The truss shown consists of 18 members and is supported by a ball-and-socket at A, two short lin

step 1 So, this is the figure unit that's book. Here, this support creates a reaction, GY. So let's dra

The truss shown consists of 18 members and is supported by a...

Key Concepts

Simple Truss

A simple truss is a structure composed entirely of straight members connected at their endpoints, typically by pin joints, so that the loads applied to the structure are transmitted exclusively as axial forces (tension or compression) within the members. This concept is foundational in structural engineering and simplifies the analysis since moments at the joints are assumed to be negligible.

Statically Determinate Structure

A statically determinate structure is one where all unknown forces and reactions can be solved entirely using the equilibrium equations of statics, without the need to consider the deformation compatibility or material properties. For trusses, checking that the number of members, joints, and support conditions balance with the available equations of equilibrium is essential to ensure the structure is fully constrained and can be analyzed using purely statics.

Support Conditions and Constraints

Support conditions define how a structure is connected to its surroundings and play a critical role in constraining the structure. In truss analysis, various types of supports (such as pin, roller, or fixed supports) offer different degrees of restraint, and the arrangement of these supports must provide the necessary reaction forces to balance the applied loads. Properly identifying and understanding these supports is vital in ensuring that the structure is stable and statically determinate.

Equilibrium Equations in Truss Analysis

The analysis of trusses is founded on the application of equilibrium equations, typically summing forces in horizontal and vertical directions as well as moments when needed. These equations are used for both the overall structure (to find reactions at the supports) and for individual joints or sections (to solve for internal member forces). The proper use of equilibrium equations is critical in determining the axial forces in each member and verifying the overall stability of the truss.

Force Determination in Truss Members

Determining the force in each truss member involves analyzing the interactions at the joints (using methods such as the method of joints) or by cutting through the truss (using the method of sections) to isolate a portion of the structure. This process allows engineers to understand how loads are distributed among the members, ensuring that each member's capacity is not exceeded and that the structure behaves as expected under the given loading.

Transcript

00:02 So, this is the figure unit that's book.

00:05 Here, this support creates a reaction, gy.

00:14 So let's draw this.

00:18 Because of this support, there will be a reaction.

00:21 Let's name it gy.

00:24 Okay, yeah.

00:25 And because of this, there will be a reaction.

00:31 Let's name it b -y.

00:35 And this, it's a reaction.

00:39 Reaction let's name it b z b z okay and here this is a principle so there will be reaction along its y and saddazis so the reaction of said that's is is named as a z and z and the y is named as a y and the y is named as a y and the x is named as a .s.

01:27 Okay? yeah.

01:29 The distance is given in feet.

01:32 So let's note your problem.

01:35 So for a space truss to be a symbol trust, it starts with a tetrahedron truss.

01:43 And any added three members attaching three existing joints with a new joint.

01:50 So in the free body diagram showing this one, let's start with tetrahedron b e fg so let's start with b e fg so i repeat for a space trust to be a simple trust it starts with a tetrahedron trust and any added three members attaching three existing joints with a new joint okay yeah so so let's start with b fg here.

02:33 First the adder members b d and e d and g and g attaching three existing joints b b and g and g and the next one be the added numbers b c and d c and g c attaching three existing joints b the and g with the new joint new joint c okay then next will be that members b and e h and g h attaching three existing joints d, e and g with new joint, new joint, h.

04:27 Okay, the last will be, the fourth one will be that members b, d, a, and e, and e, and e a, attaching three existing joints, b e d b e and d with new joint a new joint a new joint a so therefore here the trust is a simple trust like therefore trust is a simple trust is a simple trust so if the address is completely constrained from the free body diagram shown in here, we have six unknown reactions, a.

05:46 The a.

05:48 A .h.

05:49 A.

05:49 Z, a.

05:50 Z, a.

05:51 Z, a.

05:51 Z.

05:51 A.

05:51 Z.

05:52 A.

05:52 Z.

05:52 B.

05:54 Z.

05:54 B.

05:55 Z.

05:56 And g.

05:56 And the g.

05:58 So we can determine the six unknown reactions by applying equilibrium of momentaryx.

06:03 About member a, b and joined g.

06:08 Therefore, the rest is completely constrained and its reaction are statistically determine, determinate.

06:17 So first, yeah.

06:25 So first to determine the force of the six members joined at e, we need first to determine the reactions of the enderrorist.

06:33 So, force is in vectors.

06:36 So the f h will be then 275 lb along i okay minus the 240 will be around k okay and then we can find the fg hetr fg will be equal to minus of g y along j along j and you can find the vector fb will be equal to the b the y times so b y along j minus b set along k okay now then the vector f a let's find a fair also f a will be a a a a lot of i plus the a way along j minus the a z lm k okay now the summation of f equal to zero so now let's draw the diagram first so we already yeah so for joint a here let's find the components okay yeah the component at a direction to find the component at a direction we had to find summation a lot that's equal to zero that gives you 275 lb 275 lb plus a it's equal to zero so here from here therefore a x will be equal to minus of 275 lb okay and component at j direction so to find the component that direction you need to find some measure that y equal to zero that gives you a y plus b y minus gy equal to zero then the component k direction so the component at k direction means the summation along z z -dire direction equal to zero that gives you that to 40 will be minus a z minus b is it minus b is it equal to zero okay now you got two equations let's move to the momentum of ab equal to zero so this gives you sorry 9 .6 feet times gy minus 10 .08 feet times 240 lb equal to zero so here that known is gy so the value of g way will be 252 lb okay yeah no summation i know a motion of momentum of 80 equal to zero so the 11 feet into b y minus the 11 feet into g y minus 10 .0 .m .0 8 feet into the 2750 lb okay 2nd 5 0lb equal to 0 so from here we got the wire of gway substitute the wire of gway in this equation that gives you the value for b way so b way will be 5 node for nb okay so substitute b way and g y in equation one like in this equation let's name it as let's name it as equation number one and let's name this one as equation number two okay so substitute the value of b y and gy in this equation that gives you the value for a y that will be so the value for a way will be minus of 250 to lb okay and now let's find summation of momentum at cg equal to zero so 11 feet into b y minus 11 feet into a z minus 9 .6 feet into a x minus 9 .6 feet into a x equal to zero so you know the value of a h and b y so that value here that gives you the value for a is it that will be 0 and substitute the value of a is it substute the value of a is it in the equation number two that gives you the value for b is it so the b is it will be minus of 240 lb okay, we got the b is it also so after determination of the reactions use method to get the forces at members join that e so let's start from the joint a yeah so for joint a let's draw the diagram first so if this is joint a there will be f a b f a d f a e h and a y okay, let's denote the a axis...

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