Question

Consider a drive shaft design concept as shown in the accompanying figure. Assuming the shaft has a 25 mm diameter and is machined from steel with tensile and shear yield strengths of 330 MPa and 191 MPa respectively, answer the following questions: 220 mm B Ic 330 mm 100 mm 800 N 220 mm 75 mms 200 N 200 F a) Assuming a smooth transfer of power, determine the force F. [6 marks] b) In the x-- plane, determine the reaction forces at the bearings and construct the shear force and bending diagrams. [20 marks] c) In the y-- plane, determine the reaction forces at the bearings and construct the shear force and bending diagrams. [13 marks] d) Since bending moments are vector quantities, find the position of the maximum bending moment and the safety factor against yielding in bending. [9 marks] e) Determine, the position of the maximum torque and the safety factor against yielding in torsion. [8 marks] Notes: 1) Assume Force Facts at a radius of 75 mm as indicated on gear A (i.e. the pitch circle radius). ii) In the event of incorrect answers, bonus marks may be given if you interpret your answer and provide 'error checks', such as checking the dimension and comparing your answer to well-known elementary solutions. n

Consider a drive shaft design concept as shown in the accompanying figure.

Answer 1

Given

$\\ \phi=20\degree \\ d=25\ mm \\ \sigma_y=330\ MPa \\ \tau_y=191\ MPa$

a)Take $F_t$ as tangential force on gear

In the drive shaft design the torque is transfered at gear and belt drive

Torque transfered by the shaft based on belt drive tension

Torque equation with respect to z-axis

$\\ 0*\ddot{\theta}=F_t*75-T \\ F_t=T/75=60000/75=800\ N$

The gear force

F= Ft/ cos() = 800/ cos(20°) = 851.34 N

b)

The radial gear force

$\\ F_r=F_t\sin(\phi) = 291.18\ N$

Loading x-z plane

300+200=1000 N Fr 220 C ต่อ 0, S 710

Take moment balance about O in y direction

C: * 770 + 1000 * 550 = F, * 220 F, * 220 – 1000 * 550 C -631.09 N 770

Take force balance in x direction

Shear force diagram

600 631.09 N 400 200 -220 -550 770 0 -Z -200 -77.731 N -368.91 N 400

Bending diagram

- 220 550 - 770 -Z -17101 mm -20000 40000 60000 80000 -100000 -120000 - 140000 -138840.35 Nmm

c)Loading in y-z plane

220 D B ต่อ Oy Free 210

Take Moment balance about O in x direction

Taking force balance in y direction

Shear diagram

400 200 228.57 N -Z 0 -220 -550 -770 -200 -228.57 N 400 -600 -571.43 N -800

Bending diagram

-220 -550 -770 -Z -20000 40000 - 60000 -50285.714 N mm 80000 -100000 -120000 -140000 -125714.286 Năm

d)

Resultant bending moment can be calaculated as square root of sum of square bending moment in y and x direction

At A

$M_A=\sqrt{17101^2+125714.286^2}=126872.08482 \ Nmm$

At B

$M_B=\sqrt{138840.35^2+ 50285.71440^2}=147666.16356 \ Nmm$

Maximum bending momen is at B

Maximum Bending stress in the shaft

$\\ \sigma=\frac{32M_B}{\pi d^3}= \frac{147666.1635*32}{\pi* 25^3}=96.263\ MPa$

Factor of safety for bending against yielding in bending

$\\ FOS_{Bending}=\frac{\sigma_y}{\sigma}=3.4281$

d)

Torque diagram

60000 Nmm 70000 60000 50000 40000 30000 20000 10000 - C اله - 22 -550 - 770

Maximum torque is between gear and belt drive

Maximum Torsional shear

$\tau =\frac{16T}{\pi d^3}=\frac{16*60000}{\pi*25^3}= 19.557$

Factor of safety against yielding in torsion

$FOS=\frac{\tau_y}{\tau}= 9.7663$