The rigid beam is supported by the three posts A, B and C....

The rigid beam is supported by the three posts $A, B$ and C. Posts $A$ and $C$ have a diameter of $60 \mathrm{mm}$ and are made of a material for which $E=70 \mathrm{GPa}$ and $\sigma_{Y}=20 \mathrm{MPa}$ Post $B$ is made of a material for which $E^{\prime}=100 \mathrm{GPa}$ and $\sigma_{Y}^{\prime}=590 \mathrm{MPa} .$ If $P=130 \mathrm{kN},$ determine the diameter of post $B$ so that all three posts are about to yield.

Key Concepts

Yield Strength

Yield strength is the maximum stress that a material can sustain without undergoing permanent deformation. In the context of the problem, each supporting post has a defined yield strength which determines the load each can carry before yielding. This concept is critical for ensuring that all supports yield uniformly under the applied load, thereby preventing overloading of any single support and maintaining structural integrity.

Young's Modulus

Young’s modulus is a measure of the stiffness of a material, quantifying its resistance to elastic deformation under stress. In this scenario, it governs how much each post deforms under the applied load, affecting the distribution of forces among the posts. The variation in Young’s modulus between the posts plays a key role in determining the load-sharing behavior, ensuring that all posts reach their yield point simultaneously.

Load Distribution in Composite Structures

Load distribution involves determining how an external load is shared among various components of a structure based on their stiffness and strength. In the case of the rigid beam supported by multiple posts, the posts’ differing material properties and dimensions lead to a non-uniform distribution of load. Understanding this concept is fundamental to designing supports such that each post yields at the same time under a given load.

Cross-sectional Geometry and Stress Calculation

The cross-sectional geometry of a structural member directly influences its ability to resist stress. For circular posts, the diameter determines the cross-sectional area, which in turn affects the stress level in the post for a given load. Calculating the appropriate diameter to achieve simultaneous yielding involves understanding how the area relates to stress and ensuring consistency with the yield strength of the material.

Transcript

00:01 In this question it is given that an aeroplane take -offs and it travels 10 .4 km west 8 .7 km north and 2 .1 km up then we have to find that how far it is from the take -off point so first of all i am using the sign convention for this question let me draw it so let this is our initial take -off point that is o and this is our east direction which is positive x -dive and this is our vertically up direction that is and this is our west and this direction is north and this direction is south which is this is our positive y direction and this is our positive z direction so first we have given that an aeroplane take -offs and it travels 10 .4 km west so if we draw it in diagram so this is the west so from here it travels 10 .6 this is the position and now it travels in north.

01:25 So this is the north direction here and the distance is 8 .7 km and then it travels vertically up.

01:33 So this is the vertical up.

01:34 So this is the final position of the aeroplane and this is our initial takeoff.

01:39 So we have to find the displacement.

01:42 So we have to join the initial and final point.

01:53 Let this is our final point that is s here or we can say that this is and we have given that this distance is 8 .7 km this distance is 10 .4 kilometer and this distance is 2 .1 kilometer so i am writing the notation here so we take east as x direction that is plus x direction and north is our plus y direction and vertically up is our plus z direction this is our sign convention for this tuition and this is our initial takeoff point and this is the final point so if we find the displacement vector so we can say that displacement vector is given as that is s vector is equal to this is 10 .4 in minus icaf because this is in negative x direction plus this is 8 .7 in j cap plus 2 .1 of k cap.

03:33 So we can say this will come out to be minus 10 .4 i cap plus this is 8 .7 j cap plus 2 .1 k cap...

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