The truss shown is one of several supporting an advertising panel. Determine the force in each m

The truss shown is one of several supporting an advertising...

Key Concepts

Tension and Compression

Tension and compression describe the nature of the forces within structural members. Tension refers to forces that attempt to pull or elongate an element, while compression refers to forces that push or shorten it. Recognizing whether a member is in tension or compression is crucial for selecting appropriate materials and cross-sectional designs, as the failure modes, such as buckling or yielding, can vary significantly based on the type of force involved.

Static Equilibrium

Static equilibrium is a fundamental principle in mechanics stating that a body at rest must have a net force and net moment of zero. In structural engineering, ensuring static equilibrium allows designers to set up equations that accurately describe the balance of forces, leading to the determination of internal member forces and verifying that the structure will not experience unintended motion or collapse.

Method of Sections

The method of sections involves cutting through a truss to isolate a section and then applying equilibrium equations to that section. This technique is particularly useful for directly computing the forces in selected members without having to analyze the entire structure, streamlining the process of understanding how localized forces interact within a truss system.

Truss Analysis

Truss analysis is the study of the load distribution in a framework of members connected at joints, where each member is assumed to carry force only along its length. This approach simplifies structural behavior and allows engineers to predict how forces are transmitted through the truss, ensuring stability and safety under various loading conditions.

Method of Joints

The method of joints is a systematic technique used in structural analysis to resolve forces at each connection point (joint) within a truss. By applying equilibrium conditions (sum of forces in all directions equal zero) at each joint, the internal forces in the adjoining members can be determined, allowing for an accurate assessment of the load path through the structure.

Transcript

00:01 For this problem on the topic of analysis of structures, we are shown a trust in the diagram and we are told that it is supporting an advertising panel.

00:10 We want to find the force in each member of the truss for a load equivalent to the two forces shown.

00:17 We want to state whether each member is also in tension or in compression.

00:22 So if we look at the free body diagram for the entire truss and we take anticlockwise moments to be positive, if we take the sum of the moments about point f, this must equal to zero, since the system is in equilibrium.

00:40 And these moments are 800 newtons times 7 .5 meters plus 800 newtons times 300 newtons times 3 .75 meters minus a times 2 meters must equal to zero.

01:06 Which means that we can find a, and we have the magnitude of a to be positive 2 ,250 newtons.

01:16 So the vector a is 2 ,250 newton's vertically upward.

01:26 Now using this free body diagram, taking the upward direction as positive, we can also take the sum of the forces in the y direction, this is also equal to 0, since there's no net motion in the y direction, and hence we get 2 ,250 newtons, plus the y component of force f, f, fy, is equal to zero.

01:53 This means that fy is obviously minus 2 ,250 newtons, or 2 ,250 newtons vertically downward.

02:07 Lastly, if we take the right directions as positive and sum the horizontal components of the forces again we need to get zero and so we get f x or if we perform this calculation we get this to be minus 800 newtons minus 800 newtons plus the x component of f is equal to zero which means that the x component of f is simply positive 1 ,600 newtons.

02:48 And so fx is equal to 1 ,600 newtons, and that's to the right.

02:58 That's the vector.

03:00 Now let's look at joint d.

03:05 So we've drawn a free body diagram for joint d as circled here.

03:11 Now from this free body diagram for joint d, we get the following.

03:16 At joint d, we have 800 newtons divided by 8 with some simple geometry is equal to fde over 15, which is equal to fbd over 17.

03:41 This allows us to find two of the forces that we require immediately.

03:45 So we get fbd to be 1 ,700 newtons, and that's b.

03:54 Compression and we get f, de to be 1 ,500 newtons of tensile force.

04:05 So we found two of our forces that are required.

04:12 Next, we look at joint a and we'll draw a free body diagram for joint a as we have here.

04:23 And from this free body diagram for joint a, we have the following by geometry, that 2 ,250 newton's divided by 15 is equal to f .a .b.