Question 7.SP.1: In the frame shown, determine the internal forces (a) in mem......
In the frame shown, determine the internal forces (a) in member ACF at point J, (b) in member BCD at point K. This frame was previously analyzed in Sample Prob. 6.5.
STRATEGY: After isolating each member, you can cut it at the given point and treat the resulting parts as objects in equilibrium. Analysis of the equilibrium equations, as we did before in Sample Problem 6.5, will determine the internal force-couple system.
The frame shown in the question
The internal forces in member ACF at point J
The internal forces in member BCD at point K
The frame was previously analyzed in Sample Prob. 6.5
Strategy: Isolate each member and cut it at the given point, treating the resulting parts as objects in equilibrium. Analyze the equilibrium equations to determine the internal force-couple system.
Next, we can consider the forces acting in the horizontal direction. We have an unknown axial force F and the horizontal component of the force at J. By setting the sum of these forces equal to zero, we can solve for the value of F.
Finally, we can consider the forces acting in the vertical direction. We have an unknown vertical force V and the vertical component of the force at J. By setting the sum of these forces equal to zero, we can solve for the value of V.
Next, we can consider the forces acting in the horizontal direction. In this case, we have an unknown axial force F, but since there are no horizontal forces acting at K, the value of F is zero.
Finally, we can consider the forces acting in the vertical direction. We have an unknown vertical force V and the downward force of 1200 N. By setting the sum of these forces equal to zero, we can solve for the value of V.
Final Answer
MODELING: The reactions and the connection forces acting on each member of the frame were determined previously in Sample Prob. 6.5. The results are repeated in Fig. 1.
ANALYSIS:
a. Internal Forces at J. Cut member ACF at point J, obtaining the two parts shown in Fig. 2. Represent the internal forces at J by an equivalent force-couple system, which can be determined by considering the equilibrium of either part. Considering the free body AJ, you have
+↺ \Sigma M_J=0: \quad-(1800 \mathrm{~N})(1.2 \mathrm{~m})+M=0
M=+2160 \mathrm{~N} \cdot \mathrm{m} \quad \mathbf{M}=2160 \mathrm{~N} \cdot \mathrm{m} ↺
+\searrow \Sigma F_x=0: \quad F-(1800 \mathrm{~N}) \cos 41.7^{\circ}=0
F=+1344 \mathrm{~N} \quad \mathbf{F}=1344 \mathrm{~N} \searrow
+\nearrow \Sigma F_y=0: \quad-V+(1800 \mathrm{~N}) \sin 41.7^{\circ}=0
V=+1197 \mathrm{~N} \quad \mathbf{V}=1197 \mathrm{~N} \swarrow
The internal forces at J are therefore equivalent to a couple M, an axial force F, and a shearing force V. The internal force-couple system acting on part JCF is equal and opposite.
b. Internal Forces at K. Cut member BCD at K, obtaining the two parts shown in Fig. 3. Considering the free body BK, you obtain
+↺ \Sigma M_K=0: \quad(1200 \mathrm{~N})(1.5 \mathrm{~m})+M=0
M=-1800 \mathrm{~N} \cdot \mathrm{m} \quad \mathbf{M}=1800 \mathrm{~N} \cdot \mathrm{m} ⤸
\stackrel{+}{\rightarrow} \Sigma F_x=0: \quad F=0 F = 0
+\uparrow \Sigma F_y=0: \quad-1200 \mathrm{~N}-\mathrm{V}=0
V=-1200 \mathrm{~N} \quad \mathbf{V}=1200 \mathrm{~N} \uparrow
REFLECT and THINK: The mathematical techniques involved in solving a problem of this type are not new; they are simply applications of concepts presented in earlier chapters. However, the physical interpretation is new: we are now determining the internal forces and moments within a structural member. These are of central importance in the study of mechanics of materials.
