Answer:
A) micro defects are left behind on the surface of metal components during the manufacturing process. These defects, in the form of micro-cracks or pits, becomes initiation sites for crack propagation or corrosion. Removing these imperfections on the surface of metal parts by electroplating greatly improves the life of metal components.
B) it will reduce fatigue crack growth.
Dispersion hardening involves the inclusion of small, hard particles in the metal, thus restricting the movement of dislocations, and thereby raising the strength properties. In dispersion hardening it is assumed that the precipitates do not deform with the matrix and that a moving dislocation bypasses the obstacles (precipitates) by moving in the clean pieces of crystal between the precipitated particles.
C) stress concentrations such as changes in section with sharp corners caused yielding, which will typically occur first at a stress concentration. For ductile materials localised plastic deformation can cause a redistribution of stress, enabling the component to continue to carry load. Brittle materials will typically fail at the stress concentration. Repeated loading may cause a fatigue crack to initiate and slowly grow at a stress concentration leading to the failure of even ductile materials. Fatigue cracks always start at stress raisers, so removing such defects increases the fatigue strength.
Answer:
a) 2.452
b) 1.256
Explanation:
Stress due to dead weight. = 14 Ksi
Stress due to fully loaded tractor-trailer = 45Ksi
ultimate tensile strength of beam = 76 Ksi
yield strength = 50 Ksi
endurance limit = 38 Ksi
Determine the safety factor for an infinite fatigue life
a) If mean stress on fatigue strength is ignored
β = ( 45 - 14 ) / 2
= 15.5 Ksi
hence FOS ( factor of safety ) = endurance limit / β
= 38 / 15.5 = 2.452
b) When mean stress on fatigue strength is considered
β2 = 45 + 14 / 2
= 29.5 Ksi
Ratio = β / β2 = 15.5 / 29.5 = 0.5254
Next step: applying Goodman method
Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]
= 19.47 Ksi
hence the FOS ( factor of safety ) = Sa / β
= 19.47 / 15.5 = 1.256
Answer:
because burning rubber increases the grip power
Answer:
(a) The stress on the steel wire is 19,000 Psi
(b) The strain on the steel wire is 0.00063
(c) The modulus of elasticity of the steel is 30,000,000 Psi
Explanation:
Given;
length of steel wire, L = 100 ft
cross-sectional area, A = 0.0144 in²
applied force, F = 270 lb
extension of the wire, e = 0.75 in
<u>Part (A)</u> The stress on the steel wire;
δ = F/A
= 270 / 0.0144
δ = 18750 lb/in² = 19,000 Psi
<u>Part (B)</u> The strain on the steel wire;
σ = e/ L
L = 100 ft = 1200 in
σ = 0.75 / 1200
σ = 0.00063
<u>Part (C)</u> The modulus of elasticity of the steel
E = δ/σ
= 19,000 / 0.00063
E = 30,000,000 Psi
