Using the method of joints, determine the force in each member of the truss shown. State whether

Using the method of joints, determine the force in each...

Key Concepts

Static Equilibrium

Static equilibrium is the condition in which a structure remains at rest because the sum of all forces and moments acting on it equals zero. This fundamental principle underlies the analysis of trusses, ensuring that every joint or segment is balanced by counteracting forces, which allows for the systematic determination of unknown forces.

Method of Joints

The method of joints is an analytical technique used in truss analysis where each joint is isolated, and the equilibrium equations for forces in the horizontal and vertical directions are applied. This approach simplifies the process by allowing the forces in each member connected to the joint to be solved step by step.

Free-Body Diagrams

Free-body diagrams are graphical representations that isolate a component or joint of a structure and display all the forces acting upon it. These diagrams are essential in truss analysis as they help in setting up the equilibrium equations and visualizing the force distribution, making the process of solving for unknown forces more straightforward.

Tension and Compression

Tension and compression refer to the types of internal forces experienced by the members in a truss. Tension occurs when forces stretch the members, effectively pulling them apart, while compression involves forces that shorten or push the members together. Recognizing whether a member is in tension or compression is crucial for assessing the structural performance and safety of the truss.

Transcript

00:01 So here in this problem we are given this trust and a lot of magnitude 693 pound is acting at point b now we have to use a method of joint to determine the force in each member of the trust and we have to also state whether the member is in compression or in tension so first of all it will be calculating the reactions offered by the supports that is dx dy and ey for that we are taking three equilibrium conditions the first one is sigma fx equals to zero now in x direction we have only one force that is d x so d x would be equals to zero now next we are taking moment about point e that is sigma m e equals to zero now the moment in the anticlockwise direction will be taking positive and the moment in the clockwise direction will be taking negative so we'll get d y times the perpendicular distances 21 and it is making an anti -clockwise direction sorry in clockwise directions over to take a negative sign plus the next forces this 693 pound times the perpendicular distance is 5 and it is making the moment in anti -clockwise direction so we'll take a positive sign so equals to 0 so from here dy comes out to be as 163 pound next it will be considering sigma f y equals to zero and the forces in the upward direction will be taking positive and the forces in the downward direction will be taking negative so we have this positive dy this one is negative so minus 693 pound and plus ey equals to zero and i can put the values and we have 165 pound minus 693 pound plus ey equals to zero so from here we'll get the value of ey as 528 pound and the direction would be in vertically upward direction so upward and here also it would been the upward direction now next we'll be considering the free body diagram of the joint d and we are assuming that the member is in tension so the force would be like this this is f a d and this would be f cd now if the value of f ad and fcd comes out to be negative that means the member would be in compression because now we are assuming it to be in tension so this is a freeboard diagram of the joint d now we have to calculate the force f a d that is the force in the member ad and the force in the member cd for that we can use equilibrium condition the first one is sigma effects equals to 0 and the forces in this direction you will be taking positive and the forces in this direction will be taking negative so we have we'll be taking the component of f a d in the horizontal direction so it would be 5 by 13 f a d similarly the horizontal component of this force would be 4 by 5 f cd is equals to 0 now next we'll be be considering the next equilibrium condition let this be as one and that is sigma f y equals to zero and the forces in the upward direction will be taking positive and the forces in the downward direction will be taking negative so we have we'll be taking the component of this force in the vertical direction so it would be 12 by 13 f a d plus fcd the vertical component would be 3 by 5 f cd plus this form force is also acting which is d -y that we have calculated earlier to be 165 pound and it would be equals to zero so we have two equation this one and this one and we have two variables f -a -d and f -c -d so we can solve these two equations simultaneously and we'll get the force f -a -d to be minus 260 pound and f -d -c to be 125 pounds now if dc comes out to be as positive that means the assumed direction is correct and the member would be in tension now the force fad comes out to be as negative that means the assumed direction is wrong and it would be acting like this that means the for the member would be in compression so fad would be 260 pound and the member would be in…

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