A. I believe, lmk if I’m right
Answer:
The minimum diameter for each cable should be 0.65 inches.
Explanation:
Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,
(1/2)(Weight/Cross Sectional Area) = Allowable Stress
Here,
Weight = 1000 lb
Cross-sectional area = πr²
where, r = minimum radius for each cable
(1/2)(1000 lb/πr²) = 1500 psi
500 lb/1500π psi = r²
r = √1.061 in²
r = 0.325 in
Now, for diameter:
Diameter = 2(radius) = 2r
Diameter = 2(0.325 in)
<u>Diameter = 0.65 in</u>
Answer:
1561.84 MPa
Explanation:
L=20 cm
d1=0.21 cm
d2=0.25 cm
F=5500 N
a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa
lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16
longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3
(assuming a poisson's ration of 0.3)
ε_l =0.16/0.3 = 0.5333
b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)
σ_true = 1561.84 MPa
ε_true = ln( 1+ε_l)= ln(1+0.5333)
ε_true= 0.4274222
The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.
Answer:
for 5.6V 9 turns, for 12.0V 19 turns, for 480V 755 turns
Explanation:
Vp/Vs= Np/Ns
Vp: Primary voltage
Vs: Secondary Voltage
Np: number of turns on primary side
Ns: number of turns on secondary side
for output 5.6V
140/5.6= 220/Ns
Ns= 8.8 or 9 Turns
for output 12.0V
140/12= 220/Ns
Ns= 18.9 or 19 turns
for output 480V
140/480= 220/Ns
Ns= 754.3 or 755 turns
Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ




τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0