Simple Stresses and Strains in Material Strength

Explore the concept of simple stresses and strains in Strength of Materials through lectures and numerical examples. Understand statically determinate bars and beams, calculations of material elongation and dimension changes, as well as solving problems related to axial loads and bar displacements.

• Material Strength
• Stresses
• Strains
• Statically Determinate Bars
• Mechanical Engineering

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1. UNIT-01. SIMPLE STRESSES & STRAINS Lecture Number 03 Prof. S.C. SADDU MECHANICAL DEPARTMENT SIT LONAVALA Strength of Materials

2. Agenda Statically Determinate Bars Numerical Strength of Materials

3. STATICALLY DETERMINATE BEAMS Statically determinate beams are those beams reactions of the supports may be determined by the use of the equations of static equilibrium. The beams shown are examples of statically determinate beams. in which the Strength of Materials

4. TYPE A Example:1 An aluminium bar 1.8 meters long has a 25 mm square c/s over 0.6 meters of its length and 25 mm circular c/s over other 1.2 meters . How much will the bar elongate under a tensile load P=17500 N, if E = 75000 Mpa. Solution :- = PL/AE =17500*600 / (252*75000) + 17500*1200/(0.785*252*75000) =0.794 mm Strength of Materials

5. Example: 2 A steel bar having 40mm*40mm*3000mm dimension is subjected to an axial force of 128 kN. Taking E=2*105N/mm2 and = 0.3,find out change in dimensions. 40 128 kN 128 kN 3000 mm 40 Hint: New breadth B1 =B(1- ) New Depth D1 = D(1- ) Strength of Materials

6. Example: 3 A prismatic steel bar having cross sectional area of A=300 mm2 is subjected to axial load as shown in figure . Find the net increase in the length of the bar. Assume E = 2 x 10 5 MPa. ( Ans = -0.17mm) 15 kN 20 kN 15 kN C B A 1 m 1 m 2 m 15 15 A B 0 0 20 20 C = 20000*1000/(300*2x10 5)-15000*2000/(300*2 x10 5) = 0.33 - 0.5 = -0.17 mm (i.e.contraction) Strength of Materials

7. TYPE B Example: 1 A rigid bar AB, 9 m long, is supported by two vertical rods at its end and in a horizontal position under a load P as shown in figure. Find the position of the load P so that the bar AB remains horizontal. 5 m A = 445 mm 2 A = 1000 mm 2 9 m E = 2 x 10 5 3m E = 1 x 10 5 B A x P Strength of Materials

8. 5 m 9 m 3m P(9-x)/9 P(x)/9 A B x P Strength of Materials

9. For the bar to be in horizontal position, Displacements at A & B should be same, A = B (PL/AE)A =(PL/AE)B {P(x)/9}*5 {P(9-x)/9}*3 = 0.000445*2*105 (0.001*1*105) (9 - x)*3=x*5*1.1236 27-3x=5.618 x 8.618 x=27 x = 3.13 m Strength of Materials

10. Example: 2 Determine the diameter of rod 200m long hang vertically and subjected a axial pull of 325kN at its lower end if its end weight per cubic meter is 80kN and working stress is 75Mpa.Also determine the total elongation of rod. Take E=210Gpa L=200mm P=325kN Ans.: Diameter of rod:83.75mm Total elongation:63.81mm Strength of Materials

11. Example: 3 A conical bar of diameter 2m and length 3m is hang to a ceiling. Weight density of material is 75kN/m3 for rod and young s modulus of rod is 100Gpa. Determine the deformation due to its own weight. L=3m Ans.: Deformation of bar is 1.125mm Strength of Materials

12. TYPE C Example: 1 A copper rod of 40 mm diameter is surrounded tightly by a cast iron tube of 80 mm diameter, the ends being firmly fastened together. When it is subjected to a compressive load of 30 kN, what will be the load shared by each? Also determine the amount by which a compound bar shortens if it is 2 meter long. Eci=175 GN/m2,Ec= 75 GN/m2 . Cast iron 40 mm 80 mm copper Cast iron 2 meter Strength of Materials

13. Area of Copper Rod =Ac = (/4)* 0.042 = 0.0004 m2 Area of Cast Iron = Aci = ( /4)* (0.082 - 0.042) = 0.0012 m2 ci /Eci = c /Ec = 175 x 10 9 / 75 x 10 9 = 2.33 ci = 2.33 c Now, W = Wci +Wc 30 = (2.33 c ) x 0.012 + c x 0.0004 c = 2987.5 kN/m2 Strength of Materials

14. ci = 2.33 x c = 6960.8kN/m2 load shared by copper rod = Wc = c Ac = 2987.5 x 0.0004 = 3.75 kN Wci = 30 -3.75 = 26.25 kN Strain c= c / Ec = L /L L = ( c /Ec) x L = [2987.5/(75 x 10 9)] x 2 = 0.0000796 m= 0.0796 mm Decrease in length = 0.0796 mm Strength of Materials

15. Example: 2 A composite rod, 1200mm long consists of a steel tube of 50mm external diameter and 40mm internal diameter. A copper rod of 30mm diameter is placed coaxially into the steel tube .the assembly is held between two rigidly plates and is subjected to an compressive forces of 200kN.Find the stress induced in each material and the contraction produced. Take Es=200Gpa, Ec=100Gpa Copper 200kN 200kN Ans.: c =94.314Mpa S =188.628Mpa L= 1.132mm Hint: c / E c = s / E s L= c L / E c Steel tube

16. Example:3 Two copper road and one steel rod(center) together support load be shown in figure. Cross sectional area of each rod is 900 mm2 . If the stresses in copper and steel are not to exceed 50 MPa and 100 MPa respectively find safe load can be support Young s modulus of the steel is twice that of copper. Ans.: Total load P is 105kN Strength of Materials

17. Example:4 For the composite section fixed at both ends as shown in Fig., find : (i) Reactions at both ends (ii) Stresses in each part (iii) Construct axial force diagram. Assume : for Copper, AC = 4000 mm2, EC = 120 kN/mm2 for Aluminium, Aal = 6000 mm2, Eal = 70 kN/mm2 for Brass, Ab = 5500 mm2, Eb= 100 kN/mm2. Aluminium Copper Brass 30 kN 180 kN 50 kN 600 mm 500 mm 400 mm Strength of Materials