UNIT01 SIMPLE STRESSES STRAINS Lecture Number 06 Prof

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

Agenda • To study temperature stresses and temperature strain Strength of Materials

Thermal Stresses • A temperature change results in a change in length or thermal strain. There is no stress associated with 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. Strength of Materials

Temperature stresses • Extension in bar due to raise in temperature is, L= * T *L • Temperature Strain is, = * T • Temperature Stress developed is, = *E* T Strength of Materials

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 k. N is applied at the increased temperature. • For brass: Modulus of elasticity = 80 GN/m 2; Coefficient of expansion = 17 x 10 -6 /0 C • For steel: Modulus of elasticity = 210 GN/m 2; Coefficient of expansion = 11 x 10 -6 /0 C Strength of Materials

Strength of Materials

Strength of Materials

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 Strength of Materials -6 10 per 1ºC E= 2 x 10 5 N/mm 2

Solution: 1. Rise in temp. = 60 º - 10 º = 50 ºC so stress = t E =12 x 10 -6 x 50 x 2 x 10 5 = 120 MPa 2. tp x L/E = = (L t -10) = (30000 x 12 x 10 -6 x 50 -10) = 18 -10 = 8 mm tp = E /L = 8 x 2 x 10 5 /30000 = 53. 3 MPa Strength of Materials

3. If stresses are zero , Expansion allowed =(L t ) = (30000 x 12 x 10 -6 x 50) =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 )=360 C allowable temp. =10+36=460 c. Strength of Materials

Example: 3 A composite bar made up of aluminum and steel is held between two supports. The bars are stress free at 400 c. What will be the stresses in the bars when the temp. drops to 200 C, if (a) the supports are unyielding (b)the supports come nearer to each other by 0. 1 mm. Take E al =0. 7*105 N/mm 2 ; al =23. 4*10 -6 /0 C ES=2. 1*105 N/mm 2 Aal=3 cm 2 As=2 cm 2 s =11. 7*10 -6 /0 C Strength of Materials

Free contraction =Ls s t+ LAL Alt Since contraction is checked tensile stresses will be set up. Force being same in both As s= Aal al contraction of steel bar s = ( s/Es)*Ls contra. of aluminum bar al = ( al/Eal)*Lal (a) When supports are unyielding s + al = (free contraction) (b) Supports are yielding s + al = ( - 0. 1 mm) Strength of Materials

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 550 C. Calculate the stress in bolt if L=500 mm, pitch of nut is 2 mm, area of copper tube =500 sq. mm, area of steel bolt=400 sq. mm • Es=2 * 105 N/mm 2 ; s =12*10 -6 /0 C • Ec=1 * 105 N/mm 2 ; c= 17. 5*10 -6 /0 C Strength of Materials

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/As. Es + s L t) +(PL/Ac. Ec- c L t)=np Strength of Materials