Question

Problem 1 (a) (200 points) Shaft fatigue strength In the figure below, a rotating shaft (for speed reducing gearbox) is subjected to radial forces at A-400 lbrand at C-800 iht. These radial forces are caused by spur gears mounted to shar (a result of the pressure angle of the teeth, 20 degrees) that mesh with 2 other spur gears, not shown. Also, The shaft transmits a constant torque of 11110 in-Ibr (925 - due to the tangential force (s: transmitted load) at C of 2222 lbr (NOTE: there is bending in two planes). The shaft is supported by cylindrical roller bearings at Band D. The shaft is machined from AISI 4130 Q&T 1200 deg. F steel. 1) Find S., Sy, and so 2) Using Mawlab or EXCEL plot the modified Goodman tine (futigue failure criteria). 3) Using Von-Mises stresses a'. plot the load line on same plot as part 2 above) and determine and allowable (using modified Goodman line) if the fatigue factor of safety ar-2 (infinite life). 4 What are Sand Se using modified Goodman criteria? (the intersection of load line and Goodman line) 5) Determine the required shaft diameter for strength and infinite life criteria (Goodman). Note: the stress concentrations at two locations: the two keyways at each spur gear. Assume the stress concentration factors to be: K-1.7 for bending and K-2 for torsion. F-400 lb Fe=100 Wt-2222 W-1111 Profile -Keyways B 10 1 ---10" -15" -- -10" - Problem 1 (b) (100 points) Shaft deflection Once a tentative shaft diameter is determined for life and strength from Problem la, check it for deflection; Use singularity functions to find <v> and yo The deflection of the shaft, y<z>, must not exceed 0.010 inch (spur gears) at A and C. The slope of the shaft, 0<x>, must not exceed the values listed in table below (0.0012 rad.) for cylindrical roller bearings at Band D Problem 1(c) (100 points)

Answer 1

youlbf K-20" SUOLA -22221bf 0-101 Load Calalation Ibf D A B Verted looding yoolbt you 800 lbf - K Y 10' Isl loll 15" Rov RBV 2M-0 3200155 in A + B c Rov X 25 + 400x10 = fuoxi Por = 32016! RBV = 1200-20 88oo istin D C RBV (Ventreal BMD) 880 lbh 8014 6666 lytin 17776 Istin A Horizontal loadin eg 1111sf 2222154 LE IS 10 C D (Hborzontal BMD) Resultant BMD ZA RBH Ron Тz что MR = SMA 24 Mr? ethin) 11039.72626 15orin (max) [18061.73 2367 boin. = !1!! 357 RISHTROH & EMC RDHX 25= llixio B C A (our) (Acw) Torsion Diagram: T=1111* 10 =2222*5 11110 lbfiin +2222*15 (RoH 1777.6.15 RBH = !11-1777.6 RBH = -666.6 lb TRBH = 666.615th A ell0 lbfi in

to dog.f steel, As per question, Kg =1.7 , k =2.0 shaft materialin A1s1 4130 Q & T 1200 의 Quenched and tempered to Su 1200 deg.fi L 16 1ksi 138 ksi = Sy Oy E = 297 ksi at Now, / Sé=osxs, = so.skse, Ka Kay Kckd asé Machineda cold drawn And se ka 0.68 Kas surface fimesh factor 068 kg = size factor = 0.75 for d>a" (or d>somm] 161kse out Kc= Reliability factor, Let R= 95% => ke=0.868 let natch sensefivity of the shaft be 0.8, and we will felke kq ay et is greater thanks kg fatique Stres concentration factor b K4 SO A .ر0 = In this care notch sonsefreity is not requireed as a the fastique strey concentration is provided. so Se = 0.68x0.75 X0.868 XOSX 80. Sa se = 17.81787 ksi

The eque tim of Goodman's line is Um 1 Ja де vo c161000 psi 17817 87 Psi fos So ومما f 2. 161,000 17817.87 rt s. tistikooolg 1 x 17817 87 omt 1610 oog 2 1434338535 را با دو -se 17817.8t Smt 161000 Sa w Let's plot if in Matlab. (32 Bending stress is given Sot 32M xky spx 32 Mmax 32 x 1806123 236 X1.7 Solmax 703 xd3 32% 11039.72626 X1. 7 312758 Solmar Solmin EP •РУ. 191164.5381 Solman , دلى Solmax & Solmin 251961:27 Sum d3. 2 Solmax - Solmin 60796.73095 Sba 2 23.

Todsional strey is given 167 akt to and Tmax= 1110 151 7d3 16 Tmar Tmin 16x110x2 max X2 703 7d3 ²max = +3165.5307 23 Emin ginant ? min 56582.76537 so i 2 d3 manimin 56582.76537 ta 2 دلے By Von-mises Strey. 2 & Sm Som +3 in d3 15251961.272+ +3x(56582. 76537) sma 270350:346 d3, Sa = Vota +372 d3 60996.73095?+3x056582. 76537) 115330. 2671 Sa= d3 plot the line having slope o = tant ( coses in the Goodman lineaplot. © = 23.102° Sa 00 0.42657 Sm

Solving equestion ♡ lei. ( 17817.87 Smt 161000x 0.42657 Sma 1434338535 sm= 16582.5557 psi Sa =38874.17234 psi (5) line. Using Goodman .sa Sm ། fos -N f se Sy 2703 50.346 115330.267 + n 161000 x d² 17817.8703 (1.68 + 6.47.293) = Oy d3 8-152 d. 16.304 05 2.5357" (do
MatLab Code:

% Good man equation 17817.87*sm + 161000*sa=1434338535
%So here we have two points when sa=0, sm= 80500 and when sm=0, sa=8908.935
x1= 80500; x2=0;y1=0;y2=8908.935;
%Modified goodman line, for this let the line length be 8000 and theta=
%23.102 degree=0.4032 rad
L=500000;
theta = 0.4032;
x=0;y=0; % origin
x3= x+ (L*cos(theta));
y3= y+ (L*sin(theta));

figure;hold on;
line([x1,y1],[x2,y2],'Color','g','LineWidth',1.5);
line([x,x3],[y,y3],'Color','r','LineWidth',1.5);
xlabel('Mean Stress (S_m)','FontSize',20);
ylabel('Stress Amplitude(S_a)','FontSize',20);
title('Goodman Line and Modified goodman line','FontSize',20);
set(gca,'FontSize',20);
xlim([0 200000]);
ylim([0 10000]);
grid on;
hold off;

Output:

Goodman Line and Modified goodman line 10000 9000 8000 7000 6000 Stress Amplitude(sa). 5000 4000 3000 2000 1000 0.2 0.4 0.6 0.8 11 1.2 1.4 1.6 1.8 2 Mean Stress (S) x 105