Determine the force in members B C, C H, G H, and C G of the truss and state if the members are

step 1 We're looking at structural analysis. We want to consider this problem. Here we need to determin

Determine the force in members B C, C H, G H, and C G of the...

Key Concepts

Tension and Compression

Tension and compression are the two types of axial forces experienced by members in a truss. Tension refers to forces that act to elongate or stretch a member, while compression refers to forces that act to shorten or compress a member. Understanding whether a member is in tension or compression is critical because it influences the design considerations and potential failure modes of the structural elements.

Static Equilibrium

Static equilibrium is the condition in which a structure or object remains at rest because the vector sum of all forces and moments acting upon it equals zero. This principle is essential in structural analysis as it allows engineers to set up equations based on force and moment balances, ensuring that every joint or section of the truss can be analyzed for stability and internal force distribution.

Method of Joints

The Method of Joints is a fundamental analytical technique used in truss analysis. It involves isolating a joint in the truss and applying the equilibrium equations (typically in the horizontal and vertical directions) to solve for the unknown forces in the connecting members. This method is particularly useful in determining whether members are in tension or compression.

Method of Sections

The Method of Sections is another key analytical technique in truss analysis that involves cutting through the truss to expose internal forces in several members simultaneously. By applying equilibrium equations to one of the resulting sections, engineers can solve for the unknown forces in the cut members, making it a valuable approach when specific member forces are required without analyzing every joint.

Transcript

00:01 We're looking at structural analysis.

00:02 We want to consider this problem.

00:05 Here we need to determine the magnitude of the forces in the members of the structure as well as their nature.

00:14 All right.

00:15 We'll apply the method of joint to solve this problem.

00:20 But we need to look at the support reactions at a as well as at b.

00:30 So let's take moment.

00:33 We take moment about e.

00:39 Go to zero summation of moment.

00:41 So let's use arrow a, so the vertical component of the support reaction after a.

00:56 So times 16, minus 4, times 12, minus 8 times 8, 5, 5, is equal to zero so we have arrow circuit y have that reaction there vertical reaction to be equal to 8 .25 kilomuting using summation of vertical forces okay you can arrive at get this support reaction at the roller end okay which is at e so it's going to give us a arrow soft of e at that end to be equal to 8 .75 to the muting now let's take a joint let's say joint a do that we have this force coming like this one entering the joint the reaction there is that joint a there's a force going now the forces should be going out of the joint you are considered based on assumption that join a going to be that member ab this is a source with ah h going to join h from a now this angle is 36 .9 degrees using information given in the diagram of the question to solve the combine those angles.

03:19 So if we take summation of forces, summation of forces in the the horizontal direction, we are going to have f, softwit ab, ab plus f source with a h cost 36 .9 k plus does f subswritz a h cost 36 .9 degrees? equal to zero okay so that's that for that then we can also take someption of the forces in the vertical direction to do that we'll have 8 .25 that force that reaction there at a so shan sign 36 .9 degrees is equal to zero so plugging in values okay or making from you can even make this one, this force the, sort of the formula and you arrive at this value, 13, negative 13, negative 13 .74 kilo newton.

05:54 You can see that it's a member under compression, this particular member, h, okay? the magnitude is this, the sign tells us that.

06:04 Then we can take, we can look at what will, just from there we can just get our f -a -b okay from the equations that we are already in the x direction you can get our f -a -b to be equal to 10 point login values the value of f sub a h there you're muting you can see that this is positive it's giving us telling us that this member is under tension okay all right we can look at joint b so if we take joint b called it j j b we have it it looks like this joint is a joint b we have some forces coming down here if you check the diagram of the problem kilo newton okay and there's a force up there is f substrate ph going from joint b to h, we join h, the member bh, this one is b, a, and there's another force here, earth, suc, b, c, to that member b, c.

07:49 Alright, if we take summation of the forces in the y direction, you are going to arrive at f, source with bh minus 4, equal to 0, so that sosw, such with b .h.

08:21 Give you 4 kilo newton tension.

08:27 So it's a member under tension.

08:34 Then you can also take summation of forces in the horizontal direction.

08:43 Okay? take forces, some sort of forces in the horizontal direction.

08:50 If you do that, you'll have minus f with b .a.

09:01 B .a.

09:03 Plus s.

09:07 S.

09:07 S .c.

09:11 Equal to zero.

09:12 And of course this force they are going to be equal okay therefore f epsuch with b c is equal to the value is equal to 10 .99 phd you can see that it is positive it's a member under tension now we can look at joint h we look at joint h we have joint h j h have this force coming like this okay and we have this one have this force have this force okay this is our joint h okay now there's an angle here is actually inclined to our 36 .9 degree 6 .9 degree okay this one is inclined to another angle there an angle there's an angle there 26 .6 degree so use the information given in the in the diagram of the question to to figure out to get the angles so this force is f substitute hg okay now this angle is also this this angle is also here so this angle is 36 like 9 degrees so this is f subscript hc this is f, suss with h .a.

11:34 This is f, suss with hb...

Please give Ace some feedback