Answer:
a) 0.697*10³ lb.in
b) 6.352 ksi
Explanation:
a)
For cylinder AB:
Let Length of AB = 12 in
For cylinder BC:
Let Length of BC = 18 in
b) Maximum shear stress in BC
Maximum shear stress in AB
At an auction of antiques, a bidder for a particular porcelain statue would be trying to ________(a) Maximize the difference bet
1 answer:
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Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*/2*c^4
∅=2TL/G**c^4
c=∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*/2*c^4)]*c
г =[2T/(J**c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm
Answer:
The velocity at section is approximately 42.2 m/s
Explanation:
For the water flowing through the pipe, we have;
The pressure at section (1), P₁ = 300 kPa
The pressure at section (2), P₂ = 100 kPa
The diameter at section (1), D₁ = 0.1 m
The height of section (1) above section (2), D₂ = 50 m
The velocity at section (1), v₁ = 20 m/s
Let 'v₂' represent the velocity at section (2)
According to Bernoulli's equation, we have;
Where;
ρ = The density of water = 997 kg/m³
g = The acceleration due to gravity = 9.8 m/s²
z₁ = 50 m
z₂ = The reference = 0 m
By plugging in the values, we have;
50 m + 30.704358 m + 20.4081633 m = 10.234786 m +
50 m + 30.704358 m + 20.4081633 m - 10.234786 m =
90.8777353 m =
v₂² = 2 × 9.8 m/s² × 90.8777353 m
v₂² = 1,781.20361 m²/s²
v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s
The velocity at section (2), v₂ ≈ 42.2 m/s
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ = ................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm