The first-stage turbine disk for a turbofan engine has a bore radius of 5\"., a

Question

The first-stage turbine disk for a turbofan engine has a bore radius of 5"., a rim radius of 15"., and a thickness (which is assumed to be uniform from bore to nm) of 3". Attached to the disk are 40 turbine blades, each weighing I lbf (including attaching hardware). When the blades are mounted in the disk, the radius from the center of rotation to the blade centroid is 16". The maximum design rotor speed is 8000 RPM and you should allow for 10% overspeed. Find the distribution of radial and tangential stresses in the disk due to both the centrifugal forces of the blades pulling on the rim of the disk and the rotational speed of the disk. The disk material is Waspaloy, a high-temperature nickel-based alloy with a weight density of 0.286 lbf/in^3 and a Poisson's ratio of 0.30. Plot the radial and tangential stresses due to just the rotation of the disk, with no blades attached. Plot the radial and tangential stresses due to just the pull of the blades, with no rotational stresses. Plot the radial and tangential stresses due to both blade pull and disk rotation. Which effect creates greater stress on the disk? Determine the worst stress location on the disk and obtain the maximum shear stress at that location.

Explanation / Answer

solution: here given that disc has bore diameter d1=10 in and rim diameter d2=30 in

where Nmax=8000*1.1=8800 rpm

w=2*pi*Nmax/60=921.53 rad/s

material is was ploy with weight density=.286 lb/in3

hence mass m=weight density*volume/g

3) when disc is non rotating and blades are attached at moment radial stress is given as

sr=W/((d2-d1)*thickness)

total weight of blades are=n*Wb=W1=40*1=40 lbf

weight of disk=m*g=weight density*Area of disc*thickness=W2=539.097 lbf

toatl W=W1+W2=40+539.097579.097 lbf

here radial stress is

Sr=W/((d2-d1)t)=579.097/(20*3)=9.6516 lbf/in 2

where tangential stress induced due to lateral deformation along radial direction

hence e=(1/E)(Sr-poisson ratio*St)

as here strain is zero due to zero deformation we get

st=Sr/poisson ratio=9.6516/.3=32.17 lb/in2

4) disc is rotary then along with load W ,centrifugal force is also acting on disc

Fc1=m(r2)w2=1.3958*15*921.53^2=17780458.23 lb

hence W3=W+Fc1=17781037.33 lbf

sr=W3/((d2-d1)t)=296350.622 lbf/in2

and St=Sr/poisson ratio=987835.40 lbf/in2

5) when blade are rotating mounted on disc then total force is

W4=W3+Fc2b

Fc2b=Mb*Rk*w2=1*16*921.532=13587480.65 lbf

hence W4=31368517.98 lbf

where Sr=W4/t*(d2-d1)=522808.633 lbf/in2

and St=Sr/.3=1742695.443 lbf/in2

5) hence rotational effect produce high value of stresses on disc and it is limiting conditioned for selecting material for disc and blade

6) worst stress location is maximum stress location and it would be at surface of disc where blade has to attached.

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