Answer: l = 2142.8575 ft
v = 193.99 ft/min.
Explanation:
Given data:
Thickness of the slab = 3in
Length of the slab = 15ft
Width of the slab = 10in
Speed of the slab = 40ft/min
Solution:
a. After three phase
three phase = (0.2)(0.2)(0.2)(3.0)
= 0.024in.
wf = (1.03)(1.03)(1.03)(10.0)
= 10.927 in.
Using constant volume formula
= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf
Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)
= 6750 /0.2625
= 25714.28in = 2142.8575 ft
b.
vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)
= (0.12)(424.36)/0.2625
= 50.9232/0.2625
= 193.99 ft/min.
Answer:
i) σ1 = 133.5 MPa
σ2 = -2427 MPa
ii) 78.89 MPa
Explanation:
Given data:
ε1 = 0.0020 and ε2 = –0.0010
E = 71 GPa
v = 0.35
<u>i) Determine the biaxial stresses σ1 and σ2 using the relations below</u>
ε1 = σ1 / E - v (σ2 / E) -----( 1 )
ε2 = σ2 / E - v (σ1 / E) -------( 2 )
resolving equations 1 and 2
σ1 = E / 1 - v^2 { ε1 + vε2 } ---- ( 3 )
σ2 = E / 1 - v^2 { ε2 + vε1 } ----- ( 4 )
input the given data into equation 3 and equation 4
σ1 = 133.5 MPa
σ2 = -2427 MPa
<u>ii) Calculate the value of the maximum shear stress ( Zmax )</u>
Zmax = ( σ1 - σ2 ) / 2
= 133.5 - ( - 2427 ) / 2
= 78.89 MPa
Q1/A:-Determine Vo if VR2> VRI , VD = 0.
