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UNIVERSITY OF CEBU MAINCLASS ACTIVITY 2021MACHINE DESIGN 1 PROBLEM SETS(1,2,3,4,5,6,7)CLASS WORK ME415B(1701H - 2001H)12401-MACHINE DESIGN 1Instructor: Engr, Rodel NavalSTRESSES & MACHINE ELEMENTSPROBLEM SET 1INSTRUCTION: Encircle the letter that corresponds to the correct answer of your cho...

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UNIVERSITY OF CEBU MAIN CLASS ACTIVITY 2021 MACHINE DESIGN 1 PROBLEM SETS (1,2,3,4,5,6,7) CLASS WORK ME415B (1701H - 2001H)

12401-MACHINE DESIGN 1 Instructor: Engr, Rodel Naval

STRESSES & MACHINE ELEMENTS PROBLEM SET 1 INSTRUCTION: Encircle the letter that corresponds to the correct answer of your choice. MULTIPLE CHOICE: 1.

Two cylinders rolling in the opposite direction has a speed ratio of 3.

If the diameter of driver is 10 inches,

find the center distance between cylinders. A. 15 in

B. 10 inC.

25 in

D. 20 in

Given: N = 3 Di = 10 Solution: �1+�2 2

D1n1=d2n2 �1 �2

C=

�2

= �2

=

�2

10+30 2

C = 20in

3=10�� D2=30in

2. Two cylinders rolling in the same direction has speed ratio of 2.5 and center distance of 50 cm. Find the diameter of the larger cylinder. A.

56.34 in

B.

263.45 in C.

166.67 in

D.

66.67 in

�2

�2−2.5

Given: N=2.5 C=50cm Solution:

50cm= 2 d2=166.67cm

�2−�1  (eq.1) 2 �1

C=

D1( ) =d2 �2 d1(2.5) =d2

d1=

�2

 (eq. 2) 3. Four cylinders rolling in opposite directions has a speed ratio of A:B:C:D= 5:3:1:4. cylinder A and C is 70 inches, find the diameter of cylinder D. A. 18.75 inB. 26.34 in C. 12.45 in D. 16 .37 in 2.5

If center distance between

Given: Na:Nb:Nc:Nd=5:3:1:4 Cac = 70in Solution: N1/N2=D2/D1 5/3 = D2/D 3D3 = 5Da Da = 3D3/5 Db=25 N2D3=N3Dc Dc=3(25) =75 D3=Dc/3 N3Dc=N4Dd Dc=3D3 (1)(25)=(4)Dd C = Da/2 + D3 + Dc/2 Dd = 18.75in 70in=3D3/5(2) + D3+3D3/2

4. A pulley has a tangential velocity of 75 fpm. If pulley diameter is 10 inches, find the speed of the speed. A. 24.34 rpm B. 28.65 rpm C. 34.22 rpm D. 32.34 rpm

Given : V = 75fpm d=10in V=��� Solution: �� 1� x ��� 3.281��

75

1��

1�

= π (10inx12��x3.281��) N

N=28.65rpm

5. Shaft A with 12 inches diameter pulley running at 250 rpm is connected to shaft B by means of 26 inches diameter pulley. Another pulley on shaft B 16 inches diameter is connected to 10 inches diameter pulley on shaft C. Find the speed of shaft C. A.

184.61 rpm B.

284.56 rpm

C.

173.45 rpm

D.

197.44 rpm

Given : Da=12in Na=250rpm

Solution: N3=

���� �3

Nc= =

���3

(16��)(115.38���) 10��

��

12�� (250��)

D3=26in Nc=?

= 26�� N3=115.38rpm

Nc=184.61rpm

6. A steel tie rod on bridge must be withstand a pull of 6200 lbs. Find the diameter of the rod assuming a factor of safety of 4 and ultimate stress of 66,000 psi.

A.

0.234 inB.

0.534 in C.

0.691 in D.

0.734 in

Given: F=6200lbs Fs=4 Su=66,000psi d=? Solution: � �� =� �� �^2 4

66,000��� 62000��� = ��^2 4 4

D=0.692in

7.

If the ultimate shear strength of steel plate is 45,000 psi, what force is necessary to punch a 0.9 in diameter hole in a 0.5 in thick plate using a factor of safety of 3.5. A. 63,617 lbs B.

61,567 lbs

C.

65,378 lbs

D.

69,345 lbs

Given:

S=45000psi D=0.9in T=0.5 FS=3.5 Solution: � S= ���

�

45000psi=

�(0.9��)(0.5��)

F= 63,617lbs

8. A 2.5 in diameter by 1.8 in long journal bearing is to carry 5000 lbs load at 320 rpm using SAE 40 lube oil at 200oF through a single hole at 30 psi. Compute the bearing pressure.

A.

1111.11 psi

B.

Given: d=2.5in L=1.8in F=5000lbs

142.23 psi

Sb=

C. 123.34 psi D. 197.34psi

��

�� 5000��� =(2.5��)(1.8��)

Sb= 1111.11psi

N=320rpm S=30psi

9. A journal bearing has 8 cm diameter and length to diameter ratio of 3. Find the projected area in mm2. A. 19,200 B. 20,009 C. 18,058 D. 17,017 �

1�

Given: = 3 3 L=3(8)=24cm A=dl=8(24) A=192cm2

1000�� 1�

A=192cm2x100��x A= 19200mm2

10. A cable steel has a length of 100 m and stretch to 5 cm when the load is applied at both ends. If tensile stress is 50 psi, find the modulus of elasticity of the steel.

A.

100,000 psi

B.

120,000 psi

C.

110,000 psi

D.

130,000 psi

Given: L = 100m y=5cm Ss=50psi ������

Solution: E=

������

50���

0.05� 100�(

39.37�� 1� 39.37�� 1� )

E= 100,000psi

11. A shaft whose torque varies from 2200 to 6400 in-lb.

It has a diameter of 1.25 inches and yield stress of

63,000 psi.

A.

Find the variable component stress.

6524.45 psi

B.

4245.56 psi

C.

5475.95 psi

D.

7834.56 psi

Given: Tmax= 6400in-lb Tmin=2200 in-lb d=1.25in Solution: 16(����−����) Sa= 2��3

16 6400−2200 ��−��

=

2�(1.25��)

Sa=5475.95psi

12. How many 1/2 inch diameter hole that can be punch in one motion of a 1/8 inch thick plate using a force of 50 tons. The ultimate shear stress is 52 ksi and factor of safety of 3. A. 7 B. 9 C. 8 D. 10 Given: d= 1/2in t=1/8in F=50tons Su=52ksi Fs=3 Solution: �

t=

Su=���

1.22��

1/8��

��

2000�� 50����( ) 1���

52000 = �(1/2��)(�) ��

= 9.76

t=10

13. Determine the minimum diameter of a taper pin for use to fix a lever to a shaft, if it is to transmit a maximum torque of 750 in-lb. The shaft diameter is 1.5 inches and has a stressof 20,000 psi. A. 0.252 in B. 0.452 in C. 0.642 in D. 0.826 in

Given:

solution:

T=750in-llb D=1.5in

1.5 2

750in-lb=F( F=1000lbs

)

1000��� � �2 4

S= 20000psi

20000psi= D=0.252in

14. A 19mm stud bolts is used to fastened on a 250 mm diameter cylinder head of diesel engine. If there are 10 stud bolts, determine the pressure inside the cylinder if bolt stress is 50 Mpa. a. 288.8 KPa

B.

2888 KPa

C. 3426 KPa

D.

14176

Given:

solution:

Ds=19mm

50,000Kpa= �

Dc=250mm S=50Mpa

Fb=14176KN(10) � P=�

P= �(0.25)^2 4

��

S= ��2 4

4828 Kpa

��

(0.19�)

4

P=2888KPA

15. A column supports a compressive load of 250 KN. Determine the outside diameter of column if inside diameter is 185 mm and compressive stress of 50 Mpa.

A.

200.62 mm

B.

201.47 mm

C.

216.42 mm

Given: F=250KN

Solution: � F=S(4 (��2 − ��2))

do=?

Do2-di2=

di=185mm

do2=

Sc=50MPA

do=

S=� 4

�

� �

D.

208.41 mm

d=201.47mm

�

�

�(4)

+di2

�(4) 4� ��

+ ��2

4(250��)

do= �(50000���) + (0.185) 2

(��2−��2)

16. A steel hollow tube is used to carry a tensile load of 500 KN at a stress of140 Mpa. If outside diameter is 10 times the tube thickness of the tube.

A.

11.24 mm B.

107 mm C.

20.64 mm

Given: F=500KN

D.

solution: � S=� 4

S=140000Kpa

22.61 mm

(��^2−��^2) 500��

140,000Kpa= �( 10� ^2− 8� ^2) 4

do=10t

t=11.24mm

do=di+2t 10t=di+2t di=-2t+10t=8t

17. A 20 mm diameter rivet is used to fastened two 25 mm thick plate. If the shearing stress of rivet is 80 Mpa, what tensile force applied each plate to shear the bolt?

A. Given: D=0.02m t=0.025m S=80Mpa

26.35 KN B.

28.42 KN C.

30.41 KND.

25.13 KN

Solution: �

F=2513KN

S=��2 4

80,000Kpa=�

�

0.02 ^2 4

18. Two 30 mm thick plate is fastened by two bolts, 25 mm in diameter.If the plate is subjected to 50 KN tension, find the bearing stress in bolts. A. 33333.33 KPa B.

4444.44 KPa C.

5555.55 KPa D.

555555 Kpa

Given: T=30mm D=85mm F=50KN Solution: S=

� 2��

S=

50�� 2(0.03�)(0.025�)

S=33,333.33Kpa

19. What force is necessary to punch a 30 mm hole in 12.5 mm thick plate if ultimate shear stress is 410 Mpa?

A.

480 KN B.

481 KN C.

482 KN D.

483 KN

Given: D=30mm T=125mm Ss=410Mpa Solution: � Ss= ���

�

410000Kpa=�(0.03�)(0.0125�)

F=483KN

20. A 2.5 inches shaft is subjected to 3 KN.m torque. stress developed.

Find the

A. 48.62 MPaB.

52.75 MPa

C.

59.67 MPa

D.

38.64 Mpa

Given: D=2.5in T=3Kn-m Solution: 16�

S=��^3 S=

16(3��−�) 1�

�(2.5�� 39.37��) �

S=59.67Mpa

21. A shaft when subjected to pure torsion developed a stress of50Mpa.If polar moment of inertia is 6.1359 x 10-7 m4, determine the maximum torque the shaft could handle. A. 1. 23KN.mB. 1.68KN.m C. 1.84KN.m D. 2.48KN.m Given: S= 50MPA J= 6.1359x10-7 m4

D = 0.04999 m

22. A lever, secured to a 50mm round shaft by a steel tapered pin(d=10mm), has pull of 200 N at a radius of 800mm. Find S, the working stress on the pin, in Mpa. Consider double shear on the pin. A.

41

B.

43

C.

46

D.

48

Given:

Solution:

D=10mm

S=

Di=50mm F=200N R=800mm

S= 40.69Mpa S= 41Mpa

1.27(200�)(800��) (50��)(10��)^2

23. In a 2 m cantilevered beam, 2 M ton weigh is applied at free end. If the allowable stress in beam is 110Mpa,

determine the section modulus. A. 18.54in B. 21.77in

C. 26.83in

Given: L=m

D. 24.28in solution: z=1/c 39.24��.�

F=2Mton = 110000��.�^2 Sa=110Mpa =3.567x10-4(39.37in)^3 M=Fyl Z=21.77in^3 = (2x1000x0.000981)(2) M= 3924Kn.m Z= section modulus

24. A 6 mm steel wire is 5 m long and stretches 8 mm under a given load. If modulus of elasticity is 200 Gpa, find the load applied. A. 7KN B. 8KN C. 9KN D. 10KN

Given:

solution: �(5)

D=0.006m

0.008=(� 0.006 2)(200�106) 4

L=5m F=9Kn Y=8mm E=200x10^6Kpa ��

Y=��

25. A steel wire 10m long, hanging verticallysupportsatensileloadof2KN.Neglecting the weight of wire, determine the required diameter if the stress is not to exceed 140 Mpa and the total elongation is not to exceed5mm. Assume E = 200Gpa. A. 2mmB. B. 3mm C. 4mm D. 5mm

Given:

solution:

L=10m

0.005m= (� � 2)(200�106)

F=2KN

d=5mm

2��(10�)

4

S=140Mpa Y=5mm E=200GPa �� Y= ��

26. An iron rod 4 m long and 0.5 cm2 in cross section stretches 1 mm when a mass of 225 kg is hang on it. Compute the modulus of elasticity of the iron. A. 176.58GPa B. 169.81Gpa C. 160.41GPa D. 180.26 Gpa

Given:

Solution:

L=4m

0.001=

225���0.00981�� )(4�) �� 1�2 2� (0.5�� 100��2 )(�)

(

E=17658GPa

A=0.5cm^2 Y=1mm F=225kg �� Y= ��

27. A 20 m rod is stretches to a strainof0.001. Determine the deflection of the rod. 20mm B. 25mm C. 30mm D. 35mm

Given:

solution:

L=20m S=0.001 � S=

0.001=20� Y=0.02m=20mm

�

�

28. A rail having a coefficient of linear expansion of11.6x10-6m/m C increases its length when heated from 70F to 133F. Determine the strain.

A.

2.04 x10-4B. 2.77 x 10-4

6.05x10-4

C.

4.06x10-4

D.

Given: K=11.6x10-6m/m-*c T2=183 Fahrenheit T1= 70 Fahrenheit

Solution: Y=KL(t2-t1) =(11.6x10^-6)(56.111-21.11)celcius Y=4.06x10-4m/m

29. What temperature will the rails just touched if steel railroad is 10 m long are laid with clearance of 3 mm at initial temperatureof15C.Use k = 11.7 x 10-6m/mC.

A.

35.64CB.

40.56C C.45.64C D. 50.64C Given: L=10m Y=0.003m T1=15 celcius K=11.7 x 10-6m/mC.

solution: Y=KL(t2-t1) 0.003m=(11.7 x 10-6m/m C.)(10m)(t2-15) t2=45.64C

30. A hollow shaft has an inner diameter of 0.035 m and outer diameter of 0.086 m. Determine the polar moment of inertia of the hollow shaft.

A.

1.512 x10-6m4 B.

1.215 x10-6m4 C.

1.52 x 10-6m4D. 1.125 x 10-6m4

Given: Di=0.035m Do=0.086m Solution: �(��4−��4)

J= 32 J=5.229x10^-6m^4

=

� 10.086 4−(0.035)^4 32

31. It is a problem of expansion and shrink age of steel material so that the slightly smaller hole of a steel

bushing of 1.999 diameter with the following process/materials/data to apply: Coefficient of expansion of carbon steel =0.0000068in/in-F. Temperature raised by gas heating= 24.5oF Cooling media to use dry ice with boiling point of-109.3oF(-78.5oC). Shrinkage rate below boiling point is 0.00073in/in. Determine the final clearance between the expanded steel bushing hole against the shrinkage of the steel shaft. A. 0.000793B. 0.000693in C. 0.000750in D. 0.000800 in

Solution: Y=KL(t2-t1) =0.0000068in/in-F (1.999)( 24.5oF) Y=0.000333in D=1.999 + 0.000333 = 1.9993 For diameter of shaft: Y=0.00073in/in x 2= 0.00146in D=1.99854 in = 1.9993-1.99854 ¥ = 0.000793in

32. What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 m/m from a load producing a unit tensile stress of 44,000 psi?

A.

42,300 x106psi B.

41,202 x106psi C. 43,101 x106psi

D. 41,905 x 106 psi

Given: µ=0.0015m/m St=44000psi Solution: 44000��� E=0.0015�/� = 41.905x10^6 psi 33. If the weight of 6” diameter by48” longSAE1030 shafting is 174.5kg, then what will be the weight of chromium SAE 51416 of same size? A. 305.5 lbs B. 426.4lbs C. 384.6lbsD. 465.1lbs

Given:

solution:

D=6in

F=174.5kg x

L=48in F=174.5kg

F=384.77lbs

2.2�� 1��

34. Compute the maximum unit shear in a 3 inches diameter steel shafting that transmits 2400 in-lb of torque at 99rpm. A. 4530psi B. 4250 psi C. 3860psi D. 4930 psi

Given: D=3in T=2400in-lb Solution: 16�

S=��3

16(2400��−��)

S=

� 3�� 3

S= 452.71psi 35.

If the ultimate shear stress of a 1/8-inch-thick drawn steel plates is 35 ksi what force is required to punch a 1 1½ inch diameter hole?

A. 10,011lbs 20620lbs

Given: T=1/8in S=35000psi D= 1 ½ in

B.

22,322lbs

C.

11MtonsD.

Solution: �

S=��� 35000=

� 1

�(1.5��)(��) 8

F=20616.7 lbs 36. The shaft who set or que varies from 2000 to 6000in-lbs has 1½ inch in diameter and 60,000 psi yield strength.Compute for the shaft mean average stress.

A.

6036psi B.

6810 psi C.

5162psiD.

5550psi

Given: Tmin=2000in-lb Tmax=6000in-lb D=1.5in Sy=60 000 psi Solution: Save =

16(����+����) 16 2000+6000 ��−�� = 2� 1.5�� 3 2��3

Save= 6036 psi 37. link has a load factor of 0.80 the surface factor of 0.80. The surface factor is 0.92 and the endurance strength is 28000psi. Compute the alternating stress of the link if it is subjected to a reversing load. Assume a factor of safety of 3. A. 8150 B. 10920 C. 9,333 D. 7260

given: S=28000psi Fs=3

solution: 28000��� = 9333psi S= 3

38. The shaft is subjected to a steady load of 36,000in-lb at a shear stress of 10,000psi. Compute the diameter of the said shaft in inches.

A.

1 7/8

B.

2¼

C.

3

D.

2¾

Given: T=3600in-lbs Ss=10000psi Solution: 16� S= 3 ��

10000psi =

16(36000��−��) ��3

D=2.6in or 3 in 39. A shear pin is to be shear at 15 hp and 1000 rpm. The pin attaches a hub to a shaft 1.5 inches in diameter and ultimate shearing stress of50,000 psi. Find the diameter of the pin. A. 1/8in B. ¼ in C. ¾in D. 1in

Given: P=15hp N=1000rpm D=1.5in Su=50000psi Solution: P=

2���

33000 2�(

15hp=

40. A steel 0.5 inch x 1 inch steel 200 ft long is subjected to a 5000 lbs tensile load. Find the deformation along its width if poison’s ratio is 0.25.

A. 0.000007in

B.

0.0000417in

C. 0.000015in D. 0.000067in

Gvien: L=200ft

Solution: 5000��� Ey= 0.571�� 30�106

A=0.571in E=0.25

Ey=3.3333x10^4 M=Ex/Ey �� 0.25=3.3333�104 Ex=8.33325x10^-5

Ey=αb/A 8.333325x10^-5=αb/0.5 �� = 4.17x10^-5 in

41. A steel rod is stretched between two rigid wall sand carries a tensile load of 5000N at 20oC. If the allowable stress is not to exceed 130 MN/m2 at -20oC, what is the minimum diameter of the rod? Assume k = 11.7 µm/(m-oC); E = 200Gpa A.

10.22mm

B.

11.22mm

Given: F=5000N T1=20oC S=130 MN/m2 T2= -20oC

C.

12.22mm

D.

13.22mm

solution: S=µLAT = (11.7x10^-6)(200GPA)(-20-20) oC S=-90600Mpa S=-90600 – 130 000 – 36 000 mpa S=

� �

36400 = D=13.22 mm

5�� �/4�2

SHAFTING PROBLEM SET 2 INSTRUCTION: Encircle the letter that corresponds to the correct answer of your choice. MULTIPLE CHOICE:

1. A shaft is used to transmit 200 KW at 300 rpm by means of a 200 mm diameter sprocket. Determine the force tangent to the sprocket. A. 60.44 KN B. P=200 Kw N=300rpm D=200mm Fsprocket =? Solution:

60.33 KN C.

60.88 KN D.

63.66 KN

P=2πTN 200KW= 2��(

300��� 60���

)

T= 6.366KN-m T=Fr 6.366KN-m= F (

0.2�

)

2

FSprocket= 63.66 KN

2. Find the diameter of a steel shaft which will be used to operate a 110 KW motor rotating at 5 rps if torsional stress is 90 Mpa.

A.

60.20 mm B.

58.30 mm

C.

38.30 mm

D.

46.20 mm

Given: P=110 Kw N=5rps Ss=90 MPa

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