Answer:
axial stress in bar B = 25Mpa.
Deformation of bar A = 0.4mm.
Explanation:
PS: Kindly check the attached picture for the diagram showing the two bars that is to say the bar A and the bar B.
So, we are given the following data or information or parameters which we are going to use in solving this particular question or problem. Here they are;
The cross-sectional areas of Bars A and B = 400 mm2, the modulus of elasticity of bar A and bar B = 200 GPa, applied force = 10kN.
STEP ONE: The first step is to determine or calculate the axial stress in bar B. Therefore,
Axial stress in bar B = 10 × 10³ ÷ 400 × 10⁻⁶ = 25 Mpa.
STEP TWO: The second step here is to determine or calculate the deformation of bar A. Therefore,
The deformation of bar A = 20 × 10³ ×1.5 ÷ 400 × 10⁻⁶ × 200 × 10³ = 0.375 mm.
Answer:
Concentration factor will be 1.2
So option (C) will be correct answer
Explanation:
We have given outer diameter D = 1.25 in
And inner diameter d = 1 in and fillet ratio r = 0.2 in
So 
ratio will be 
And 
ratio will be 
Now from the graph in shaft vs torsion the value of concentration factor will be 1.2
So concentration factor will be 1.2
So option (C) will be correct answer.
Answer with Explanation:
There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:
1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.
2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.
3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.
4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.
Answer:
22.90 × 10⁸ kg
Explanation:
Given:
Diameter, d = 0.02 m
ωₙ = 0.95 rad/sec
Time period, T = 0.35 sec
Now, we know
T= 
where, L is the length of the steel cable
g is the acceleration due to gravity
0.35= 
or
L = 0.0304 m
Now,
The stiffness, K is given as:
K = 
Where, A is the area
E is the elastic modulus of the steel = 2 × 10¹¹ N/m²
or
K = 
or
K = 20.66 × 10⁸ N
Also,
Natural frequency, ωₙ = 
or
mass, m = 
or
mass, m = 
mass, m = 22.90 × 10⁸ kg
