Answer:
81.76 N/mm² ( MPa), 1.71233
Explanation:
Modulus of elasticity = stress / strain
stress = modulus of elastic × strain
strain = ΔL / L = 250.28 mm - 250 mm / 250 mm = 0.00112
Modulus of elasticity E = 73 GPa = 73 × 10³ MPa where 1 MPa = 1 N/mm²
E = 73 × 10³N/mm²
stress = 73 × 10³N/mm²× 0.00112 = 81.76 N/mm² ( MPa)
b) Factor of safety = maximum allowable stress / induced stress = 140 MPa / 81.76 MPa = 1.71233
Answer:
see explaination for all the answers and full working.
Explanation:
deflection=8P*DN/Gd^4
G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2
for outer spring,
deflection=8*3*50^3*5/(70*9^4)=32.66mm
for inner spring
deflection=8*3*30^3*10/(70*5^4)=148.11mm
max stress=k*8*P*C/(3.14*d^2)
for outer spring
c=50/9=5.55
k=(4c-1/4c-4)+.615/c=1.2768
max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2
for inner spring
c=6
k=1.2525
max stress=2.29KN/mm^2
Answer:
Tire inflation can be caused by temperature and speed :)
Answer:
Option B
Select the Marketing User checkbox in the user record
Explanation:
The user settings may limit users who can view and set up advanced campaigns. In case there's no keyboard shortcut already programmed to create a new campaign and also the button for creating new campaign isn't visible, the best way to go about it is by selecting the marketing user checkbox in the user record and create a new campaign,
Answer:
μb = 0.096
μc = 0.073
Explanation:
member AB:
-800( 4/3 ) + Nb (2) = 0
Nb (2) = 3200/3
Nb = 533.3N
Post BC:
summation of force along the y axis=0
Nc + Nb + 150(3/5 ) -50(9.81)=0
Nc + 533.3 + 150(3/5 ) -50(9.81)=0
Nc = 933.83 N
Also (-4/5)(150)(3) + Fb(0.7)= 0
Fb = (4/5)(150)(3)/0.7 = 51.429 N
Likewise alog the x axis,
4/5(150) - Fc -Fb = 0
4/5(150) - Fc -51.429 = 0
Fc = 4/5(150) -51.429 =68.571 N
μb = Fb/Nb = 51.429/533.3 = 0.096
μc = Fc/Nc = 68.571 / 933.83 = 0.073