Shaft Deflection

8/13/2019 Shaft Deflection

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Sumali, Shaft Deflection Design 1

ABE 330 Design of Machin e Elements

Design Of Shafts for Def lect ions

ProblemDesign an idler shaft for an underdrive mechanism in a tractor. The shaft is labeled SIIin Fig. 1.

SpecificationsPower transmitted, hp= 80 hp.

Speed of shaft SII, Na2= 600 rpm;

Radius of gear G2= 4 in. Radius of gear G3= 3.5 in.

The pressure angle is 20o. This means radial force = tan 20

o* tangential force.

G1is a right-hand helical gear. (Verify that the axial force on G2is in the negative xdirection.)

The helix angle is 15o. This means axial force = tan 15

o* tangential force.

Assume the indicated distances in Fig. 1 are between centers of gears or bearings.Distance between bearings B3and B4= 12 in; x3and x5minimum = 3.5 in.

Maximum deflection slope at bearings B3and B4is 1o.

Maximum deflection at gear G3is 1/100 in.

SI, SII,SIIIare shafts. G1, G2, and G3are helical gears. B1, B2, B6are deep-groove ball bearings.

Figure 1 Tractor underdrive gears, shafts and bearings.

Include in report

1. Free-body diagram of shaft SII. Calculate all forces and moments.a. Isometric (3-D) sketch.b. In x-y plane. Replace gears and bearings with forces and concentrated moments. Assume axial

components of bearing forces B3y= B4y.

c. In x-z plane.2. Bending moment diagrams of shaft SII:

a. In x-y plane.b. In x-z plane.

3. (Most important result) Sketch of shaft SIIwith its dimensions and positions of gears G2and G3.

SI

B1

B2

G1

G2

SII

B3

B4

G3

SIII

G4B5

B6

x3

4

From

engine

To

load

x5

zx

y


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a. Pick x3and x5according to the specifications.

b. Calculate the diameter so that the defl ections are with in specif ications. Assume uniform diameterfor the whole shaft. If all forces and moments in one plane are much higher than forces and moments

in the other plane, use only the high-load plane for deflection calculations.

c. Round up the diameter to the nearest bearing internal diameter in Hamrock Table 13.6.

4. Calculate the safety factor against yielding.

a. Calculate the torsion in the shaft between G2and G3.

b. At G2, and G3, combine the bending moments vectorially, e.g. at G3,2

32

33 zy MMM += .

c. Calculate the bending (normal) stress at G2, and G3. (If one bending moment is much higher than theother, don't worry about calculating the smaller bending stress).

d. Calculate the torsional (shear) stress at G2or G3(same value).e. Use Mohr's circle formula to obtain the principal stresses at G2or at G3, whichever has a higher

bending moment.

f. Calculate the von Mises equivalent stresse. Safety factor against yielding = yield strength/e.

5. Plot the deflections of the shaft as a function of x:a. In the x-y plane.

b. In the x-z plane.


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Possible Solution to the shaft deflection design problem

Free-Body Diagram of shaft-gear assembly

Calculate tangential force F2yusing torque-power-rpm relationship.

Calculate vertical force F2z= F2ytan(pressure angle).Calculate axial force F2x= F2ytan(helix angle).

Calculate the reaction forces using static equilibrium equations.

G3

zx

y

F2y


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%deflab.m

%design of shaft based on deflection

%sumali Feb 2000

clf; clear

F2y=2101;

F2z=765;

F3y=2401;

F3z=874;

R2y=788;

R2z=797;R3y=1088;

R3z=842;

F2x=563;

F3x=643;

R2=4;

R3=3.5;

M2=F2x*R2;

M3=F3x*R3;

dx=0.1; Imax=121;

%Bending Moment in x-z plane

for i=1:Imax;

x(i)=(i-1)*dx;

My(i)=-R3z*(x(i));end

for i=36:Imax; My(i)=My(i)+F3z*(x(i)-3.5)-M3;end

for i=86:Imax; My(i)=My(i)+F2z*(x(i)-8.5)+M2;end

plot(x,My);

grid; title('Bending Moment in x-z plane');

xlabel('x (in)'); ylabel('M (lb.in)');

%Deflection in x-z plane

C1=(-R3z/6*12^3+F3z/6*8.5^3-M3/2*8.5^2+F2z/6*3.5^3+M2/2*3.5^2)/12;

for i=1:Imax;

z(i)=R3z/6*(x(i))^3+C1*x(i);

end

for i=36:Imax; z(i)=z(i)-F3z/6*(x(i)-3.5)^3+M3/2*(x(i)-3.5)^2;end

for i=86:Imax; z(i)=z(i)-F2z/6*(x(i)-8.5)^3-M2/2*(x(i)-8.5)^2;end

EI1=(R3z/6*3.5^3+C1*3.5)/(-0.01);

E=30e6;

I1=EI1/E;

d1=2*(4*I1/pi)^0.25;

%d turns out to be 1.4 in. Round it up to 40 mm to fit bearing

d=40/25.4;

I=pi/4*(d/2)^4;

EI=E*I;

z1=z/EI1;

z=z/EI;

figure

plot(x,z,'-',x,z1,':');legend('d=1.575in','d=1.40in');

grid; title('Deflection in x-z plane');

xlabel('x (in)'); ylabel('z (in)');

%Slope in x-z plane

for i=1:Imax;

thetay(i)=R3z/2*(x(i))^2+C1;

end

for i=36:Imax; thetay(i)=thetay(i)-F3z/2*(x(i)-3.5)^2+M3*(x(i)-3.5);end

for i=86:Imax; thetay(i)=thetay(i)-F2z/2*(x(i)-8.5)^2-M2*(x(i)-8.5);end

thetay=thetay/EI;


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figure

plot(x,thetay*180/pi);

grid; title('Slope in x-z plane');

xlabel('x (in)'); ylabel('Theta y (deg)');

%Bending Moment in x-y plane

for i=1:Imax;

Mz(i)=-R3y*(x(i));

end

for i=36:Imax; Mz(i)=Mz(i)+F3y*(x(i)-3.5);end

for i=86:Imax; Mz(i)=Mz(i)-F2y*(x(i)-8.5);end

figureplot(x,Mz);

grid; title('Bending Moment in x-y plane');

xlabel('x (in)'); ylabel('M (lb.in)');

%Deflection in x-y plane

C1y=(-R3y/6*12^3+F3y/6*8.5^3-F2y/6*3.5^3)/12;

for i=1:Imax;

y(i)=R3y/6*(x(i))^3+C1y*x(i);

end

for i=36:Imax; y(i)=y(i)-F3y/6*(x(i)-3.5)^3;end

for i=86:Imax; y(i)=y(i)+F2y/6*(x(i)-8.5)^3;end

EI2=(R3y/6*3.5^3+C1y*3.5)/(-0.01);

E=30e6;I2=EI2/E;

d2=2*(4*I2/pi)^0.25;

%d turns out to be 1.026 in. So keep d=40 mm from the x-z plane calculation

y=y/EI;

figure

plot(x,y);

grid; title('Deflection in x-y plane');

xlabel('x (in)'); ylabel('y (in)');

%Slope in x-y plane

for i=1:Imax;

thetaz(i)=R3y/2*(x(i))^2+C1y;

end

for i=36:Imax; thetaz(i)=thetaz(i)-F3y/2*(x(i)-3.5)^2;endfor i=86:Imax; thetaz(i)=thetaz(i)+F2y/2*(x(i)-8.5)^2;end

thetaz=thetaz/EI;

figure

plot(x,thetaz*180/pi);

grid; title('Slope in x-y plane');

xlabel('x (in)'); ylabel('Theta z (deg)');


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Shaft Deflection

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