Determine the force in each member of the Warren bridge truss shown. State whether each member i

Determine the force in each member of the Warren bridge...

Key Concepts

Static Equilibrium

This concept refers to the condition where all the forces and moments acting on a structure balance so that the structure remains at rest. In structural analysis, especially for trusses, it is essential to apply the equilibrium equations (sum of forces in the horizontal and vertical directions, and moments) to ensure that all forces calculated accurately reflect a state of balance.

Free Body Diagram

A free body diagram (FBD) is a simplified representation of a structure or a section of a structure isolated from its surroundings, which shows all the applied forces, reactions, and moments. For truss analysis, drawing the FBD helps in visualizing the individual forces acting on each joint or member, making it easier to apply equilibrium equations.

Method of Joints

The method of joints involves analyzing each joint of the truss separately to solve for the forces in the connected members. By ensuring that the sum of forces in both horizontal and vertical directions at each joint equals zero, one can determine whether a member is in tension or compression. This method is critical for understanding how loads are transferred through the structure.

Method of Sections

The method of sections involves cutting through the truss to expose the internal forces of a limited number of members and then applying equilibrium equations to this cut section. This approach is especially useful when only a few member forces are required, allowing for a more direct calculation without having to analyze every joint.

Tension vs. Compression

Tension and compression are types of axial forces acting within structural members. A member is in tension when it is being pulled apart, leading to elongation, while it is in compression when it is being pushed together, leading to shortening. Identifying whether each member is in tension or compression is crucial for assessing the stability and load-bearing capacity of a truss.

Truss Geometry

Truss geometry refers to the arrangement and configuration of the truss elements, including the shape and size of the individual members and the pattern of their connections. For instance, the Warren truss is characterized by a series of equilateral or isosceles triangles that distribute loads efficiently. Understanding the geometry is essential for determining the force paths and accurately calculating the forces in each member.

Transcript

00:01 So here in this problem we are given this truss 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 to solve this problem will be first calculating the reactions that is a x a y and gy so for that we'll be using equilibrium condition the first one is sigma fx is equals to zero now we have only one force in x direction this one is x direction and this one is the wide edge so we have a x only and that would be equals to zero next i will be considering the equilibrium forces in y direction so sigma fy equals to zero now the forces in the upward direction will be taking positive and the forces in the downward direction will be taking negative so we have a y plus gy minus six minus six is equals to zero so from here y plus gy would be equals to 12 and we can take moment about point a so sigma m a should be equals to zero and the moment in the anti -clockwise direction will be taking positive and the moment in the clockwise direction will be taking negative so we can take the moment now so this is this is making moment in the clockwise direction so minus six times the perpendicular distance is 18 again minus six times 18 plus 18 is 36 this total is 36 now plus gy times 18 into 3 is equal to 0 so from here gy comes out to be as six gifts and we can plug gy here to calculate a y and from here a y would be equals to 12 minus 6 is 6 now we have calculated all the reactions now next it will be considering the free body diagram of the joint a so the this is the free boarded diagram of the joint a now we have to calculate the force in the member ab and ac now we have assumed that the forces acting at joint a would be like this that means a member would be in tension now if the value comes out to be as negative that means our assumed direction is wrong and the member would been compression so let's find out the values for that will be considered the equilibrium condition the first one is sigma f y equals to zero and we'll be following the same sign convention as we followed earlier so we can write we have take the vertical component of the force fab so it would be four by five fab plus six is equals to zero so from here fab comes out to be as minus 7 .5 now here minus and notes that our assumed direction is wrong the direction would be like this that mean the member would be in compression so we can say that if a b will be equals to 7 .5 and the member would be in compression now next i will be calculating fac for that we can use one equilibrium condition that is sigma fx is equals to zero and the forces in this direction will be taking positive and the forces in this direction will be taking negative so we have f a b we have to take the what the original component of this force so 3 by 5 plus f a c would be equals to 0 so 3 by 5 f a b we have calculated as 7 .5 means 7 minus 7 .5 plus f ac would be equals to 0 so from here if ac comes out to be as 4 .5 kips and since the value comes out to be as positive so the member would be intention and our assumed direction is correct so these are the two forces in the member ab and the next one is the member ac and the direction would be like this in ac it was correct and in ab it would be like this now next i will be considering the free body diagram of the joint to be so this is the free body diagram of the joint b now we know that force in the member a b it was compression the member would been compression so force would be acting like this now next we have assumed that the force in the member b would be like this the member would be in tension and here also we assume that it to be like this that means the member would be in tension now here if the value comes out to be as negative that means we can conclude that our assumed direction is wrong and the force would be the member would be in compression so let's start for that again we'll be using the same equivalent condition the first one is sigma f y equals to zero so we have to take the component of fab in the vertical direction so it would be 4 by 5 times fab is 7 .5 kips then minus 4 by 5 f bc is equals to 0 so from here fbc comes out to be as 7 .5 and since the value comes out to be as positive so we can say that the member would be in tension next we'll be considering the next equilibrium condition that is sigma f x equals to zero so we have f bd plus 3 by 5 f bc plus 3 by 5 f bc plus 3 by 5 f a b that is 7 .5 is equals to 0 and f bc we have 7 .5 so we can plug it here and we have only one variable now fbd and from here fbd comes out to be as 9 kips and since the value is in negative that means the member would been compression so f bd would be equals to nine kips and the member would be in compression therefore the direction of force in the member bd would be like this now next we'll be considering the force on the on the on the joint c for that it will be making the free body diagram of the joint c so this is a free board diagram of the joint c now here also will be using the equivalent condition the first one is sigma if y equals to zero now we'll be taking forces in the y direction so we have 4 by 5 times 7 .5 plus 4 by 5 times fcd that we have to calculate minus 6 would be equals to 0 so from here fcd comes out to be as 0 now next we'll be considering the forces in x direction so sigma f x equals to 0 so we have f c e in the x direction minus 4 .5 minus 3 by 5 times 7 .5 is equals to 0 so from here the force in the member c e will get this value as 9 kips and since the value comes out to be as positive that means the member would be in tension so now next i will be considering this section now if you see this section carefully we can say that the trust is symmetric about this section and thus we can write that the force in the member ab would be equals to the force in the member fg and this thing we have calculated earlier as 7 .50 kips and the member would be in compression...

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