Question

7 30.10 Required information A rotating step shaft is loaded as shown, where the forces FA and Fg are constant ot 570 lbf and 285 lbf, respectively, and the torque alternates from 0 to 1800 lbf in. The shaft is to be considered simply supported at points and Cand is made of AISI 1045 CD steel with a fully corrected endurance limit of So" 40 kpsi. Let Kr 21 and Kis -17 Take the value of design factor to be 25. Files T. Determine the minimal acceptable diameter of section BC using the DE Gerber criterion The minimal acceptable diameter of section BC is in

Answer 1

In x-y plane

E70 Iь 1 1 RA RB 0 18 (in) 6 Let the left support have a vertical reaction RA and the right support have a vertical reaction RB. Sum of forces along the y-axis is equal to zero for static equilibrium: +1 EF,=0 RA+RB - 570 = 0 RA + RB - 570 = 0 RA+RB = +5701b ----(1) Sum of moments about the left support is equal to zero for static equilibrium: + M =0 RB (18 – 0)+(-570) (6) = 0 RB (18 - 0) + (-570) (+6)=0 Rp = +1901b ---(2) Substitute (2) into (1): RA+RB = +570 RA+190 =+570 BA=+380lb

Shear for 0 sxs6 1 Vis) 380 lb 0 18 (in) Take a cut for 0 <<<6: ++ SF,=0 +380 - V1 = 0 V1 (2) = +380 ..V(2) = +380 for 0<x<6

Shear for 6 <x18 670 lb V2() 380 lb 0 18 (in) 6 Take a cut for 6 <I<18: ++ SFy=0 +380 – 570 - 12) = 0 V22) = -190 ..V2(3) = -190 for 6<< < 18

Moment for 0 5x56 M) 380 lb 18 (in) Take a cut for 0 <2<6: +b2M=0 +(380) (2 - 0) - M (2) = 0 M(C) = +3800 ..M(2) = +380 for 0<x<6

Moment for 6 SX S18 670 lb M.() 380 lb 18 (in) Take a cut for 6 <r <18: +) SM = 0 +(380) (2 - 0)+(-570) (1-6) - M2(T) = 0 M22) = +3420 - 190x ... M2(x) = +3420 - 190.2 for 6<< 18

(lb) Shear Force 380 0 -190 2 6 18

Bending (lb-in) Moment 2280 2 1 1 0 ! + 6 18

In zx plane

285 lb + 1 RA RB 0 18 (in) 12 Let the left support have a vertical reaction RA and the right support have a vertical reaction RB. Sum of forces along the y-axis is equal to zero for static equilibrium: + 1 EF, = 0 RA+RB - 285 = 0 RA+RB - 285 = 0 RA+RB = +2851b ---(1) Sum of moments about the left support is equal to zero for static equilibrium: +OSM=0 RB (18 - 0) + (-285) (12) = 0 RB (18 – 0)+(-285) (+12) = 0 Rp = +1901b (2) Substitute (2) into (1): RA +RB = +285 RA+190 =+285 RA=+95lb

Shear for 0 <x< 12 VIO) 95 lb 0 18 1 in) T Take a cut for 0 <I< 12 : ++ SFy=0 +95 - V1 (2) = 0 V1 (2) = +95 ..V(3) = +95 for 0<< 12

Shear for 12 5xs18 285 lb VE) 95 lb 0 00 (in) 12 Take a cut for 12 <I<18: +1 EF, = 0 +95 – 285 - 12(x) = 0 V2(x) = -190 ..122) = -190 for 12 << 18

Moment for 0 SX512 M () 3 95 lb 0 18 I (in) Take a cut for 0 <I< 12: +0 SM,=0 +(95) (2-0) - M (T) = 0 M(C) = +950 ..M (2) = +952 for 0 << 12

Moment for 12 5x5 18 285 lb M (E) 95 lb 0 18 (in) 12 H Take a cut for 12 <I<18: + 2M =0 +(95) (2-0)+(-285) (2 - 12) - M22) = 0 M22) = +3420 – 1902 ..M2(x) = +3420 – 190:0 for 12 << < 18

(lb) Shear Force 95 0 -190 2 12 18

Bending (lb-in) Moment 1140 2 0 12 18

For AISI 1045 CD steel:

$\small \\\\\text{Ultimate tensile strength }S_{ut}=91 \ ksi \\\\\text{Yield strength }S_{y}=77 \ ksi \\\\\text{corrected endurance limit }S_{e}=40 \ ksi$

Mean torque, Tm TO + T. min 2 1800+ 0 = 900 lb-in 2 Alternating torque, T, = Tmax – Tmin 2 1800 - 0 = 900 lb-in 2 161 Mean shear stress, T, πd 16x900 psi πd 16T. Alternating shear stress, To = πd 16x900 Ad psi

T K; ta.n=1 +K Kiš T eg

2 ksi + TA 1 х 2.1 16x900 x 10-3 1.7x16x900 x10- 2.5 πd πd 40 92.65 0.48688 = 1 ( tica. (ds)? d 92.65 ksi d' (di -0.48688)

for bending stress

$\small \\\\\sigma_b=\frac{32M}{\pi d^3}=\frac{32*\sqrt{1140^2+1140^2}}{\pi d^3} \\\\\sigma_b=\frac{51590.5}{\pi d^3}$

$\small \\\\\sigma_b=\frac{51.5905}{\pi d^3} \ ksi$

For combined shear stress

$\small \\\\\sigma_b^2+4\tau_{eq}^2 \le \frac{s_yt^2}{n^2} \\\\\left (\frac{51.5905}{\pi d^3} \right )^2+4\left ( \frac{92.65}{d^3(d^3-0.48688)} \right ) \le \frac{77^2}{2.5^2} \\\\\left (\frac{51.5905}{\pi d^3} \right )^2+4\left ( \frac{92.65}{d^3(d^3-0.48688)} \right )=948.64 \\\\\frac{269.98}{d^6}+\frac{370.6}{d^9-0.48688d^3}=948.64 \\\\\text{by solving this } \\\\d=1.0036849 \ in \ \ \ \& \ \ d=0.62617 \ in$

So the minimum diameter required is 1.0036849 in