Question 6.SP.5: Determine the components of the forces acting on each member......

Determine the components of the forces acting on each member of the frame shown.

STRATEGY: The approach to this analysis is to consider the entire frame as a free body to determine the reactions, and then consider separate members. However, in this case, you will not be able to determine forces on one member without analyzing a second member at the same time.

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Step 1:
We start by considering the free-body diagram of the entire frame. We apply the principle of moments to find the moment equation. The equation is set to zero since the frame is in equilibrium. Solving the moment equation, we find the value of one of the unknowns, which is the horizontal component of the reaction at one of the supports.
Step 2:
Next, we apply the principle of equilibrium in the vertical direction by considering the sum of forces in the y-direction. Again, we set the equation to zero and solve for the unknown vertical component of the reaction at the other support.
Step 3:
Finally, we apply the principle of equilibrium in the horizontal direction by considering the sum of forces in the x-direction. Since the frame is in equilibrium, the horizontal component of the reaction at the other support is found to be zero.
Step 4:
Now, we dismember the frame and analyze each member separately. We consider the free-body diagram of member BCD and apply the principle of moments to find the vertical components of the forces at the joints.
Step 5:
We continue by analyzing member ABE. We apply the principle of moments to find the horizontal component of the force at one of the joints. Then, we apply the principles of equilibrium in both the x and y-directions to find the remaining components of the forces at the joints.
Step 6:
After finding all the unknown components of the forces at the joints, we reflect and think about the results. We can verify the equilibrium of member ACF by applying the principle of moments and checking if the sum of moments is zero.
By following these steps, we are able to determine all the unknown reactions at the supports and verify the equilibrium of the frame.

Final Answer

MODELING and ANALYSIS: The external reactions involve only three unknowns, so compute the reactions by considering the free-body diagram of the entire frame (Fig. 1).

+↺ \Sigma M_E=0: \quad-(2400 \mathrm{~N})(3.6 \mathrm{~m})+F(4.8 \mathrm{~m})=0

F=+1800 \mathrm{~N} \quad \mathrm{~F}=1800 \mathrm{~N} \uparrow

+\uparrow \Sigma F_y=0: \quad-2400 \mathrm{~N}+1800 \mathrm{~N}+E_y=0

E_y=+600 \mathrm{~N} \quad \mathbf{E}_y=600 \mathrm{~N} \uparrow

\stackrel{+}{\rightarrow} \Sigma F_x=0: \quad \mathbf{E}_x=0

Now dismember the frame. Since only two members are connected at each joint, force components are equal and opposite on each member at each joint (Fig. 2).

Free Body: Member BCD.

+↺ \Sigma M_B=0: \quad-(2400 \mathrm{~N})(3.6 \mathrm{~m})+C_y(2.4 \mathrm{~m})=0 \quad C_y=+3600 \mathrm{~N}

+↺ \Sigma M_C=0: \quad-(2400 \mathrm{~N})(1.2 \mathrm{~m})+B_y(2.4 \mathrm{~m})=0 \quad B_y=+1200 \mathrm{~N}

\stackrel{+}{\rightarrow} \Sigma F_x=0: \quad-B_x+C_x=0

Neither B_x \text { nor } C_x can be obtained by considering only member BCD; you need to look at member ABE. The positive values obtained for B_y \text { and } C_y indicate that the force components B_y \text { and } C_y are directed as assumed.

Free Body: Member ABE.

+↺ \Sigma M_A=0: \quad B_x(2.7 \mathrm{~m})=0 \quad B_x=0

\stackrel{+}{\rightarrow} \Sigma F_x=0: \quad+B_x-A_x=0 \quad A_x=0

+\uparrow \Sigma F_y=0: \quad-A_y+B_y+600 \mathrm{~N}=0

-A_y+1200 \mathrm{~N}+600 \mathrm{~N}=0 \quad A_y=+1800 \mathrm{~N}

Free Body: Member BCD. Returning now to member BCD, you have

\stackrel{+}{\rightarrow} \Sigma F_x=0: \quad-B_x+C_x=0 \quad 0+C_x=0 \quad C_x=0

REFLECT and THINK: All unknown components have now been found. To check the results, you can verify that member ACF is in equilibrium.

+↺ \Sigma M_C=(1800 \mathrm{~N})(2.4 \mathrm{~m})-A_y(2.4 \mathrm{~m})-A_x(2.7 \mathrm{~m})

=(1800 \mathrm{~N})(2.4 \mathrm{~m})-(1800 \mathrm{~N})(2.4 \mathrm{~m})-0=0 (checks)

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