The maximum allowable tensile force in the members of the truss is (Ft)max=5 kN, and the maximum

step 1 We'll be looking at structural analysis. For the problem at hand, we want to determine at the hi

The maximum allowable tensile force in the members of the...

Key Concepts

Truss Analysis

Truss analysis involves the study of structures composed of slender members connected at joints, where the forces are carried axially. This methodology typically assumes that joints behave as ideal hinges, so only axial tensile or compressive forces are present. The analysis determines the internal force distribution in the members when loads are applied at various points on the structure.

Member Force Limits

Each member in a truss is designed to withstand only a certain maximum axial force, whether in tension or compression. These limits are crucial to prevent material failure or structural instability. The design and analysis ensure that the computed forces in each member remain within these allowable limits, which are determined based on material properties and safety considerations.

Equilibrium Conditions

The principle of equilibrium is central to structural analysis, where the sum of forces and moments in a system must equal zero. In a truss, applying equilibrium conditions at each joint or using methods such as the method of joints or sections allows the computation of the axial forces in the individual members caused by the external loads.

Limit State Design

Limit state design is an approach used in structural engineering to ensure that structures perform adequately under maximum expected loads without reaching failure. This involves verifying that the forces produced by applied loads do not exceed the maximum allowable compressive or tensile forces in any of the truss members, thereby ensuring safety and structural integrity.

Transcript

00:01 We'll be looking at structural analysis.

00:02 For the problem at hand, we want to determine at the highest value of p that can be applied to this structure.

00:11 Now to solve this problem, we will apply the method of joints to solve the method of joint.

00:18 But before, then let us look at the support reaction.

00:21 So if we take summation of forces of moment rather at b, it will be equal to 0, they are going to have minus p, times 3 plus arrow a reaction at a times 2 minus p times 1 so if we solve this out we are going to have a reaction at day to be equal to 2 so if we consider joint a joint a we have this you have this have this one that is these are joint a we have this force that is coming up reaction that we determine which is a 2p r a r a a short of 2p they will have these uh so the forces should be going out of the joint except to the lamatory coming in like our reaction at a there so this is force f so should be going from joint b to d to b sorry so if we take summation of forces in the take assumption of forces in the horizontal direction they are going to have minus apsuit a d to be as a sign now that angle there is inclined to 30 degrees okay this place is but now member this particular force is inclined to 30 degree there okay so take note of take note of that to trace it out like this sync line to 30 degrees so we use the information given in the diagram to solve that to determine the angles okay so sign 30 degrees plus f a sub which ab f a source which a b is equal to zero so when we do this we are going to arrive at f source with ab equal to 0 .5 f source with ad.

03:44 Okay, so we can continue for take summation of forces.

04:00 Summation of forces in the horizontal, in the vertical direction, you are going to have f substrates ad, or 30 degrees, plus r.

04:25 The reaction at that reaction we got to that place so we solve this art you are going to have find this art i'm going to have sos with ad equals to 2 .31 2 .31 2 .31 3 1 2 .31 p or my 2 .31 p so you are going to have to be 2 .31 p it's a member on sorry, kilo newton, sorry, kilo newton, kilo, kilo, kilo, kilo.

05:25 Now this is a force, the member under compression, fine.

05:41 So let us look at, now, haven't gotten this, we can determine epsop with a, b, using that equation, this equation as, it's going to be half of this e, d, so it's going to give us equal to 1 .16p, okay, kilo newton.

06:11 Kilo -leutin, actually it's negative.

06:18 So it's going to give us 0 .16, 1 .16p to compressive force.

06:44 So if we take a summation of, sorry, we'll look at another joint.

06:50 This will look at joint d, call it jd.

06:56 We are going to have this force b coming down, it's there at joint d, this force.

07:06 So the forces will be going out of the joint assumption and then have this one, f substrate, db, we have this one, f sos 3.

07:28 This is 30 degrees, using formation in the diagram question to solve, remind these ones.

07:40 This is f source with d .a.

07:47 Okay.

07:49 Of course this is joint d that we're under consideration.

07:53 So if we look at some function of forces in the horizontal direction, okay? we are going to arrive of half.

08:11 So if we take some function of forces in that direction, okay? forces in that direction.

08:25 We're going to arrive at f source with...

08:35 D a sign 30 degrees plus f source with d b cause 30 degrees plus plus f with dc to zero so you can also look at solution of forces in the vertical direction solution of forces in the vertical direction assumption of forces in the vertical direction if we'll do that we're going to have minus f source with d a cost 38 sorry 30 degrees plus f source with d b minus p is equal to zero so if we solve this making f subp b the software the formula you can get s of d a to be equal to uh 2p plug in your values of cosine sine 2p kilo newton and you can see that it's positive extension under the member under tension then we can look for epssotc using the relation we've got in earlier on as of dc to be equal to 0 .5 the f sosr d -a minus 0 .866 so you just plug plugging the values of your sign and plusine or ever into our area equation.

11:48 They're going to have this one to be equal to 2.

11:54 Sorry, this one to be equal to minus 2 .889p.

12:12 Okay? so you're going to have it to be 0 .89.

12:28 You can see that it's a member under tension it's positive okay sorry so there's no need to fix this one we're gonna have a positive value in the end so we can also use this same sorry f um so with dc okay this sos with dc x sos with dc sos with d c sos with d b a, s with dc, okay, we had an exercise with d, b, or s with d, a was, okay, yes, that's the value then our sotc was 2 .8, 2 .89p, okay, 2 .89p...

Please give Ace some feedback