Answer:
Change in length = 0.1257 mm
Change in diameter= -0.03771mm
Explanation:
Given
Diameter, d = 15 mm
Length of rod, L = 200mm
F = Force= 300N
d = 0.015m
Ep=2.70 GPa, np=0.4.
First, we have to calculate the normal stress using
σ = F/A where F = Force acting on the Cross-sectional area
A = Area
Area is calculated as πd²/4 where d = 0.015m
A = 22/7 * 0.015²/4
A = 0.000176785714285m²
A = 1.768E-4m²
So, stress. σ = 300N/1.768E-4m²
σ = 1696832.579185520Pa
σ = 1.697MPa
Calculating E(long)
E(long) = σ /Ep
E(long) = 1.697E-3/2.70
E(long) = 0.0006285
At this point, we fan now calculate the change in length of the element;
∆L = E(long) * L
∆L = 0.0006285 * 200mm
∆L = 0.1257mm
Calculating E(lat)
E(lat) = -np * E(long)
E(lat) = -4 * 0.0006285
E(lat) = -0.002514
At this point, we can now calculate the change in diameter of the element;
∆D = E(lat) * D
∆L = -0.002514 * 15mm
∆L = -0.03771mm
Answer:
Matlab code with step by step explanation and output results are given below
Explanation:
We have to construct a Matlab function that creates a row vector "countValues" with elements 1 to endValue. That means it starts from 1 and ends at the value provided by the user (endValue).
function countValues = CreateArray(endValue)
% Here we construct a row vector countValues from 1:endValue
countValues = 1:endValue;
% then we transpose this row vector into column vector
countValues = countValues';
end
Output:
Calling this function with the endValue=11 returns following output
CreateArray(11)
ans =
1
2
3
4
5
6
7
8
9
10
11
Hence the function works correctly. It creates a row vector then transposes it and makes it a column vector.
Answer:
Mass of earth will be 
Explanation:
We have given acceleration due to gravity 
Radius of earth = 6370 km =
Gravitational constant 
We know that acceleration due to gravity is given by
, here G is gravitational constant, M is mass of earth and R is radius of earth
So 

So mass of earth will be 
Answer:
correct option is (A) 0.5
Explanation:
given data
axial column load = 250 kN per meter
footing placed = 0.5 m
cohesion = 25 kPa
internal friction angle = 5°
solution
we know angle of internal friction is 5° that is near to 0°
so it means the soil is almost cohesive soil.
and for a pure cohesive soil
= 0
and we know formula for
is
= (Nq - 1 ) × tan(Ф) ..................1
so here Ф is very less
should be nearest to zero
and its value can be 0.5
so correct option is (A) 0.5