# Problems on Simple Stress(Lec 4b)

#### Short Description

Machine Design 1...

#### Description

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

113097.34600 

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 2d t  225mm16mm 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  25mm12.5mm

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 2512.5 mm 33333.33 N S2   106.67 or MPa 2 2512.5 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 25mm12.5mm 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 ;   2N P  2NT 2 160 3 N m 100 x10  T s 60 T  5968.31 N  m  mean torque For Tmax  1.25Tmean  1.255968.31  7460.39 N  m : Tr 16Tmax Ss   J d 3

167460.39 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  2NT 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 165172.54 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  2NTmean

N  m 2 110 600 x10  Tmean s 60 Tmean  52087.07 N  m 3

Tmax  1.2052087.07   62504.48 N  m

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

1662504.48 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 

3172.7   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    

3175.5 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 447123.89  48000  d 2 3 d  1.94 in

F d2

Su N

4 447123.89 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 1000F : 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. 163151.25 0.6133000   3 16T Su s d 3  3 d  0.85in d N 163151.25 0.75152000  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  20001.5  30002.5 R1  1285.7lbs

 Fv  0 1285.7  R2  3000  2000 R2  3714.3lbs

Ma  0

M b  0  1285.71  1285.7

M c  1285.7   1714.32.5  3000 M d  3000  20001.5  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 b2b  12 b  1.86in h  21.86   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 36 above 6

Convert the following to their preferred sizes : 1. D  0.213" x D  ; x  0.21316   3.408  4 16 D4  1 " 4. h  4.62" 16 4 2. b  0.172" x h  4 ; x  0.624   2.48  3 x 4 b  ; x  0.17232   5.504  6 32 h 43 " 4 6 3 D  " 5. b  6.58" 32 16 3. d  1.123" x b  6 ; x  0.582   1.16  2 x 2 d  1 ; x  0.1238  0.984  1 8 b  6 2  7" 2 1 d 1 " 8