Mechanical Engineering Design 2 (MECH3110) Assignment Solution, 2019

This document presents a detailed solution to a Mechanical Engineering Design 2 (MECH3110) minor assignment. The assignment addresses two primary engineering design problems. The first problem focuses on determining the minimum weld size required for a sling connected to two cantilever PFCs, considering two different welding scenarios and checking for tearing stress. The second problem involves analyzing a plate held by four bolts, calculating the bolt size required to withstand a horizontal load, and subsequently, a fatigue analysis of a shaft under combined bending and torsion, using Soderberg's method, stress concentration factors, and calculating equivalent bending and twisting moments to determine the required shaft diameter to prevent fatigue failure. The solution includes detailed calculations, formulas, and intermediate steps to arrive at the final answers, making it a valuable resource for students studying mechanical engineering design.

Solution
Q2)
Bolt group
∑ x2=2∗2002=80,000
∑ y2 =2∗2002 =80,000
∑ x2 +∑ y2=80,000+80,000=160,000
r1 = √ 160,000 = 400
e=275
Load due to the moment = FT = 275∗10∗400
160,000 = 6.875 kN
Load due to shear = 10.4 = 2.5
Where,
cos ∅ = 100
141.42 = 0.7071
100 141.42
100
Resultant load = FR = ¿ = 8.82 kN
F . S= ultimate stress
Permissible stress
Ultimate stress as per SAE 4.6 = 60 %∗400 MPa=240 N /m m2
Permissible stress = 240/3 = 80 N/mm2
But,
FR = Area * permissible stress

8.82 = π d2
4 ∗80∗10−3
d2=140.375
d = 11.85 mm
Q3a) Soderberg’s
Sut =850 Mpa
Sy=700 Mpa
T2
C
O T3 = 800 N A B
T4 = 160 N
T1
T2 = 0.2T1
T 1
T 2 = T 3
T 4
Torgue on the pulley A
= (T3 – T4)RA
= (800 -160)*0.55
= 352 N.m
Since Torque on the both pulley A and C are same
(T1 – T2)RC = 352
T1 – T2 = 352
0.3 =1173.33
T1 – 0.2T1 = 1173.33
0.8T1 = 1173.33

T1 = 1466.67 N
T2 = 0.2*1466.67 = 293.33 N
Vertical load on pulley C = T1 + T2
Wc = 1760 N
Horizontal force on pulley A = T3 + T4
WA = 960 N
∑ M 0 =0
Rv0 RvB = 960
RVB * 0.85 = 960 * 0.5
RVB = 564.71 N
RVO = 960 – 564.71 = 395.29 N
Moment at O and A
MOV = MAV = 0
B.M at A
MAv = 395.29 * 0.5 = 197.65 N.m
B.M at C
MCv = 564.71 * 0.35 = 197.65 N.m
ROH + RBH = 1760 N
MOH + MAH = 0
1760 * 0.325 = RHO*0.85
RHO = 672.94 N
RHB = 1760 – 672.94
= 1087.06
MAH = 672.94 * 0.5 = 336.47 N.m
B .M at C

MB = 1087.06 * 0.325 = 353.29 N
Resultant bending at A
MB = √ 197.652+ 336.472
= 390.23 N.m
Resultant bending at C
MB = √197.652+ 353.292
= 404.82 N.m
Maximum bending moment = 404.82 N.m
Twisting moment (Torque)
τ a= √m2+T 2
= √ 404.822 +293.332
= 499.92 N.m
Equivalent bending moment
Me = 1
2 ( 404.82+ 499.92 ) =452.37 N . m
τ e1= √ m2 +T2
= √390.232+1466.672
= 1517.70 N.m
Midrange of bending moment and torsion respectively
Mm = 390.23+404.82
2 = 397.523 N.m
τ m= 499.92+1517.70
2 = 1008.81 N.m
Stress concentration factor for bending
Kf = 404.82
390.23 =1.04

Stress concentration factor for torsion
Kf = 1517.70
499.52 =3.04
d=¿ ¿
Note that
Se=0.5∗Sut
= 0.5 * 850 = 425 Mpa
Substituting the values
d=¿ ¿
= 52.43 mm
b) Goodman fatigue
d=¿ ¿
Substituting the values
d=¿ ¿
= 50.71 mm