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.

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

ReplyDeleteTry to look at p.96 of Strength of Materials book Singer-Pytel 3rd Edition.

ReplyDeletethere 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.

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

ReplyDeleteyup you are right...i'm wrong at this one...sorry I haven't notice it for a long time...the formula should have been

ReplyDeleteks=1+1/2C.

The choices are wrong.. I will try to fix that today...

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.

ReplyDeleteThanks and God bless!