The A-36 steel pipe has a 6061-T6 aluminum core. It is subjected to a tensile force of 200 kN. D

The A-36 steel pipe has a 6061-T6 aluminum core. It is...

Key Concepts

Tensile Force

A tensile force is a pulling force that acts to elongate a material or structural member. It is a fundamental concept in mechanics where the applied force is used to assess how structures behave under loads that pull the material apart.

Axial (Normal) Stress

Axial or normal stress is defined as the internal force distributed over the cross-sectional area of a material when subjected to an external load. It is calculated by dividing the tensile force by the area, and it is a key parameter in evaluating the material’s ability to withstand applied loads without failure.

Cross-Sectional Area Calculation

The calculation of cross-sectional area is crucial in determining the stress experienced by a material. For circular or annular shapes, the area is derived from geometric formulas that consider the dimensions of the component, ensuring the stress is properly computed based on the effective area that resists the applied load.

Composite Materials Under Load

Composite materials combine two or more different substances to achieve unique mechanical properties. When subjected to a load, each constituent material carries a portion of the overall stress based on its individual cross-sectional area and inherent material properties, making it essential to understand how the total load is distributed in the analysis of composite structures.

Stress Distribution in Composite Sections

Stress distribution in composite sections involves analyzing how different materials within a composite structure share and resist applied loads. This concept is important because the materials typically have different strengths and stiffnesses, and the distribution is determined by both the geometric layout and the mechanical properties of each material.

Transcript

00:01 In this cution it is told that two blocks are connected with strings and they are slides down on an inclined plane which is making an angle of 30 degrees.

00:09 So first of all i am drawing the diagram for this given cution let this is the inclined plane which makes an angle of 30 degree so we have given this angle as 30 degree and this is the first block and let this is the second block both are connected via a string let this is the string and let this mass is m1 and and this mass is m2.

00:35 Now i'm going to show the forces acting on this block.

00:40 So one is the weight of this block that is m1g which is acting vertical downward and weight of this block also vertical downward that is m2g.

00:51 So component of this weight in this direction is m1g sine 30 and this will be in this direction this will given by m2g sign 30.

01:07 And let the tension in this string is t and there is a friction force between the block and this surface let this is f1 this friction force is given by f1 and the friction between this block and this surface is let this is f2 so we have given that m1 is the mass of the block that is 4 kg m2 is 8 kg we have given the value of kinetic friction that is mu is 0 .25 and mu2 is 0 .35 so first of all i'm going to write the force balance so we can say that for block m1 we can say that net force m1g into sine alpha minus frictional force minus tension force that is t is equals to m1a now we can say that this will be written as m1g x x xm 1 into m1 g kos alpha alpha minus t is equal to m1a alpha is this angle so we can say the alpha is 30 degree here now put all the data in this equation so this is 4 this is g into sine 30 minus this is 0 .25 into 4 into this is g into cost 30 this is minus t is equal to this is 4 into a after solving this we will get this is 11 .11 minus t is equal to 4 a let this is our equation first similarly we can write this for block m2 we can say that net force is given by that is m2g sine alpha minus frictional force which is mu to m2g into cos alpha is equal to this is sorry tension force also there so this is plus t is m2a now put all the data in this equation so this is 8g of sine 30 minus 0 .35 into 8 this is 8 this is 8 this is 8 this is to this is 8a now i'm going to solve this and this will come out to be 15 .44 plus t into is equal to 8a this is our equation second now i'm going to add both of the equation so add equation 1 and equation 2 we will get this is 26 .55 is equal to 12a or i can say acceleration a is given by 2 .21 meter per second square so this is the acceleration of the both the blocks and now we have to find the tension force so i'm going to put this value of acceleration in equation first so we can say that from equation first we get this is 11 .11 minus t is equal to this is 4 into acceleration is 2 .21 and after solving this we get tension that is 2 .27 newton so this is the value of tension in the string this is the solution of our part a...

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