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PROBLEMS ON STRESS ME 561P/ME 147P

MACHINE DESIGN I

SIMPLE STRESSES 1. A cast iron link, as shown in the figure below, is required to transmit a steady tensile load of 45 kN. Find the tensile stress induced in the link material at sections A-A and B-B. [Ans:50 MPa, 64.3 Mpa]

2. The piston rod of a steam engine is 50 mm in diameter and 600 mm long. The diameter of the piston is 400 mm and the maximum steam pressure is 0.9 N/mm2. Find the induced stress and the compression of the piston rod if the Young's modulus for the material of the piston rod is 210 kN/mm2. 2 F pApiston 0.9 N / mm2 400mm 4

F 113097.34 N F S cro d Arod

113097.34 N

2 50mm 4

57.6 N

mm2 FL From : Arod E

orMPa

113097.34600

2 50 210 1000 4

0.165mm

3. A pull of 80 kN is transmitted from a bar X to the bar Y through a pin as shown. If the maximum permissible tensile stress in the bars is 100 N/mm2 and the permissible shear stress in the pin is 80 N/mm2, find the diameter of bars and of the pin.

4. Two plates 16 mm thick are joined by a double riveted lap joint as shown. The rivets are 25 mm in diameter. Find the crushing stress induced between the plates and the rivet, if the maximum tensile load on the joint is 48 kN.

P 48000 N N Sb 60 or MPa 2 2d t 225mm16mm mm

5. A mild steel rod of 12 mm diameter was tested for tensile strength with the gauge length of 60 mm. Following observations were recorded : Final length = 80 mm; Final diameter = 7 mm; Yield load = 3.4 kN and Ultimate load = 6.1 kN. Calculate : 1. yield stress, 2. ultimate tensile stress, 3. percentage reduction in area, and 4. percentage elongation.

Ans: 1. 30.1 MPa; 2. 54 MPa; 3. 66%; 4. 25%

STRESS IN COMPOSITE BARS A composite bar may be defined as a bar made up of two or more different materials, joined together, in such a manner that the system extends or contracts as one unit, equally, when subjected to tension or compression. In case of composite bars, the following points should be kept in view: 1. The extension or contraction of the bar being equal, the strain i.e. deformation per unit length is also equal. 2. The total external load on the bar is equal to the sum of the loads carried by different materials.

6. A bar 3 m long is made of two bars, one of copper having E = 105 GN/m2 and the other of steel having E = 210 GN/m2. Each bar is 25 mm broad and 12.5 mm thick. This compound bar is stretched by a load of 50 kN. Find the stress produced in the steel and copper and the increase in length of the compound bar. The length of copper as well as of steel bar is 3 m each.

δcopper δsteel δ1 δ2 F1 L FL 2 A1 E1 A2 E2 F1 F2

A1 E1 A2 E2

F F1 F2 F2

A1 E1 F2 A2 E2

A1 E1 A1 E1 A2 E2 F F2 1 F2 A2 E2 A2 E2

A2 E2 F2 F A1 E1 A2 E2 Similarly :

We know that : F1 L FL 2 A1 E1 A2 E2

A1 E1 F1 F A1 E1 A2 E2

S1 S 2 E1 E2

A1 E1 F1 F A1 E1 A2 E2 but, A1 A2 25mm12.5mm

E1 E2 S1 S 2 and S 2 S1 E2 E1 F F1 F2 S1 A1 S 2 A2

105 Load shared by the copper bar F1 50000 N 16666.67 N 105 210 F2 F F1 50000 16666.67 33333.33 N Load shared by the steel bar 16666.67 N 53.33 or MPa 2 2512.5 mm 33333.33 N S2 106.67 or MPa 2 2512.5 mm S1

stress produced in copper bar

stress produced in steel bar

Elongation for both bars are equal.

1 2

F1 L F2 L AE1 AE 2

16666.67 N 3000mm 1.52mm 2 25mm12.5mm 105000 N / mm

7. A central steel rod 18 mm diameter passes through a copper tube 24 mm inside and 40 mm outside diameter, as shown in the figure. It is provided with nuts and washers at each end. The nuts are tightened until a stress of 10 MPa is set up in the steel. The whole assembly is then placed in a lathe and turned along half the length of the tube removing the copper to a depth of 1.5 mm. Calculate the stress now existing in the steel. Take Es = 2Ec.

8. A shaft is transmitting 100 kW at 160 r.p.m. Find a suitable diameter for the shaft, if the maximum torque transmitted exceeds the mean by 25%. Take maximum allowable shear stress as 70 MPa. P T ; 2N P 2NT 2 160 3 N m 100 x10 T s 60 T 5968.31 N m mean torque For Tmax 1.25Tmean 1.255968.31 7460.39 N m : Tr 16Tmax Ss J d 3

167460.39 N m 1000mm N 1m 70 mm2 d 3 d 81.57 mm

9. A shaft is transmitting 97.5 kW at 180 rpm. If the allowable shear stress in the material is 60 MPa, find the suitable diameter for the shaft. The shaft is not to twist more that 1° in a length of 3 metres. Take G = 80 GPa.

P 2NT N m 2 180 97.5 x103 T s 60 T 5172.54 N m

Compute for diameter of the shaft based on strength and stiffness. Considerin g strength of the shaft : Tr 16T Ss 3 J d 165172.54 N m 1000mm N 1m 60 mm2 d 3 d 76 mm

Considerin g the stiffness or rigidity of the shaft : TL JG 2 1000 mm 5172.54 N m 3m m 1 d 4 80000 N 180 32 mm2 d 103.15 mm

Use the larger value for d. Therefore, d=103.15 mm

10. A hollow shaft is required to transmit 600 kW at 110 rpm, the maximum torque being 20% greater than the mean. The shear stress is not to exceed 63 MPa and twist in a length of 3 metres not to exceed 1.4 degrees. Find the external diameter of the shaft, if the internal diameter to the external diameter is 3/8. Take modulus of rigidity as 84 GPa.

P 2NTmean

N m 2 110 600 x10 Tmean s 60 Tmean 52087.07 N m 3

Tmax 1.2052087.07 62504.48 N m

Considerin g strength of the shaft : Tr 16Tmax d 0 Ss J d 04 d i 4

1662504.48 N m 1000mm d 0 N 1m 63 mm2 d 04 d i 4

di 3 3d 0 But di d0 8 8

1000mm d 16 62504 . 48 N m 0 N 1 m 63 2 mm 4 3d 0 4 d 0 8 d 0 172.7 mm di

3172.7 64.8mm 8

Considerin g the stiffness or rigidity of the shaft : TL JG 2 1000 mm 62504.48 N m 3m m 1.4 180 d 04 d i4 84000 N 32 mm2 d 3d 3 But i d i 0 d0 8 8

62504.48 N m 3m 1000mm

1.4 180

d 0 175.5mm

2

m

4 3d 0 4 d0 84000 N 2 32 mm 8

3175.5 di 65.8mm 8

Use the larger value for d0. Therefore, d0 =175.5 mm and di=65.8 mm

11. The piston rod of a gas engine is to carry a load based on a maximum pressure of 150 psi. The load is repeated but uni-directional application. The cylinder of the engine is 20 inches in diameter. If the material of the piston rod is AISI C1020 as rolled, recommend its diameter based on the following: a) Yield Strength; and b) Ultimate Strength.

Properties of AISI C1020, as rolled : from AT 7, p.576 S y 48ksi 48000 psi Su 65ksi 65000 psi Factor of Safety (from p.20, Table 1.1)

N

Sy

rep ,1dir

N

Su

rep ,1dir

3 6

Analysis : Sind S0 S d Strength of M aterials S 0 nominal stress S d design stress Su F Tr Mc S y or or or A J I N N From : p

F Aapplication

; Aapplication

4

2 Dcyl

lb 2 20 in 47123.89lbs 2 in 4 Sd b. Sind S d

F 150 a. Sind

F d2

Sy N

4 447123.89 48000 d 2 3 d 1.94 in

F d2

Su N

4 447123.89 65000 2 d 6 d 2.35 in

12. A transmission shaft is to transmit 25 hp at 500 rpm. It is made of AISI 3140 OQT 1000°F. The load is repeated but not reversed. Determine the diameter of shaft based on yield and ultimate strengths.

Properties of AISI 3140 OQT 1000F : from Fig. AF2, p.573 S y 133000 psi Su 152000 psi Factor of Safety (from p.20, Table 1.1)

N

Sy

rep ,1dir

N

Su

rep ,1dir

3 6

a. Sind S d ; load is torsional. Tr S d ; since torsional stress is a shear stress J Generally, from AT 7, Td S 2 ys S ys 0.6S y d4 N 32 Sus 0.75Su 16T S ys 3 d N P 25hp 63025 3151.25in lbs But T N 500rpm b. Sind S d ; load is torsional. 163151.25 0.6133000 3 16T Su s d 3 3 d 0.85in d N 163151.25 0.75152000 3 d 6 d 0.95in

13. A rectangular beam is loaded as shown. If the beam is made of ASTM 50 cast iron and load is steady, determine the dimensions h and b, where h=2b.

Mc 0

R1 3.5 20001.5 30002.5 R1 1285.7lbs

Fv 0 1285.7 R2 3000 2000 R2 3714.3lbs

Ma 0

M b 0 1285.71 1285.7

M c 1285.7 1714.32.5 3000 M d 3000 20001.5 0

M max 3000 ft lbs 36000in lbs

Properties of ASTM 50(cast iron) - brittle material From AT 6, p.570 S0 Sd Su 50ksi 50000 psi Mc S u p .20 S : N S y no value steadyload 5 to 6; say 6 f I N

h 36000in lbs 2 50000 ; but h 2b bh 3 6 12 2b 36000in lbs 2 50000 3 6 b2b 12 b 1.86in h 21.86 3.72in

Preferred Sizes • For many machine elements, there are standardized sizes such as bolts, keys, I-beams, which means that such sizes are more readily available in the market and are also cheaper. • The designer always uses standard items and standard proportions unless he feels strongly that some custom design is desirable. • If there are no such standard sizes, refer below for the preferred dimensions:

Increment 1 64 1 32 1 16 1 8 1 4 1 2

Range of S izes 1 1 0.0156 - 0.03125 64 32 1 3 0.03125 - 0.1875 32 16 3 7 0.1875 - 0.875 16 8 7 3 0.875 - 3 8 36 above 6

Convert the following to their preferred sizes : 1. D 0.213" x D ; x 0.21316 3.408 4 16 D4 1 " 4. h 4.62" 16 4 2. b 0.172" x h 4 ; x 0.624 2.48 3 x 4 b ; x 0.17232 5.504 6 32 h 43 " 4 6 3 D " 5. b 6.58" 32 16 3. d 1.123" x b 6 ; x 0.582 1.16 2 x 2 d 1 ; x 0.1238 0.984 1 8 b 6 2 7" 2 1 d 1 " 8

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MACHINE DESIGN I

SIMPLE STRESSES 1. A cast iron link, as shown in the figure below, is required to transmit a steady tensile load of 45 kN. Find the tensile stress induced in the link material at sections A-A and B-B. [Ans:50 MPa, 64.3 Mpa]

2. The piston rod of a steam engine is 50 mm in diameter and 600 mm long. The diameter of the piston is 400 mm and the maximum steam pressure is 0.9 N/mm2. Find the induced stress and the compression of the piston rod if the Young's modulus for the material of the piston rod is 210 kN/mm2. 2 F pApiston 0.9 N / mm2 400mm 4

F 113097.34 N F S cro d Arod

113097.34 N

2 50mm 4

57.6 N

mm2 FL From : Arod E

orMPa

113097.34600

2 50 210 1000 4

0.165mm

3. A pull of 80 kN is transmitted from a bar X to the bar Y through a pin as shown. If the maximum permissible tensile stress in the bars is 100 N/mm2 and the permissible shear stress in the pin is 80 N/mm2, find the diameter of bars and of the pin.

4. Two plates 16 mm thick are joined by a double riveted lap joint as shown. The rivets are 25 mm in diameter. Find the crushing stress induced between the plates and the rivet, if the maximum tensile load on the joint is 48 kN.

P 48000 N N Sb 60 or MPa 2 2d t 225mm16mm mm

5. A mild steel rod of 12 mm diameter was tested for tensile strength with the gauge length of 60 mm. Following observations were recorded : Final length = 80 mm; Final diameter = 7 mm; Yield load = 3.4 kN and Ultimate load = 6.1 kN. Calculate : 1. yield stress, 2. ultimate tensile stress, 3. percentage reduction in area, and 4. percentage elongation.

Ans: 1. 30.1 MPa; 2. 54 MPa; 3. 66%; 4. 25%

STRESS IN COMPOSITE BARS A composite bar may be defined as a bar made up of two or more different materials, joined together, in such a manner that the system extends or contracts as one unit, equally, when subjected to tension or compression. In case of composite bars, the following points should be kept in view: 1. The extension or contraction of the bar being equal, the strain i.e. deformation per unit length is also equal. 2. The total external load on the bar is equal to the sum of the loads carried by different materials.

6. A bar 3 m long is made of two bars, one of copper having E = 105 GN/m2 and the other of steel having E = 210 GN/m2. Each bar is 25 mm broad and 12.5 mm thick. This compound bar is stretched by a load of 50 kN. Find the stress produced in the steel and copper and the increase in length of the compound bar. The length of copper as well as of steel bar is 3 m each.

δcopper δsteel δ1 δ2 F1 L FL 2 A1 E1 A2 E2 F1 F2

A1 E1 A2 E2

F F1 F2 F2

A1 E1 F2 A2 E2

A1 E1 A1 E1 A2 E2 F F2 1 F2 A2 E2 A2 E2

A2 E2 F2 F A1 E1 A2 E2 Similarly :

We know that : F1 L FL 2 A1 E1 A2 E2

A1 E1 F1 F A1 E1 A2 E2

S1 S 2 E1 E2

A1 E1 F1 F A1 E1 A2 E2 but, A1 A2 25mm12.5mm

E1 E2 S1 S 2 and S 2 S1 E2 E1 F F1 F2 S1 A1 S 2 A2

105 Load shared by the copper bar F1 50000 N 16666.67 N 105 210 F2 F F1 50000 16666.67 33333.33 N Load shared by the steel bar 16666.67 N 53.33 or MPa 2 2512.5 mm 33333.33 N S2 106.67 or MPa 2 2512.5 mm S1

stress produced in copper bar

stress produced in steel bar

Elongation for both bars are equal.

1 2

F1 L F2 L AE1 AE 2

16666.67 N 3000mm 1.52mm 2 25mm12.5mm 105000 N / mm

7. A central steel rod 18 mm diameter passes through a copper tube 24 mm inside and 40 mm outside diameter, as shown in the figure. It is provided with nuts and washers at each end. The nuts are tightened until a stress of 10 MPa is set up in the steel. The whole assembly is then placed in a lathe and turned along half the length of the tube removing the copper to a depth of 1.5 mm. Calculate the stress now existing in the steel. Take Es = 2Ec.

8. A shaft is transmitting 100 kW at 160 r.p.m. Find a suitable diameter for the shaft, if the maximum torque transmitted exceeds the mean by 25%. Take maximum allowable shear stress as 70 MPa. P T ; 2N P 2NT 2 160 3 N m 100 x10 T s 60 T 5968.31 N m mean torque For Tmax 1.25Tmean 1.255968.31 7460.39 N m : Tr 16Tmax Ss J d 3

167460.39 N m 1000mm N 1m 70 mm2 d 3 d 81.57 mm

9. A shaft is transmitting 97.5 kW at 180 rpm. If the allowable shear stress in the material is 60 MPa, find the suitable diameter for the shaft. The shaft is not to twist more that 1° in a length of 3 metres. Take G = 80 GPa.

P 2NT N m 2 180 97.5 x103 T s 60 T 5172.54 N m

Compute for diameter of the shaft based on strength and stiffness. Considerin g strength of the shaft : Tr 16T Ss 3 J d 165172.54 N m 1000mm N 1m 60 mm2 d 3 d 76 mm

Considerin g the stiffness or rigidity of the shaft : TL JG 2 1000 mm 5172.54 N m 3m m 1 d 4 80000 N 180 32 mm2 d 103.15 mm

Use the larger value for d. Therefore, d=103.15 mm

10. A hollow shaft is required to transmit 600 kW at 110 rpm, the maximum torque being 20% greater than the mean. The shear stress is not to exceed 63 MPa and twist in a length of 3 metres not to exceed 1.4 degrees. Find the external diameter of the shaft, if the internal diameter to the external diameter is 3/8. Take modulus of rigidity as 84 GPa.

P 2NTmean

N m 2 110 600 x10 Tmean s 60 Tmean 52087.07 N m 3

Tmax 1.2052087.07 62504.48 N m

Considerin g strength of the shaft : Tr 16Tmax d 0 Ss J d 04 d i 4

1662504.48 N m 1000mm d 0 N 1m 63 mm2 d 04 d i 4

di 3 3d 0 But di d0 8 8

1000mm d 16 62504 . 48 N m 0 N 1 m 63 2 mm 4 3d 0 4 d 0 8 d 0 172.7 mm di

3172.7 64.8mm 8

Considerin g the stiffness or rigidity of the shaft : TL JG 2 1000 mm 62504.48 N m 3m m 1.4 180 d 04 d i4 84000 N 32 mm2 d 3d 3 But i d i 0 d0 8 8

62504.48 N m 3m 1000mm

1.4 180

d 0 175.5mm

2

m

4 3d 0 4 d0 84000 N 2 32 mm 8

3175.5 di 65.8mm 8

Use the larger value for d0. Therefore, d0 =175.5 mm and di=65.8 mm

11. The piston rod of a gas engine is to carry a load based on a maximum pressure of 150 psi. The load is repeated but uni-directional application. The cylinder of the engine is 20 inches in diameter. If the material of the piston rod is AISI C1020 as rolled, recommend its diameter based on the following: a) Yield Strength; and b) Ultimate Strength.

Properties of AISI C1020, as rolled : from AT 7, p.576 S y 48ksi 48000 psi Su 65ksi 65000 psi Factor of Safety (from p.20, Table 1.1)

N

Sy

rep ,1dir

N

Su

rep ,1dir

3 6

Analysis : Sind S0 S d Strength of M aterials S 0 nominal stress S d design stress Su F Tr Mc S y or or or A J I N N From : p

F Aapplication

; Aapplication

4

2 Dcyl

lb 2 20 in 47123.89lbs 2 in 4 Sd b. Sind S d

F 150 a. Sind

F d2

Sy N

4 447123.89 48000 d 2 3 d 1.94 in

F d2

Su N

4 447123.89 65000 2 d 6 d 2.35 in

12. A transmission shaft is to transmit 25 hp at 500 rpm. It is made of AISI 3140 OQT 1000°F. The load is repeated but not reversed. Determine the diameter of shaft based on yield and ultimate strengths.

Properties of AISI 3140 OQT 1000F : from Fig. AF2, p.573 S y 133000 psi Su 152000 psi Factor of Safety (from p.20, Table 1.1)

N

Sy

rep ,1dir

N

Su

rep ,1dir

3 6

a. Sind S d ; load is torsional. Tr S d ; since torsional stress is a shear stress J Generally, from AT 7, Td S 2 ys S ys 0.6S y d4 N 32 Sus 0.75Su 16T S ys 3 d N P 25hp 63025 3151.25in lbs But T N 500rpm b. Sind S d ; load is torsional. 163151.25 0.6133000 3 16T Su s d 3 3 d 0.85in d N 163151.25 0.75152000 3 d 6 d 0.95in

13. A rectangular beam is loaded as shown. If the beam is made of ASTM 50 cast iron and load is steady, determine the dimensions h and b, where h=2b.

Mc 0

R1 3.5 20001.5 30002.5 R1 1285.7lbs

Fv 0 1285.7 R2 3000 2000 R2 3714.3lbs

Ma 0

M b 0 1285.71 1285.7

M c 1285.7 1714.32.5 3000 M d 3000 20001.5 0

M max 3000 ft lbs 36000in lbs

Properties of ASTM 50(cast iron) - brittle material From AT 6, p.570 S0 Sd Su 50ksi 50000 psi Mc S u p .20 S : N S y no value steadyload 5 to 6; say 6 f I N

h 36000in lbs 2 50000 ; but h 2b bh 3 6 12 2b 36000in lbs 2 50000 3 6 b2b 12 b 1.86in h 21.86 3.72in

Preferred Sizes • For many machine elements, there are standardized sizes such as bolts, keys, I-beams, which means that such sizes are more readily available in the market and are also cheaper. • The designer always uses standard items and standard proportions unless he feels strongly that some custom design is desirable. • If there are no such standard sizes, refer below for the preferred dimensions:

Increment 1 64 1 32 1 16 1 8 1 4 1 2

Range of S izes 1 1 0.0156 - 0.03125 64 32 1 3 0.03125 - 0.1875 32 16 3 7 0.1875 - 0.875 16 8 7 3 0.875 - 3 8 36 above 6

Convert the following to their preferred sizes : 1. D 0.213" x D ; x 0.21316 3.408 4 16 D4 1 " 4. h 4.62" 16 4 2. b 0.172" x h 4 ; x 0.624 2.48 3 x 4 b ; x 0.17232 5.504 6 32 h 43 " 4 6 3 D " 5. b 6.58" 32 16 3. d 1.123" x b 6 ; x 0.582 1.16 2 x 2 d 1 ; x 0.1238 0.984 1 8 b 6 2 7" 2 1 d 1 " 8

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