Determine the force in each member of the Gambrel roof truss shown. State whether each member is

Determine the force in each member of the Gambrel roof truss...

Key Concepts

Static Equilibrium

In any structural analysis, particularly in trusses, the principle of static equilibrium is fundamental. This concept states that the sum of all forces and moments acting on a body must be zero, ensuring that the structure is at rest or moving with a constant velocity. This principle is used to set up equations for the forces in each member of the truss, ensuring that at every joint the forces balance in both horizontal and vertical directions.

Method of Joints

The method of joints is a common technique used to analyze truss structures. It involves isolating each joint and applying the conditions of static equilibrium to solve for the unknown forces in the connecting members. This approach simplifies complex truss problems into a series of two-dimensional equilibrium problems at each joint, making it easier to determine whether each member is subjected to tension or compression.

Method of Sections

The method of sections is another technique used for analyzing trusses. Instead of solving for forces at every joint, this method cuts through the truss to isolate a section and then applies the equilibrium equations (sum of forces and moments) to that section. This method is particularly useful for finding forces in specific members without having to analyze the entire structure.

Tension and Compression

Understanding the difference between tension and compression is crucial in truss analysis. Tension refers to forces that pull or stretch a member, working to elongate it, while compression refers to forces that push or compress a member, working to shorten it. Identifying whether a member is in tension or compression helps in determining the appropriate design and reinforcement needed for the structural members.

Transcript

00:01 Now this is the question this diagram is given in the test book so here there is two things we need to note in point h so in point h we can see there a support and we know that it is a pin support so we know the special to pin support will produce creates two forms, two reactions card, you can call it's reactions.

00:44 That is a vertical and a horizontal.

00:52 So let's call the vertical as rh and this create horizontal also.

01:05 So let's call the horizontal say it's it's.

01:13 Edge and the next is point a there is a support and we can see that support is a roller support and we know the specialty of roller support that also creates some reactions and due to this roller support there will be a vertical force so there will be a vertical force and let's call that vertical force as r a as the point is here let's call it as r a now let's denote the force here itself therefore we can complete the free body diagram so we know to calculate the forces in each member of the trust we need to draw a free body diagram of end up and apply equilibrium equations.

02:22 So let's draw the forces.

02:25 In h, there will be vertical force rh and as well as a horizontal force, say h, and in a, because this is a lower support, because of this support, there will be a vertical force or a okay now let's calculate the support reaction so from the diagram we can see this is a symmetric one and all the forces everything is symmetric here the 300 lp is acting vertically the same likewise a 300 will be sanctimatically downward here also so it's almost symmetry so we can say the r a ring will be equal to rh and we can calculate this that will be half into 3 into 600 and 4 into 300 that is 1 ,200 kilo newton and both rh and r is adding inward.

04:13 That is that will create a false compression.

04:18 That will create a compression and i denoted the compression by the letter c.

04:25 Now let's go to each joint, consider each joint and so.

04:34 So the entire truss.

04:40 Now to get forces in each member of trust, let's start by solving joint a.

04:49 Let's take it right for joint a.

04:52 So for joint a, for joint a, we can draw the free body diagram.

05:06 Let's consider this point as joint a.

05:09 There will be a force r a that thing vertically upward and there will be a 600 force 600 this is 600 and there will be fab acting downward and there will be a fac fac and we know the distance here it's given 8 feet and this is given as 6 feet we are going to just take the ratio that is this will be 4 and this will be 3 that makes the solving more easier so here it will be 4 and here it will be 3 that makes this hypotenuse it's 5 let's solve for joining.

06:33 Let's take the summation along why.

06:38 This will give us r .a minus fab, fab, sin theta.

06:49 If it's sine theta, we know the sign is opposite by hypotenuse.

06:56 Here, the opposite is 3 and hypotenuse is 5.

07:00 So we can write it as 3 by 5 and there will be minus 300 also that is here there will be 600 and 300 there will be a 300 and my axis so let's see and we know the value of ra that is we calculated the value of ra that is we calculated the value of ra that we got as 1 ,200 so therefore 1 ,200 minus f a p.

07:59 3 by 5 minus 300 will be equal to 0 so i are just equating this to zero so from here we can calculate and we will get the value of fap they can write the value fab as 1 ,500 kilo and fab is going inward so that creates a compression let's denote it by the letter c now let's take the summation along x long x there is f a c minus fab cost theta we know the cost theta we know the cost theta is archeson by hypotenuse so here there will be 4 by 5 equal to 0 we just calculator the value of fab here that is 1 ,500 so from here we will get the value of fac as 1 ,200 kilo and you can see fac is going outward from point a that creates a tension and add in order to the letter t.

09:50 Now let's move to the next joint.

09:54 Let's take green color and move to next joint c.

10:02 So for joint c, let's draw the body diagram.

10:14 Consider this is joint c.

10:15 There will be f -w's, f .c, matching this direction and there will be a force fca acting this direction and there will be a force f .c .b.

10:34 I think this direction and there will be a force f .c .b.

10:41 I think vertically upward.

10:45 Now that's all.

10:47 Let's take summation of l of y and we can see there is no force else.

10:55 Else than so there is only fcb along my axis so this will be fcb equal to zero so here we found fcb is equal to zero because bc is a zero force member so like that makes bc is a zero force now let's take x -axis.

11:43 Along at such is there is fce and fac or f -a or the f -c -a -o the f -c -a -m -t -a -m -t -a will be equal to zero and we just found the value of facc that is 1 ,200 kilo, and therefore we will get the value of fce, fce will also be 1 ,200 kilo and we can see fce is going outward from the point that creates a tension and it in order by the letter t.

12:37 Now let's move to the next joint and we can see then next join let's take it as join b so for join b let's take the color do yeah join b let's draw the diagram for join b auto force in there is it's its center will be writing downward and fba and fba fba okay there will be f .c acting over and f.

14:00 F .e.

14:02 Making some anger and at last fpd...

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