Determine the force developed in members F E, E B and B C of the truss and state if these member

Determine the force developed in members F E, E B and B C of...

Key Concepts

Axial Force in Truss Members

Axial forces refer to the forces acting along the length of a truss member. Determining whether these forces are tensile (pulling apart) or compressive (pushing together) is critical in assessing the performance and safety of a truss structure.

Method of Joints

The method of joints is a technique used in truss analysis where each joint is isolated and the equilibrium equations are applied. This method systematically solves for the forces in each member by considering the balance of forces in the horizontal and vertical directions.

Tension and Compression

Tension and compression are types of axial forces encountered in truss members. Tension occurs when a member is being stretched, while compression occurs when a member is being shortened or squashed. Identifying these forces helps in selecting appropriate materials and designing efficient truss systems.

Truss Analysis

Truss analysis involves studying the behavior and internal forces within a framework of members connected at joints. It is fundamental in structural engineering to ensure that the structural components support expected loads safely and efficiently.

Equilibrium Equations

The equilibrium equations, based on Newton's First Law, state that the sum of forces and moments acting on any joint or member must be zero. These equations are essential for determining the unknown forces in a truss system.

Transcript

0:00 General analysis.

00:02 For the problem at hand, we want to determine the magnitude of the forces in the members of the structure as well as their nature, okay, to solve this problem, we will apply the method of joints.

00:15 But apart from that, let us look at the support reactions before we can commence with any joint.

00:26 Now let's look at the key moment about the indeed.

00:34 About d point d equal to zero summation of moment so if we take it about that point we'll have this one times 5 .5 minus 11 times 3 .5 minus 22 times 2 is equal to 0 so we will now have arrow here the reaction at to the upward reaction at a vertical reaction and the using summation of vertical forces you can also determine the reaction at the roller end equal to 18 kilo newtine okay so let's look at joint a let's call it g a we have this force these are this is a force diagram for the joint.

02:09 Now based on the information given in the question, you can determine, you can figure out the angles.

02:14 This is 45 degrees.

02:16 So you try that.

02:19 So this force is going from joint a to b.

02:25 Because this will be is at joint a.

02:31 And this is the reaction that will calculated at e that has support to be 15 kilos.

02:42 New thing okay all right so okay this force is f substrate a actually going to f to join f so let's take some function of forces in the horizontal direction if we take some action of the forces in the horizontal direction they're going to have f substitute a b plus f substitute a f substitute a f cost 45 degrees is equal to zero.

03:40 So we can also look at summation of forces in the vertical direction.

03:52 If you do that, we're going to have 15 plus f and soft with af force, the force going from joint a to f of that member af, sign 45 degrees, is equal to zero.

04:38 So that we can now make this one the circle of the formula a soft with f to give us 21 which are a negative 21 point to 1 kilo mutine okay you can see that this this is the magnitude but there's a negative sign there tells us that is a member under compression okay so this is 15 so we can look at what we some other first give us, based on what we've got in now, we can get our sos with ab from there to plug it into values into the equations.

05:49 We're going to have this one to be equal to 14, from that horizontal fx, f of x, equal to the equation in the horizontal direction, we're going to have it to be to the 14.

06:11 Point 99 okay 14 .99 sorry i said that don't mean f of as i meant a sumption of forces in the horizontal direction effects okay summation of f x equal to zero okay all right so that is this force okay then you can see that this force is under tension sorry the the the member is under tension so we can look at another joint.

07:02 Let's see we take joint, joint f.

07:12 We are going to have this force coming like that.

07:23 So this is f.

07:24 These are joint f.

07:31 This f substrate fb.

07:48 This f substrote f .a...

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