Determine the force in each member of the truss shown. State whether each member is in tension o

step 1 In this question we are given this trust. So we are asked to find a force in each member and sta

Determine the force in each member of the truss shown. State...

Key Concepts

Tension and Compression

Tension and compression are the two primary types of forces experienced by truss members. A member in tension is being pulled apart, while a member in compression is being pushed together. Identifying whether a member is in tension or compression is crucial for understanding its behavior under load and for designing safe, efficient structures.

Static Equilibrium

Static equilibrium is a fundamental concept in structural analysis where the sum of all forces and moments acting on a system is zero. This principle underpins the calculations in truss analysis, ensuring that the structure remains stable and that each member's force contributes appropriately to maintaining this balance.

Truss Analysis

Truss analysis involves evaluating statically determinate structures composed of straight members connected at joints, usually assumed to be pin-connected. This analysis is used to calculate the internal forces acting in each member when the structure is subject to loads, thereby ensuring the structural integrity under various loading conditions.

Method of Joints

The method of joints is a technique for analyzing the forces in each member of a truss by isolating a joint and applying conditions of equilibrium. By ensuring the sum of forces in both the horizontal and vertical directions equals zero at each joint, the forces in the connected members can be systematically determined.

Method of Sections

The method of sections involves cutting through a truss to isolate a section of the structure and applying equilibrium equations. This technique is particularly useful for determining the forces in specific members without having to analyze the entire truss, making the process more efficient for large structures.

Transcript

00:01 In this question we are given this trust.

00:03 So we are asked to find a force in each member and state whether each member is in tension or compression.

00:13 So we are analyzing the free body diagram for each point or the whole trust for this whole problem.

00:25 Okay, so the first thing we will look at the free body diagram of the end.

00:31 Entire trust.

00:35 We want to determine the force at h and f.

00:40 Okay, so there will be a force pointing up at h, and then there will be a force at f, f, y, and f, y, and f x going to the right.

00:56 Okay, so and the problem will be solved by looking at summation of forces along the x and along the y to be zero and also the moments okay okay so if you look at summation of forces along the x to be zero okay just tell us that f x okay would be zero okay and then summation of forces along the moment at point fx it should be 0 so we have h to be 5 kilo nton upward k yeah so and then we have some general forces along the y equals to 0 which says that uh f y or f y plus h is equal to 5 kilo newton since we found that h is 5 kilo newton upward.

02:13 So fy will be 0.

02:20 So there is, there's no force at f, along the x and along the y direction.

02:28 Okay, next we look at the freebole diagram for joint c.

02:41 So this is how the diagram looks like at c.

02:44 There's a 5 kilo newton force down, and then c is connected to b.

02:51 And it's also connected to e.

02:55 Okay, and this whole thing is in equilibrium.

02:57 So, okay, so we are going to have f -c -b that points to the left and f -c -e that points towards c, okay? and then this is 5 kilo newton, and then this is 45 degrees, okay? alright, so by looking at the diagram, you can tell that fcb, because of the 45 degrees, right? yeah, so fcb is equal to 5 kilo newton.

03:39 So this one is in tension.

03:43 Yeah, because c is pulling on the, so there's a force that points towards b.

03:50 So means that at point c is pointing c is pulling the rod okay so fcb is in tension okay then fce okay if you use pyagoras theorem 5 root 2 and then you get 7 .07 kilo -newtons and this one is in compression okay right so we have a few answers already okay okay so you move on by symmetry okay about the horizontal center line okay what means is that so we have cb and here is h and g hey by symmetry right we have f hg to be f h g to be 5 kilo newton yeah this is a center line okay intention and then f h .e is equal to 7 .07 k.

05:47 Newton in compression.

06:05 Okay.

06:06 Next for joint e.

06:09 Okay.

06:15 So members line to intersecting trade lines.

06:34 Okay, we have f .e .g.

06:38 It goes to fec.

06:42 It goes to 7 .07, kload newton's compression.

06:50 Then feb equals to feh equals to 7 .07 kilo -newtons and also in compression...