1. UNIT-01. SIMPLE STRESSES & STRAINS Lecture Number - 06 Prof. S.C.SADDU MECHANICAL DEPARTMENT SIT LONAVALA Strength of Materials

2. Agenda To study temperature stresses and temperature strain Strength of Materials

3. Thermal Stresses Strength of Materials A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports. Treat the additional support as redundant and apply the principle of superposition. The thermal deformation and the deformation from the redundant support must be compatible.

4. Temperature stresses Extension in bar due to raise in temperature is,  L=  *  T *L Strength of Materials Temperature Strain is,  =  *  T Temperature Stress developed is,  =  *E*  T

5. Example: 1 A steel tube having an external diameter of 36 mm and an internal diameter of 30 mm has a brass rod of 20 mm diameter inside it, the two materials being joined rigidly at their ends when the ambient temperature is 18 0 C. Determine the stresses in the two materials: (a) when the temperature is raised to 68 0 C (b) when a compressive load of 20 kN is applied at the increased temperature. For brass: Modulus of elasticity = 80 GN/m2; Coefficient of expansion = 17 x 10 -6 /0C For steel: Modulus of elasticity = 210 GN/m2; Coefficient of expansion = 11 x 10 -6 /0C Strength of Materials

8. Example: 2 A railway is laid so that there is no stress in rail at 10º C. If rails are 30 m long Calculate, 1.The stress in rails at 60 º C if there is no allowance for expansion. 2. The stress in the rails at 60 º C if there is an expansion allowance of 10 mm per rail. 3. The expansion allowance if the stress in the rail is to be zero when temperature is 60 º C. 4. The maximum temp. to have no stress in the rails if the expansion allowance is 13 mm/rail.Take  = 12 x 10 -6 per 1ºC E= 2 x 10 5 N/mm 2 Strength of Materials

9. Solution: 1. Rise in temp. = 60 º - 10 º = 50 ºC so stress =  t E =12 x 10 -6 x50x 2 x 10 5 = 120 MPa 2.  tp x L/E =  = (L  t -10) = (30000 x 12 x 10 -6 x50-10) = 18 -10 = 8 mm  tp =  E /L = 8x 2 x 10 5 /30000 = 53.3 MPa Strength of Materials

10. 3. If stresses are zero, Expansion allowed =(L  t ) = (30000 x 12 x 10 -6 x50) =18 mm 4. If stresses are zero  tp =E /L*(L  t -13)=0 L  t=13 so t=13/ (30000 x 12 x 10 -6 )=36 0 C allowable temp.=10+36=46 0 c. Strength of Materials

11. Example: 3 A composite bar made up of aluminum and steel is held between two supports. The bars are stress free at 40 0 c. What will be the stresses in the bars when the temp. drops to 20 0 C, if (a) the supports are unyielding (b)the supports come nearer to each other by 0.1 mm. Take E al =0.7*10 5 N/mm 2 ;  al =23.4*10 -6 / 0 C E S =2.1*10 5 N/mm 2  s =11.7*10 -6 / 0 C A al =3 cm 2 A s =2 cm 2 Strength of Materials

12. Free contraction  =L s  s t+ L AL  Al t Since contraction is checked tensile stresses will be set up. Force being same in both A s  s = A al  al contraction of steel bar  s  = (  s /E s )*L s contra.of aluminum bar  al  = (  al /E al )*L al (a) When supports are unyielding  s  +  al  =  (free contraction) (b) Supports are yielding  s  +  al  = (  - 0.1mm) Strength of Materials

13. Example: 4 A steel bolt of length L passes through a copper tube of the same length, and the nut at the end is turned up just snug at room temp. Subsequently the nut is turned by 1/4 turn and the entire assembly is raised by temp 55 0 C. Calculate the stress in bolt if L=500mm,pitch of nut is 2mm, area of copper tube =500sq.mm,area of steel bolt=400sq.mm Es=2 * 10 5 N/mm 2 ;  s =12*10 -6 / 0 C Ec=1 * 10 5 N/mm 2 ;  c = 17.5*10 -6 / 0 C Strength of Materials

14. Solution: Two effects (i) tightening of nut (ii)raising temp. tensile stress in steel = compressive force in copper [Total extension of bolt +Total compression of tube] =Movement of Nut [  s+  c] = np ( where p = pitch of nut) PL/A s E s +  s L t) +(PL/A c E c -  c L t)=np Strength of Materials