n = 2.6101
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 D M P D T
eq = S_y/n - 4/(pi * D^3) * ((8 * M + P * D)^2 + 64*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)
% 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)
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 
.
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
There is a bending moment and torsion i.e. M and T.
Determine reaction forces, bending moment and torques from the information given.
% 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]
% Calculate D using maximum exergy 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])
D = eval(solve(eq, D, 'Real', true)) %[m]
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
% Bending moment at any point from the left is given by the equation,
% maximum will be at the centre
% M = x(F_a - x*W/2)
M = 1.5 * (3 - 1.5 * 2 / 2) %at midpoint
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])
S_y = eval(solve(eq, S_y)) %[MPa]
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, 1 HP = 0.7457 W
P = 100 * 745.7 %[W]
% Calculate angular velocity
omega = N * 2 * pi / 60 %[rad/s]
% Calculate torque
% P = T * omega
T = P / omega %[Nm]
% Calculate force on the shaft surface
% T = F * r
F = T/r %[N]
% Calculate the key length required
% L = (2 * F) / (S_y * w)
L = (2 * F * n)/(S_y * w) %[m]