especially when dealing with kip and kip-ft.
- Miscalculating the moment arms for distributed loads, which can lead to incorrect moment calculations.
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