The truss is made of three A-36 steel members, each having a cross-sectional area of 400 mm^2. D

The truss is made of three A-36 steel members, each having a...

Key Concepts

Truss Analysis

Truss analysis involves evaluating structures composed of slender members connected at joints assumed to be frictionless. This method uses equilibrium equations to determine the forces in each member, helping to ensure that the overall structure remains stable under applied loads.

Axial Deformation

Axial deformation refers to the change in length of a structural member when subjected to an axial load. It is primarily calculated using the relationship ? = PL/(AE), where P is the axial force, L is the original length, A is the cross-sectional area, and E is the Young's modulus of the material.

Material Properties

Material properties, such as Young's modulus and yield strength, are essential in predicting how materials behave under different loads and deformations. These properties ensure that the calculated deformations and stresses remain within safe and acceptable limits for the intended application.

Load-Displacement Relationship

The load-displacement relationship quantifies how an applied force causes a displacement in a structure. This concept is critical when designing structural systems to achieve the required performance, ensuring that displacements do not exceed serviceability limits while maintaining overall structural integrity.

Transcript

00:01 In this question we have given that there are three forces that are applied to the wheel and it is told that one force is perpendicular to the rim one is the tangent to it and the other one is at 40 degree angle with the radius.

00:14 We have to find the net torque on the wheel.

00:17 So first of all i am drawing the diagram for this given cution so let this is our wheel and we have given the radius of this wheel and this is the center of the wheel and we have given three force one force is given to us that is perpendicular this is the force that is f1 one force is given to us that is tangential and let this force is this that is f3 and one force is given to us that is at an angle of 40 degrees this is the radius and this is a force is given to us that is f2 and this is at angle of 40 degree now first of all i am writing given data for this question so f1 is given to us that is 11 .9 newton f2 is given to us that is 14 .6 newton and f3 is given to us that is 8 .5 newton we have given the radius of this wheel that is 35 semi -overa i can say 0 .35 meter now torque due to force f1 is given by that is ttow 1 is equal to this is r1 f1 into sine theta and we can see that these are opposite to each other so here angle is 180 so this is r1 f1 into sine 180 degree and this is 0 .35 into f1 is 11 .9 and this is 0 so from here we get f to talk due to this force f1 is 0 this is the torque due to force f1 similarly we can find the torque due to force f2 and this will be written as this is tau 2 is equal to r2 f2 into sine theta 2 now put all the data so this is 0 .32 into this is f2 is given to that is 14 .6 into this is this is 14 .6 into sign the angle is 360 minus this is 40 from here we get this is minus 3 .28 newton meter and now finally we have to find the tour due to the the force f3 at the center and this is 2 or 3 is equal to this is r3 f3 into sine theta 3 now radius is same so this is 0 .35 and the force is 8 .5 into the angle is 90 degree so from here we get this is 2 .97 5 newton meter now we have to find net torque so we can say that net torque is given by torque 1 plus torque due to force 2 and torque due to force 3 so after putting this value we get this is minus 0 .31 newton meter...

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