Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
Answer:
σ = 391.2 MPa
Explanation:
The relation between true stress and true strain is given as:
σ = k εⁿ
where,
σ = true stress = 365 MPa
k = constant
ε = true strain = Change in Length/Original Length
ε = (61.8 - 54.8)/54.8 = 0.128
n = strain hardening exponent = 0.2
Therefore,
365 MPa = K (0.128)^0.2
K = 365 MPa/(0.128)^0.2
k = 550.62 MPa
Now, we have the following data:
σ = true stress = ?
k = constant = 550.62 MPa
ε = true strain = Change in Length/Original Length
ε = (64.7 - 54.8)/54.8 = 0.181
n = strain hardening exponent = 0.2
Therefore,
σ = (550.62 MPa)(0.181)^0.2
<u>σ = 391.2 MPa</u>
Answer:
a, c
Explanation:
As the problem statement tells you, the phasor technique cannot be used when the frequencies are different. The frequencies are different when the coefficients of t are different. The different ones are highlighted.
a. 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b. 25 cos(50t + 160°) + 15 cos(50t +70°)
c. 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d. 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
