Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length 
for the deformation, we use the following relation;
= [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
= 3161.025289 / 7240
= 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m
Answer:
The critical length of surface flaw = 6.176 mm
Explanation:
Given data-
Plane strain fracture toughness Kc = 29.6 MPa-m1/2
Yield Strength = 545 MPa
Design stress. =0.3 × yield strength
= 0.3 × 545
= 163.5 MPa
Dimensionless parameter. Y = 1.3
The critical length of surface flaw is given by
= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2
Now putting values in above equation we get,
= 1/3.14( 29.6 / 1.3 × 163.5)^2
=6.176 × 10^-3 m
=6.176 mm
Answer:
Qcd=0.01507rad
QT= 0.10509rad
Explanation:
The full details of the procedure and answer is attached.
Answer:
778.4°C
Explanation:
I = 700
R = 6x10⁻⁴
we first calculate the rate of heat that is being transferred by the current
q = I²R
q = 700²(6x10⁻⁴)
= 490000x0.0006
= 294 W/M
we calculate the surface temperature
Ts = T∞ + 
Ts = 
The surface temperature is therefore 778.4°C if the cable is bare
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:
Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:
Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:
Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero
