. The assembly consists of two posts A B and C D each made...

. The assembly consists of two posts $A B$ and $C D$ each made from material 1 having a modulus of elasticity of $E_1$ and a cross-sectional area $A_1$, and a central post $E F$ made from material 2 having a modulus of elasticity $E_2$ and a cross-sectional area $A_2$. If posts $A B$ and $C D$ are to be replaced by those having a material 2 , determine the required cross-sectional area of these new posts so that both assemblies deform the same amount when loaded.

Key Concepts

Material Substitution and Equivalent Section Properties

When substituting components made from one material with those made from another, engineers must calculate equivalent section properties to achieve the same deformation under load. This involves using material properties such as the modulus of elasticity along with geometry (cross-sectional area and length) to find the design that will mirror the performance of the original configuration, ensuring that the overall assembly behaves as intended under load.

Deformation Compatibility in Assemblies

Deformation compatibility ensures that different components of a structural assembly deform by the same amount under applied loads. When replacing one or more components in an assembly, it is necessary to adjust parameters (such as cross-sectional area) for the new material so that the deformations of the replaced components match the original assembly, thereby ensuring proper load transfer and overall structural integrity.

Modulus of Elasticity

The modulus of elasticity (E) is a measure of a material's stiffness, defining the relationship between stress and strain in the elastic region of the material. In structural analysis, especially where deformation compatibility is required, the modulus of elasticity is used to calculate how much a given structural member deforms under a load, making it a fundamental property for comparing different materials.

Axial Stiffness

Axial stiffness is the product of a member’s modulus of elasticity (E) and its cross?sectional area (A) divided by its length (L). It quantifies the resistance of a structural member to deformation under axial loads. In problems involving axial deformation, ensuring equivalent stiffness across different members or assemblies is crucial to maintain similar deformation responses under identical load conditions.

Transcript

00:01 In this question it is given that the speed of the aeroplane decreases at a constant rate from 180 meter per second to 160 meter per second and we have to find the magnitude of the abrupt change in the force that exited on a 90 kilogram passenger as the plane passes point b so first of all i am writing the given data for this question so we have given that speed of the aeroplane at a is 180 meter per second and speed of the aeroplane at point b is given to us that is 160 meter per second and we have given the mass of the passenger that is 90 kilograms so let this is the track we have given and this is the radius of this track and this is that p is equal to 6 kilometers this is the point a this is point b and here point c we have given this distance as 0 .8 kilometers and here is our aeroplane and we have given the angle that is theta is equal to 8 degree so we can say that the distance from a to b is that is dab can be written as the radius into theta that means p into theta and this is in radian so we can say that it will be written as this is 6 kilometers so 6 into 10 to the power 3 and this is 8 into pi by 180 and this will be the distance of ab now we have given the distance between point a and c that is dac is or i can say that i am writing here below so dac will be that is distance between a to b plus distance b to c and this can be written as distance a to b plus this is 800 meters so distance of a to c is we have calculated.

02:28 Now we know that acceleration is written as that is vc square minus va square upon this is 2 times of dacc and this will be also written as like this is vb square minus va square upon this is 2 times of d ab.

02:57 Now put all the data in this equation and we get the value of vb so i am going to put the data so vc is given to us that is sorry this is here we have to given that this is the velocity at point c this is vc we have given 160 meter per second so we can find the velocity of point b so this is 160 square minus 180 square upon this is 2 into distance of ac is equals to this is b b square minus this is 180 square upon two times of distance from ab.

03:40 After solving this and after putting this distance we find the value of bb and this will come out to be 170 .1 meter per second.

03:50 We can clearly see that speed decreases from a to b and b to c.

03:56 Now we have to find the force at b so we can say that force exerted at point b is given by that is mass into acceleration that is normal acceleration so this will come out to m into vb square upon this is p.

04:14 Now put all the data in this equation...

Please give Ace some feedback