The portion of truss shown represents the upper part of a power transmission line tower. For the

The portion of truss shown represents the upper part of a...

Key Concepts

Static Equilibrium

In structural analysis, static equilibrium is the fundamental concept that requires the sum of all forces and the sum of all moments acting on a system to be zero. This principle is used to analyze structures like trusses, ensuring that the forces in each member balance out so the structure remains stable under applied loads.

Method of Joints

The method of joints is a technique in truss analysis where each joint (or node) is isolated and the equilibrium equations are applied to determine the forces in the connected members. This method is particularly useful for finding forces in individual members when the loads and reactions at the joints are known.

Method of Sections

The method of sections is an alternative approach used in the analysis of trusses that involves cutting through the structure to create a section. By applying equilibrium equations to this section, one can efficiently compute the internal forces in multiple members simultaneously, especially when the member of interest is not easily accessible through the method of joints.

Tension and Compression

Understanding whether a member is in tension or compression is crucial in truss analysis. Tension occurs when a member is being pulled apart, while compression occurs when a member is being pushed together. Determining the state of a member helps in assessing its suitability and safety under the given loading conditions.

Transcript

00:01 In this problem on the topic of analysis of structures, we are given the portion of a truss shown in the diagram that represents the upper part of a power transmission line tower.

00:13 Now we are given a loading and asked to determine the force in each of the members in the truss located above point ahj, and we want to state if either member or each member is in tension or compression.

00:27 Now for the free body diagram of joint a, which we have as follows in the diagram, we can see that f .a .b over 2 .29, using some geometry, is equal to fac over 2 .29, and this is equal to 1 .2 kilonutons over 1 .2 .2.

00:59 So immediately we can find the force on each of the members, fab and fac.

01:09 So we get fab to be 2 .29 kiloons and this is tension.

01:23 Fac is 2 .29 kiloons and this is compression.

01:32 So we found the force for the first two members.

01:41 Now we look at the free body diagram of joint f.

01:48 So we're looking at the first free body diagram that we've drawn here.

01:52 So for joint f, we have the force f, df over to .29 is equal to f ef over 2 .2 .2.

02:16 29 and this is 1 .2 kiloons over 1 .2 .2.

02:25 So we can immediately find the force of the two members, f, df, which is 2 .29 kiloons, and this is a tensile force, and f, ef is equal to 2 .29 kiloons, and this is a tensile force, and f, ef is equal to 2 .29 kiloons, which is a compressive force.

02:53 So from that free body diagram we've found two more of the members forces.

03:01 Next we're going to joint d.

03:05 So we've drawn the free body diagram for joint d as above.

03:14 Now using this free body diagram we have that the force fbd over 2 .21 is the same as force de over 0 .6 and this is equal to 2 .29 kiloons over 2 .29 .29.

03:42 So we can see that fbd is equal to 2 .21 kilonutons.

03:55 And this is a tensile force.

04:00 Fde is simply 0 .6 kilo -neutons and this is a compressive force.

04:10 So from joint d we found two more of the forces.

04:17 Now we move on to joint b which in a diagram, now free body diagram looks as follows.

04:31 Now in joint b if we take the right direction to be positive and we take the right direction to be positive and we take the sum of all the x component of the forces, this must equal to zero for equilibrium.

04:49 And this gives us 4 over 5 fbe plus 2 .21 kiloons minus 2 .21 over 2 .29 into 2 .29 kiloons.

05:14 And this must equal to 0.

05:17 So from here we can see that f, b, e must equal to 0.

05:27 Now, still on this free body diagram, if we take now the upward direction to be positive and look at the y components of all the forces, again, the sum of these components must be zero...

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