One shaft two diameters angle of deformation

[togo]

Homework Statement

Compute the angle of twist of the free end relative to the fixed end of the steel bar:
200 N*m, 80 x 10^9 GPa (shear modulus of elasticity)

(Length 1: 1.2 m, dia of .040 m on left, length of .4 m dia of .020 m, on right)

Homework Equations

angle = (torque)(length) / (shear modulus of elasticity)(J of shaft)

The Attempt at a Solution

(200 n*m)(1.2 m) / (80 x 10^9)(25.1 x 10^8) = 1st section of shaft
(200 n*m)(0.4 m) / (80 x 10^9)(1.5 x 10^8) = 2nd section of shaft

= 0.012 + 0.0064 = 0.018 rad

Book answer: 0.0756 rad

Thanks!

 

[togo]

no idea?

 

[nvn]

togo: Exponent or units on G (shear modulus of elasticity) listed wrong in section 1 of post 1, but listed correctly in section 3. Exponent on J listed wrong in post 1. J values rounded or truncated too much. Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. Unit symbol for Newtons is uppercase N, not lowercase n. Lowercase n means nano. Numbers less than 1 must always have a zero before the decimal point. E.g., 0.4 m, not .4 m. Try again.

 

[togo]

rounded or truncated too much. Did you run the calculation yourself? I've been using about 4 sig figs throughout the book with accurate results up to this point. G is correct afaik.

ps.
.040^4 = 25.1 x 10^-8 according to my calculator with no additional figures
.020^4 = 16 x 10^-8 also, no additional figures

 

[nvn]

togo: That is not the correct formula for J. Check your book to find the relevant equation for J. Try again.

 

[togo]

J = (pi x d^4) / 32

 

[nvn]

Right.

 

[togo]

that's the formula I used, thanks

answer is still wrong

 

[nvn]

togo: Try it again, correcting the mistakes explained in post 3. Show your work.

 

[togo]

.040^4 = 25.1 x 10^-8
.020^4 = 16 x 10^-8

J1 = .04 x 10^-8 x (pi/32) = 2.513 x 10^-7
J2 = .02 x 10^-8 x (pi/32) = 1.571 x 10^-8

(200 n*m)(1.2 m) / (80 x 10^9)(2.513 x 10^-7) = 1st section of shaft = 0.0119 rad
(200 n*m)(0.4 m) / (80 x 10^9)(1.571 x 10^-8) = 2nd section of shaft = 0.0636 rad

=0.0756 rad

 

[nvn]

togo: The formula you listed for J1 and J2 in post 10 is currently wrong, and does not match the right-hand side. Also, review the information in post 3.

(1) The unit symbol for Newton is uppercase N, not lowercase n. Lowercase n means nano.

(2) Numbers less than 1 must always have a zero before the decimal point. E.g., 0.04, not .04. See the international standard[/color] (ISO 31-0[/color]). Or see any credible textbook.​

 

[togo]

thanks for that nvn. please don't respond to my help threads anymore

I will be seeing an instructor about this one

you will note that the correct numbers are in the final formula and you've neglected getting to the heart of the matter for half a month now