UNIT01 SIMPLE STRESSES STRAINS Lecture Number 03 Prof
UNIT-01. SIMPLE STRESSES & STRAINS Lecture Number – 03 Prof. S. C. SADDU MECHANICAL DEPARTMENT SIT LONAVALA Strength of Materials
Agenda • Statically Determinate Bars • Numerical Strength of Materials
STATICALLY DETERMINATE BEAMS Statically determinate beams are those beams in which the 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. Strength of Materials
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) + 2*75000) =0. 794 mm 17500*1200/(0. 785*25 Strength of Materials
• Example: 2 A steel bar having 40 mm*3000 mm dimension is subjected to an axial force of 128 k. N. Taking E=2*105 N/mm 2 and = 0. 3, find out change in dimensions. 40 128 k. N 3000 mm 40 128 k. N Hint: New breadth B 1 =B(1 - ) New Depth D 1 = D(1 - ) Strength of Materials
Example: 3 A prismatic steel bar having cross sectional area of A=300 mm 2 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. 17 mm) 20 k. N C B A 1 m 1 m 2 m A 15 0 20 C B 15 k. N 15 0 20 = 20000*1000/(300*2 x 10 5)-15000*2000/(300*2 x 10 5) = 0. 33 - 0. 5 = -0. 17 mm Strength (i. e. contraction) of Materials
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. A = 1000 mm 2 E = 1 x 10 9 m 3 m 5 A x 5 m A = 445 mm 2 E = 2 x 10 5 B P Strength of Materials
5 m 9 m 3 m P(9 -x)/9 P(x)/9 A B x P Strength of Materials
For the bar to be in horizontal position, Displacements at A & B should be same, A = B (PL/AE)A =(PL/AE)B {P(9 -x)/9}*3 (0. 001*1*105) {P(x)/9}*5 = 0. 000445*2*105 (9 - x)*3=x*5*1. 1236 27 -3 x=5. 618 x 8. 618 x=27 x = 3. 13 m Strength of Materials
• Example: 2 Determine the diameter of rod 200 m long hang vertically and subjected a axial pull of 325 k. N at its lower end if its end weight per cubic meter is 80 k. N and working stress is 75 Mpa. Also determine the total elongation of rod. Take E=210 Gpa L=200 mm P=325 k. N Strength of Materials Ans. : Diameter of rod: 83. 75 mm Total elongation: 63. 81 mm
Example: 3 A conical bar of diameter 2 m and length 3 m is hang to a ceiling. Weight density of material is 75 k. N/m 3 for rod and young’s modulus of rod is 100 Gpa. Determine the deformation due to its own weight. L=3 m Ans. : Deformation of bar is 1. 125 mm Strength of Materials
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 k. N, 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/m 2, Ec= 75 GN/m 2. Cast iron 40 mm copper Cast iron Strength of Materials 2 meter 80 mm
• Area of Copper Rod =Ac = ( /4)* 0. 042 = 0. 0004 m 2 • Area of Cast Iron 0. 0012 m 2 = Aci = ( /4)* (0. 082 - 0. 042) = 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 k. N/m 2 Strength of Materials
• • ci = 2. 33 x c = 6960. 8 k. N/m 2 load shared by copper rod = Wc = c Ac = 2987. 5 x 0. 0004 • = 3. 75 k. N • • Wci = 30 -3. 75 = 26. 25 k. N • 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
• Example: 2 A composite rod, 1200 mm long consists of a steel tube of 50 mm external diameter and 40 mm internal diameter. A copper rod of 30 mm diameter is placed coaxially into the steel tube. the assembly is held between two rigidly plates and is subjected to an compressive forces of 200 k. N. Find the stress induced in each material and the contraction produced. Take Es=200 Gpa, Ec=100 Gpa Copper 200 k. N Steel tube 200 k. N Hint: c / E c = s / E s L= c L / E c Ans. : c =94. 314 Mpa S =188. 628 Mpa L= 1. 132 mm
• 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 mm 2. 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 105 k. N Strength of Materials
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 mm 2, EC = 120 k. N/mm 2 for Aluminium, Aal = 6000 mm 2, Eal = 70 k. N/mm 2 for Brass, Ab = 5500 mm 2, Eb = 100 k. N/mm 2. Copper 180 k. N 500 mm Aluminium 30 k. N 50 k. N 400 mm Brass 600 mm Strength of Materials
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