Determine the force in each member of the truss and state if the members are in tension or compr

Determine the force in each member of the truss and state if...

Key Concepts

Truss Analysis

Truss analysis involves studying the framework of a truss, which is a structure made of straight members connected at joints. The objective is to determine the internal forces such as tension and compression in each member, ensuring that the structure is both stable and efficient. This understanding is crucial in identifying how loads are distributed and which members potentially need reinforcement.

Free-Body Diagram

A free-body diagram is a simplified illustration of a structure or a joint, showing all the external forces and moments acting on it while isolating it from its environment. In truss analysis, drawing free-body diagrams for joints or sections helps visualize and set up the equilibrium equations required to determine unknown forces.

Static Equilibrium

Static equilibrium is the condition in which a structure or object is at rest, meaning the sum of all forces and moments acting on it is zero. This core principle allows engineers to apply balance equations (?F_x = 0, ?F_y = 0, and ?M = 0) to solve for unknown internal forces in truss members.

Method of Joints

The method of joints is an analytical approach used in truss analysis where each joint is considered separately. By applying the equations of static equilibrium to these joints, the forces in the connected members can be sequentially solved, identifying whether they are in tension or compression.

Method of Sections

The method of sections involves cutting through the truss to isolate a section of the structure, enabling the analysis of several member forces simultaneously using equilibrium equations. This method can be more efficient than the method of joints when only specific member forces are required.

Tension vs Compression

In truss structures, members can experience either tension, where forces pull the member apart, or compression, where forces push the member together. Determining whether a member is in tension or compression is critical for appropriate design and material selection, as each type of force has different implications for the member's behavior and failure modes.

Transcript

00:01 Looking at structural analysis, consider the structure where we want to determine the magnitude of the process in the member of the structure as well as their nature.

00:13 To do this, we are going to apply what we call the method of joint to solve the problem.

00:19 But before that, let us find the support reactions at a and d.

00:26 To do that, we can apply summation of movement about d.

00:31 0 so that r a the reaction at a okay at that support a times 12 minus six times eight minus nine okay minus nine times four will give us equal zero so when you solve this out you're going to have a reaction at a equal to eight sorry seven kilo newton upward and the downward forces or vertical forces to be summed to be subpoenaed, and you sum them, they are going to arrive at rd to be equal to 8 kilo -newt that is adding roller support there and d.

01:33 Then we can consider joints, okay? joint e.

01:39 Consider joint a, we are having this force coming like this, okay, and another force coming like this, sorry, have this force coming like this okay the forces should be going out of the joint except do that are coming in naturally like this seven kilo mutine the reaction this reaction and is support others going out of the joint that is our assumption we're going to this angle is 45 degree so use the dimensions given in the information given the question to find the angle to undermine the angles this is f -a -e, the force f, f, a, going out of the joint a.

02:32 So if you take summation of forces in the horizontal direction, sorry, let us name this force.

02:43 This force is f -a -b, okay, going towards a.

02:50 So some sort of forces in the horizontal direction is zero.

02:55 So if we do that, then they'll arrive at, we bring it down here, f -a -e.

03:05 Cost 45 degrees plus f .a .e.

03:13 Plus f .a .b.

03:19 F, subscript a, b.

03:22 It will be equal to zero.

03:25 To be equal to zero.

03:27 So we can, when we serve this out, you're going to arrive at, just serve it.

03:32 F .a .e.

03:34 It will give you equal to minus 7 over sign 45.

03:43 Degrees okay is equal to uh 9 .9 minus 9 .9 kilo newton okay now that's 9 .9 kilo newtine 9 .9 okay now this one is saying the same thing not saying is equal to uh 9 .9 kilo newton is a member under compression it's a member under compression okay so haven't gotten that we can look for plugging our values into and get fab so fab is gonna give you equal to just plug it into the equation that is already there we have fab it's gonna give you sorry 9 .9 cost for 5 degrees so when you solve this out i'm going to arrive at 8 .27 kilo milting so it's a member under tension so we can continue and look at joint b look at joint b and you have these something like this okay we have this force coming here we have this first going there okay call this one this is our joint b, here, call this on f, b, c.

06:14 And coming from b to c, going to c.

06:20 This is force here coming down to.

06:25 This is our 6 -killonutin force that is given in the question.

06:33 Why this one is our f, b, a, going from b to a...

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