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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
Load 50 grams
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.

Comments

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

    ReplyDelete
  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.

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

    ReplyDelete
  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...

    ReplyDelete
  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!

    ReplyDelete

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