Untitled

Question 1

A solid steel shaft of diameter D, carries loads P, M, T. Determine the safety factor n, assuming that failure occurs according to the following criteria;

Maximum sheet stress
Maximum energy of distortion

Maximum shear stress criterion

Maximum energy of distortion

% Solution - Maximum sheer stress criterion
syms S_y n M P D T
 
eq = S_y/n - 4/(pi * D^3) * ((8 * M + P * D)^2 + (8 * T)^2)^(1/2);
 
 
n = solve(eq, n);
n = subs(n, [S_y, D, P, M, T], [260e6, 100/1000, 50e3, 5e3, 8e3]);
n = eval(n)
n = 2.6101
 
% Solution - Maximum energy of distortion
syms S_y n
                        D M P D
                            T
 
eq = S_y/n - 4/(pi*D^3)*((8*M+P*D)^2+(48*T)^2)^(1/2);
 
n = solve(eq, n);
n = subs(n, [S_y, D, P, M, T], [260e6, 100/1000, 50e3, 5e3, 8e3]);
eval(n)
ans =
                        0.5282

Question 2

A solid steel shaft carries belt tensions (at an angle

from the y axis in the yz plane) at pulley C. For

and a factor of safety of n, design the shaft according to the following failure criteria, given

and

a. Maximum sheer stress

b. Maximum energy of distortion.

Solution

% calculate D using maximum shear stress

syms S_y n D M T

eq = S_y/n - 32/(pi * D^3) * (M^2 + T^2)^(1/2);

eq = subs(eq, [S_y, n, M, T], [250e6, 1.5, 1.125e3, 0.6e3]);

D = eval(solve(eq, D, 'Real', true)) %[m]

D = 0.0427

% Calculate D using maximum energy of distortion

syms S_y n D M T

eq = S_y/n - 32/(pi * D^3) * (M^2 + 3/4*T^2)^(1/2);

eq = subs(eq, [S_y, n, M, T], [250e6, 1.5, 1.125e3, 0.6e3])

eq =

D = eval(solve(eq, D, 'Real', true)) %[m]

D =
                        0.0423

Question 3

Given W = 2000 N/m and L = 3m calculate the yield strength for a safety factor of 3 and a material with a diameter of 200mm with maximum shear stress theory of failure.

Solution

%To find the bending moment, take moments about x

% sum(M_x) = M + W*x*x/2 - F_a*x = 0

% M = x(F_a - x*W/2)

M = 1.5 * (3 - 1.5 * 2 / 2) %at midpoint

M =
                        2.2500

syms S_y n D M T

eq = S_y/n - 32/(pi * D^3) * (M^2 + T^2)^(1/2);

eq = subs(eq, [D, n, M, T], [200/1000, 3, 2.25e3, 0])

eq =

S_y = eval(solve(eq, S_y)) %[MPa]

S_y =
                        8.5944e+06

Question 4

A shaft of diameter D rotates at 600 rpm and transmits 100 hp through a gear. A square key of width w is to be used as a mounting part. Determine the required length of the key.

Given: D = 50 mm, w = 12 mm.

Design Decisions: The shaft and key will be made of AISI 1035 cold-drawn. The transmitted power produces intermittent minor shocks and a factor of safety of n = 2.5 is used.

% Solution
                
 
% Write out what is known
P_hp = 100; %[hp]
N = 600; %[rpm]
S_y = 460e6; %[Pa]
r = 25/1000; %[m] shaft radius
n = 2.5; %[] safety factor
 
% key known dimensions
D = 50/1000; %[m]
w = 12/1000; %[m]
 
% Convert HP to SI units, 1HP = 0.7457 W
P = 100 * 745.7 %[W]
P =
                        74570
 
% Calculate angular velocity
omega = N * 2 * pi / 60 %[rad/s]
omega
                        = 62.8319
 
% Calculate torque
% P = T * omega
 
T = P / omega %[Nm]
T = 1.1868e+03
 
% Calculate force on the shaft surface
% T = F * r
                
 
F = T/r %[N]
F =
                        4.7473e+04
 
% Calculate the key length required
% L = (2 * F) / (S_y * w)
 
L = (2 * F * n)/(S_y * w) %[m]
L =
                        0.0430