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

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

Key Concepts

Truss Structures

Trusses are frameworks composed of straight members connected at joints to form a rigid structure. They are designed to efficiently distribute loads through triangular units, providing stability and strength in bridges, roofs, and other structures.

Static Equilibrium

This principle states that for a system in equilibrium, the sum of forces and moments acting on any part of the structure must be zero. Applying equilibrium equations (?Fx = 0 and ?Fy = 0) at each joint or section is fundamental to determining the forces in individual truss members.

Method of Joints

The method of joints involves isolating each joint in the truss and applying the equilibrium conditions to solve for unknown forces in the connecting members. This method is particularly useful for systematically finding the force in every member by working through one joint at a time.

Method of Sections

An alternative analytical approach that involves cutting through the truss to analyze a section and directly calculate the forces in the members of interest. This method uses equilibrium conditions on the isolated section and is often more efficient when only a few member forces need to be determined.

Tension and Compression

Determining whether a truss member is in tension or compression is crucial as it affects the design and behavior of the structure. Members in tension are being pulled apart, while members in compression are being pushed together, and each condition requires different design considerations.

Transcript

00:01 So here on this problem we are given this trust and we have to calculate the force in each member and we have to also state whether the member is in tension or compression so for that i will be calculating the angle this angle theta and alpha first so angle theta would be equals to if you consider the triangle a d e so by using this triangle the angle theta would be equals to tangent inverse this distance which is 2 .4 divide the triangle by this total distance which is 7 meter so from here theta comes out to be as nearly 18 .92 degrees alpha would be equal to for that will be considering the triangle c -d -e so alpha would be equals to tangent inverse this is 2 .4 divided by the base which is 3 .2 so from here the angle alpha comes out to be as 36 .86 degrees next will be caliq the reactions so that will be using the equilibrium condition the first one is sigma fx equals to zero now and on the trust there is only one force which is acting in the horizontal direction that is x a x so a x would be equals to zero now due to symmetricity of the truss and the loading we can say that the reaction force a y should be equals to the reaction force gy and it should be equals to one by two of the total loading which is acting so this is 5 .7 plus 10 .5 plus 9 .6 plus 10 .5 plus 5 .7.

01:52 So from here, the reaction force, a .y, which is equals to gy and it comes out to be as 21 kilo newton.

02:02 Now we have calculated the reaction forces and the angles theta and alpha.

02:07 Now next we'll be considering the free body diagram of the joint a.

02:11 So this is a free border diagram of the joint a and we have considered that the force acting in the member a b would be acting like this that means the member would be in compression and here we are assuming it to be in this direction and if the value comes out to be as negative that means our assumed direction would be wrong it would be acting in the opposite direction so we'll be calculating the forces now for that we can use the equivalent condition the first one is sigma f5 is equals to zero so we'll be equating the forces the y direction and the forces in the vertically upward direction will be taking positive and in downward direction will be taking negative similarly in x direction in this way we'll be taking positive and is this direction will be taking negative now 21 is acting in the upward direction so we'll take 21 as positive this is acting the downward direction so minus 5 .6 here if you take the component of this force in the vertical direction and that would be acting in the upward direction so we will take a positive sign so fab sign theta so theta we have 18 .92 and it is equals to zero so from here fab that is the force in the member ab comes out to be as negative 47 .18 kilo newton that means our assumed direction is wrong the force in the member ab would be acting like this that means a member would be in compression so we can say that the force in the member ab is 47 .18 kilo newton and the member is in compression so is the force and the member ab next i will be calculating the force in the member ac for that we can use equilibrium condition sigma fx is equals to zero now we'll be creating the forces in the x direction so we have f a b cosine 18 .92 plus f ac is equal to zero we have just followed this sign convention now can put the values this one is minus 47 .18 cosine 18 .92 plus f ac is equals to zero so from here the force in the member ac comes out to be as 44 .6 kiluniton and since the force comes out to be as that means our assumed direction is correct and the member would be in tension now next i will be considering the free body diagram of the joint b and here do we know that the member the force in this member would be acting like this and in this member we are assuming that the force is acting like this and in this member we are assuming that the force is acting like this so this is a free border diagram of the joint b now we'll be using the equivalent condition that is sigma fx equals to 0 so let me write this sigma fx is equals to 0 and will be following the same sign convention so we have f bd cosine theta so theta we have 18 .92 degrees plus 47 .18 cosine 18 .92 is equals to 0 so from here the force in the member bd comes out to be as negative 47 .18.

05:54 Means our assumed direction is wrong the force in this member would be acting like this or we can say that the force in member bd is 47 .18 and the member is in compression now next we'll be using sigma fy equals to zero so sigma fy equals to zero and we'll be following the same sign convention so we have a minus 10 .5 because it is acting in the downward direction so we've taken a sign similarly minus f b c so plus we have to take the vertical component of this force so plus f bd cosine theta and theta we have 18 point 92 degrees and then plus f a b cosine for sorry this is 18 .92 degrees and it is equals to zero now you can put the values here if bd we have calculated as minus 47 .18 so we can put it here this is minus 47 .18 and fab we have this much so this is 47 .18 so from here we can calculate the force in the member ab no sorry bc and it comes out to be as negative 10 .5 kilo newton therefore our assumed direction is wrong and the member would be in compression and the force would be acting like this so this is a force in the member bc or we can write in this also like force in member b c is equal to 10 .5 and the member would be in compression now next will be considering the freeboard diagram of the joint c so this is a freeboard diagram of the joint c and here we have assumed that the force in the member cd would be acting like this and in c e it would be acting like this and in this member we know it is acting like this now if the value comes out to to be as negative that means our assumed direction is wrong otherwise our assumed direction is correct so we'll be using the same equilibrium condition the first one is sigma f5 equals to zero so we're balancing the force in the y direction so we have f cd this is sign alpha minus 10 .5 is equals to zero and alpha we know it is 36 .86 so this is 36 .86 degrees now from here the force in the member cd it comes out to be as 17 .50 kilo newton and since the value comes out to be as positive that means our assumed direction is correct and the member would be in tension now next we'll be considering the forces in the x direction so sigma f x is equals to zero and will be following the same sign convention so we have f c e plus 17 .5 cosine alpha minus 44 .625 is equal to 0 so from here in alpha we can plug it here this is 36 .86 so from here the force in the member c comes out to be as nearly 30 .625 kilo newton and since the value comes out to be as positive that means our assumed direction is correct and the member would be in tension.

09:40 Now next we'll be considering the free body diagram of the joint so this is a freeboard diagram of the joint a and i will be calculating the force in the member d .e and since there is only one force which is acting in the in the vertical direction the force in the member d .e would be equals to zero...

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