## Friday, March 12, 2010

### Board Exam Refresher: Machine Design (Answers)

Here are the answers and solution to the previously posted set of questions in Strength of materials topic.

1. A steel rod 40 mm in diameter 800 mm long cannot extend longer than 41.5 mm. Find the maximum allowable load it can carry in Newtons. E for steel is 200 GPa.
A. 471240
B. 600000
C. 188495
D. 712440

Solution:
Y=PL/EA
P=YEA/L
P=1.5mm (200000N/mm2)(3.1416)(40mm)^2/(4)(800mm)
P=471240N

2. A simply supported beam 3m long has a load of 3KN 1 m from the left support. Calculate the maximum moment.
Mmax=Fab/L
Mmax=3KN(1m)(2m)/3m
Mmax=2KN-m

A. 2.25 kN-m
B. 2.0 kN-m
C. 4.5 kN-m
D. 6.0 kN-m

3. Calculate the maximum deflection of a 12 m beam with fixed ends if the load applied on it is 3KN located at its center. (I=4,200,000 E=200 GPa).
Y=PL^3/192EI
Y=3000N (12,000 mm) ^3/192 (200,000N/mm2) (4,200,000mm4)
Y=32.14 mm

A. 128.6 mm
B. 32.14 mm
C. 0.17 mm
D. 0.27 mm

4. A 40mm short shaft has is rotating at 300RPM. Calculate the power capacity.
For short shafts: (PME Code)
P=D^3*N/38
P-in hp
D-inches
N-rpm
P= (40/25.4) ^3(300)/38
P=30.83hp (0.746hp/kw)
P=23 KW

A. 30.83 kW
B. 23 kW
C. 16.34 kW
D. 10.93 kW

5. Calculate the wall thickness required for a 1m spherical thin walled vessel carrying 30Mpa of fluid pressure. The maximum allowable stress is 150Mpa.
S=PD/4t
150MPa=(30N/mm2)(1000mm)/4(t)
t=50 mm

A. 100mm
B. 200 mm
C. 150 mm
D. 50 mm

6. A 33 mm long single leaf spring is carrying a load of 50 grams 3 mm from its tip. If the material used in making the spring is a 0.3 mm sheet metal, calculate the required width if the allowable stress is not to exceed 250 Mpa.
Leaf spring is just a cantilever beam. h=0.3 mm
S=Mc/I
M=PL= (50g) (30mm) (1kg/1000g)(9.81N/kg)=14.715N -mm
250= (14.715N -mm) (0.3/2)/ (b*0.3^3)/12
b=3.924mm

A. 4.12 mm
B. 0.71 mm
C. 3.92 mm
D. 1.18 mm

7. A compression spring has the following data:
Wire diameter: 0.5 mm
Solid height: 5mm
OD: 4 mm
Calculate the Wahl’s factor.
K=(4C-1)/(4C-4)+0.615/C
C=(OD-WD)/WD=(4-0.5)/0.5
C=7
K= (4*7-1)/ (4*7-4) +0.615/7
K=1.213

A. 1.213
B. 1.184
C. 1.145
D. 1.162

8. A spring has a Wahl’s factor of 1.212, a free length of 8 mm, and an index of 7 calculate the curvature factor.
K=KsKc
Ks=direct shear factor
Kc=curvature factor
Ks=1+1/2C
Ks=1.071
Kc=1.212/1.071
Kc=0.942

A. 1.125
B. 1.051
C. 0.962
D. 1.131

9. A 20 mm diameter, 700 cm long rod elongates 2 mm at a load of 15KN. Calculate the modulus of elasticity of the material used.
E=s/e
E=deformation/orig. length
e=2mm/7000mm=0.000286
s=F/A=15000N/ (0.25*3.1416*20^2); A=0.25pi (D^2)
s=47.75 MPa
E=47.75 MPa/0.000286
E=166945Mpa
E=167 Gpa

A. 167 Gpa
B. 668 GPa
C. 743 GPa
D. 334 GPa

10. Calculate the force required to punch a 10 mm diameter hole into a 2 mm thick sheet metal with the these properties: Sy=155 Mpa Su=300 Mpa
S=F/piDt
300N/mm2=F/3.1416(10mm)(2mm); use Su because you are going to cut the steel.
F=18849.6 N
F=18.85kN

A. 11.31 kN
B. 9.74 kN
C. 18.85 kN
D. 14.14 kN

If there's any error in one of the questions or answers, please let me know.

1. Ks=1+2/C, sir where did you got this formula?

2. Try to look at p.96 of Strength of Materials book Singer-Pytel 3rd Edition.
there you can see the quantity (1+d/4R) d is the wire dia of spring while R is the spring mean radius.
d/R is actually equal 2/C coz C=2R/d
That's the factor for direct stresses w/o accounting for the curvature stresses experienced by the wire while it is being coiled during the manufacturing process.

3. sir if you substitute d/R=2/C in (1+d/4R), it will become (1+(1/2C)) not (1+(2/C))

4. yup you are right...i'm wrong at this one...sorry I haven't notice it for a long time...the formula should have been
ks=1+1/2C.
The choices are wrong.. I will try to fix that today...

5. the question has been fixed...thanks for the feedback...I'm trying to make it as error free as I can but sometimes, simple mistakes happen. I really appreciate my readers comments.
Thanks and God bless!