## Active forum topics

- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
- Sight Distance of Vertical Parabolic Curve
- Application of Differential Equation: Newton's Law of Cooling

## New forum topics

- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
- Eliminate the Arbitrary Constants
- Required diameter of solid shaft

## Recent comments

- Use integration by parts for…1 week 5 days ago
- need answer1 week 5 days ago
- Yes you are absolutely right…2 weeks 1 day ago
- I think what is ask is the…2 weeks 1 day ago
- $\cos \theta = \dfrac{2}{…2 weeks 2 days ago
- Why did you use (1/SQ root 5…2 weeks 2 days ago
- How did you get the 300 000pi2 weeks 2 days ago
- It is not necessary to…2 weeks 3 days ago
- Draw a horizontal time line…3 weeks ago
- Mali po ang equation mo…4 weeks 2 days ago

## Given:

Given:

P = 370 kW = 370 x103 watts ((N-m)/s)= 3.7 〖x10〗^8 ((N-m)/s)

f = 120 rev/min x 1min/60s=2 rev/s

τ=92 N/mm^2

Solution:

1. Calculate Twisting Moment/ Torsion, T

P=2πT f

3.7 〖x10〗^8 ((N-mm)/s)=2 π T (2 rev/s)

T = 2.944〖 x 10〗^7 N-mm

2. Solve for diameter, D of solid shaft using the following formulas:

τ = Tr/J

J= π/32 D^4

r= D/2

τs= Tr/J

92 N/mm^2 = (2.944〖 x 10〗^7 N-mm (D/2))/(π/32 D)

D=117.69mm of 4.72inches