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 Hi everyone today we are going to solve problem number 6 .8 here first we had to draw the free body diagram of the entire trust and use this diagram to determine the reactions at the supports so this one is the free body diagram of the entire structure point a point b point c point d then we have to find the equilibrium right down the equilibrium equation taking the moment about point c so first we have to write total horizontal force acting is equal to zero so you'll get cx plus 2 into 5 kilo newton which will be equal to 0 so cx can be minus 10 kilo newton so its direction is like opposite to that way to have seemed and moment about point c is equal to 0 so we can write d into 2 plus 5 into 8 and again d into 2 and 5 into 4 okay and 5 into 4 correct d into 2 minus 5 kilo newton into 8 meter minus 5 kilo in 4 meter which is equal to 0 so we can find d equals 30 kilo newton similarly equilibrium of vertical forces acting will be equal to 0 c y plus 30 kilo newton will be equal to 0 c y equals minus 30 kilo newton its direction is opposite to what we have assumed then we will find a joint a so fad and fab are the two unknown forces acting on point a so we can draw the force triangle like this one by assuming the directions so we can find fab divided by four which is equal to fad divided by root 17 by using hypotenous triangle which is equal to fad divided by root 70 by using which is known value divided by 1 so we will get fab and fad from this we can find fab as 20 kilo newton and fadd as 5 root 17 kilo newton and fad as 5 root 17 kilo that join the b similarly the unknown force unknown components are bc and bd here so we can find at like equilibrium point total vertical forces and total horizontal forces that thing will be equal to zero.

04:26 So we will get 5 kilo newton plus 1 by root 5 of fbd that is the horizontal component of bd force bd which will be equal to 0.

04:42 So f bd will be equal to minus bd .d will be equal to minus...

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