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(19) Figure Q19 shows a framework consisting of horizontal members 3 m long and vertical members 4 m long. The framework is loaded at joints J and L with downward load forces of 2 kN. The applied forces cause a vertical reaction forces at A and G and no horizontal reaction force. You are asked to find the internal force in member JK - what would be your approach to solve this problem? Explain your solution process and some of your results by filling in the blanks below. 2 kN 2 kN H RAY RAX A K M N B C D E F 3 m 1 RGY 4m Fill in the multiple blanks. Figure Q19 Finding the vertical reactions is the starting point which can be done by taking moments at A and G but since this is symmetrical loading case the vertical reactions can simply be calculated by halving the total loading 4 kN. Ideally, we can solve the problem using the Method of cutting through the members JK, DJ and It would be sensible to select the left-hand side of the diagram as there are less full members and only one force from the reaction at node A. . This will expose the internal forces which can be labelled with the names of the members themselves. Since we are required to find JK, examining the framework shows it is not a straight-forward matter, and we will require finding all three unknown internal forces. The easiest internal force to find is Next, we can take moments at node as we can resolve forces in the vertical direction. in order to find the internal force JK and find that it has value of kN (use the convention that compressive internal forces are negative values).

(19) Figure Q19 shows a framework consisting of horizontal members 3 m long and vertical members 4 m long. The framework is loaded at joints J and L with downward load forces of 2 kN. The applied forces cause a vertical reaction forces at A and G and no horizontal reaction force. You are asked to find the internal force in member JK - what would be your approach to solve this problem? Explain your solution process and some of your results by filling in the blanks below. 2 kN 2 kN H RAY RAX A K M N B C D E F 3 m 1 RGY 4m Fill in the multiple blanks. Figure Q19 Finding the vertical reactions is the starting point which can be done by taking moments at A and G but since this is symmetrical loading case the vertical reactions can simply be calculated by halving the total loading 4 kN. Ideally, we can solve the problem using the Method of cutting through the members JK, DJ and It would be sensible to select the left-hand side of the diagram as there are less full members and only one force from the reaction at node A. . This will expose the internal forces which can be labelled with the names of the members themselves. Since we are required to find JK, examining the framework shows it is not a straight-forward matter, and we will require finding all three unknown internal forces. The easiest internal force to find is Next, we can take moments at node as we can resolve forces in the vertical direction. in order to find the internal force JK and find that it has value of kN (use the convention that compressive internal forces are negative values).

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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(19) Figure Q19 shows a framework consisting of horizontal members 3 m long and vertical members 4 m long. The framework is loaded at joints J and L with downward load forces of 2 kN. The applied forces cause a vertical reaction forces at A and G and no horizontal reaction force. You are asked to find the internal force in member JK - what would be your approach to solve this problem? Explain your solution process and some of your results by filling in the blanks below. 2 kN 2 kN H RAY RAX A K M N B C D E F 3 m 1 RGY 4m Fill in the multiple blanks. Figure Q19 Finding the vertical reactions is the starting point which can be done by taking moments at A and G but since this is symmetrical loading case the vertical reactions can simply be calculated by halving the total loading 4 kN. Ideally, we can solve the problem using the Method of cutting through the members JK, DJ and It would be sensible to select the left-hand side of the diagram as there are less full members and only one force from the reaction at node A. . This will expose the internal forces which can be labelled with the names of the members themselves. Since we are required to find JK, examining the framework shows it is not a straight-forward matter, and we will require finding all three unknown internal forces. The easiest internal force to find is Next, we can take moments at node as we can resolve forces in the vertical direction. in order to find the internal force JK and find that it has value of kN (use the convention that compressive internal forces are negative values).
Transcribed Image Text:(19) Figure Q19 shows a framework consisting of horizontal members 3 m long and vertical members 4 m long. The framework is loaded at joints J and L with downward load forces of 2 kN. The applied forces cause a vertical reaction forces at A and G and no horizontal reaction force. You are asked to find the internal force in member JK - what would be your approach to solve this problem? Explain your solution process and some of your results by filling in the blanks below. 2 kN 2 kN H RAY RAX A K M N B C D E F 3 m 1 RGY 4m Fill in the multiple blanks. Figure Q19 Finding the vertical reactions is the starting point which can be done by taking moments at A and G but since this is symmetrical loading case the vertical reactions can simply be calculated by halving the total loading 4 kN. Ideally, we can solve the problem using the Method of cutting through the members JK, DJ and It would be sensible to select the left-hand side of the diagram as there are less full members and only one force from the reaction at node A. . This will expose the internal forces which can be labelled with the names of the members themselves. Since we are required to find JK, examining the framework shows it is not a straight-forward matter, and we will require finding all three unknown internal forces. The easiest internal force to find is Next, we can take moments at node as we can resolve forces in the vertical direction. in order to find the internal force JK and find that it has value of kN (use the convention that compressive internal forces are negative values).
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