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

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.

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.

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.

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.

Transcript

00:01 So here in this problem we are given this trust and we have to calculate the forces in each member by using the method of joints so this point is a this point is c and this point is b now first of all it will be calculating the reactions given by the support this is cx a x and a y so for that we'll be using the equilibrium condition sigma fi equals to zero so in vertical direction we have only two forces a y and this 2 .8 kilo newton so we can write and the force in the downward direction will be taking negative and the force in the upward direction will be taking positive so these two forces are in the downward direction so we can write minus a y minus 2 .8 kilo newton is equals to 0 so from here a y will be equals to minus 2 .8 kilo newton that means a y is 2 .8 kilo newton that means a .y is 2 .8 kilo newton but in the upward direction.

01:11 Now next will be taking moment about this point about the point a, that is sigma m a is equals to 0 and the moment in the anticlockwise direction will be taking positive and the moment in the clockwise direction and will be taking negative.

01:30 So at this point we are at about this point we are taking moment.

01:33 So there are only two forces that we have to consider this one cx and this one because these two forces are passing through point a only so this more the moment due to these forces would be 0 so we have a cx times this distance is 1 .4 so cx times 1 .4 minus 2 .8 kilo newton times this distance is 0 .75 equals to 0 so from here we can easily calculate cx and its value comes out to be as 1 .5 kilo newton and its direction is in this direction now next it will be calculating a x for that we can use one equilibrium condition that is sigma fx equals to zero now we have only two forces in the x direction that is cx and a x so and the forces in this direction will be taking positive and in this direction will be taking negative so we have cx plus a x is equals to zero so from here a x would be equals to negative which is minus 1 .5 kilo newton.

02:47 Therefore, a x would be equals to 1 .5 kilo newton, but the direction would be this side.

02:54 Now next, we'll be considering the free border diagram of the joint c.

02:58 So this is a free bullet diagram of the joint c.

03:02 Now, we'll be calculating this angle theta first, this angle, which can be calculated from this triangle, this one.

03:12 So this angle would also be theta...

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