Unit I Stress and Strain - PDF Free Download

Unit I Stress and Strain Stress and strain at a point Tension, Compression, Shear Stress Hooke s Law Relationship among elastic constants Stress Strain Diagram for Mild Steel, TOR steel, Concrete Ultimate Stress Yield Stress Factor of safety Thermal Stresses Compound Bars Thin Cylinders and Shells Strain Energy due to Axial Force Resilience Stresses due to impact and Suddenly Applied Load 1

References 1. Rajput.R.K. Strength of Materials, S.Chand and Co, New Delhi, 2007. Bhavikatti. S., "Solid Mechanics", Vikas publishing house Pvt. Ltd, New Delhi, 2010. Junnarkar.S.B. andshah.h.j, Mechanics of Structures, Vol I, Charotar Publishing House, New Delhi,1997. 2

Stress: The resistance offered by a body against deformation is called stress. The external force acting on the body is called the load. The load P is applied on the body while the stress f is induced in the material of the body. Stress, f = P (Load) A (Area) Saint Venant s Principle: The principle states that except at the extreme ends of a bar carrying direct loading, the stress distribution over the cross section is uniform. 3

Saint Venant s Principle (contd ): P b 2 3 1 b P b/2 1 2 1.5 b f av 3 b/2 f av f max =1.387f av b f max =1.027f av f av 1.5 b f max = f av 4

Strain: Strain is measure of the deformation caused due to external loading Strain, e = δl l Hook s Law: Stress is proportional to strain within elastic limit. f e within elastic limit. f = Ee P A = E δl l δl = Pl AE 5

Problem: When a rod of diameter 20 mm is subjected to a tensile force of 40 kn, the extension is measured as 250 divisions in 200 mm extension meter. Find the modulus of elasticity if each division is equal to 0.001 mm. 6

Solution: Diameter, d= 20 mm Tensile force, P= 40 kn Extension, δl =250 0.001 = 0.250 mm Length, l =200 mm. Modulus of elasticity, E=? As per hooke s law, δl = Pl AE E = Pl = 40 103 200 π = 1.0186 A δl 4 202 0.250 105 N/mm 2 7

Principle of superposition: When a number of loads are acting on a body, the resulting strain according to the principle of superposition will be the algebraic sum of the strains caused by the individual forces. P 1 P 2 P 3 P 4 l 1 l 2 l 3 For equilibrium, P 1 +P 3 = P 2 + P 4 δl = δl 1 + δl 2 + δl 3 8

Principle of superposition (contd ): P 1 P 2 P 3 P 4 l 1 l 2 l 3 P 1 P 2 P 3 + P 4 If P 1 > P 2, l 1 P 1 P 2 P 4 P 3 l 2 P 3 + P 1 P 2 P 4 l 3 9

Principle of superposition (contd ): P 1 P 2 P 3 + P 4 l 1 P 1 P 2 P 4 P 3 l 2 P 3 + P 1 P 2 P 4 l 3 As per principle of superposition deformation of whole bar, δl = δl AB + δl BC + δl CD = P 1l 1 + (P 1 P 2 )l 2 A 1 E A 2 E + P 4l 3 A 3 E 10

Problem 1: Stress and Strain at a point A steel bar 3.5 m long is acted upon by forces as shown in Figure. Determine the value of P and the total elongation of the bar. Take E=210 kn/mm 2 40 mm 30 mm 30 mm 45 kn P 30 kn 55 kn 1.0 m 1.5 m 1.0 m 11

Solution: Stress and Strain at a point 40 mm 30 mm 30 mm A B C 45 kn P 30 kn D 55 kn 1.0 m 1.5 m 1.0 m For equilibrium, As per principle of superposition, δl = δl AB + δl BC + δl CD = 45 + 30 = P + 55 P = 20 kn 12

Solution: δl AB = Pl Stress and Strain at a point = 45 1.0 103 π AE AB 4 302 210 =0.303 mm 30 mm A 45 kn 1.0 m B 45 kn δl BC = Pl = 25 1.5 103 π AE BC 4 402 210 =0.142 mm B 40 mm C 25 kn (45-20) 25 kn (55-30) δl CD = Pl = 55 1.0 103 π AE CD 4 302 210 =0.371 mm Total elongation, δl = δl AB + δl BC + δl CD δl = 0.303 + 0.142 + 0.371 δl =0.816 mm. 55 kn (45+30-20) 1.5 m 30 mm C 1.0 m D 55 kn 13

Problem 2: Stress and Strain at a point Find the value of P and the change in length of each component and the total change in length of the bar shown in Figure. E=200 kn/mm 2 30 mm 25 mm 20 mm 130 kn P 120 kn 50 kn 0.8 m 1.6 m 0.4 m 14

30 mm 25 mm 20 mm B A C 130 kn 120 kn P D 50 kn 0.8 m 1.6 m 0.4 m For equilibrium, 130 + 120 = P + 50 P = 200 kn As per principle of superposition, δl = δl AB + δl BC + δl CD = 15

Solution: δl AB = Pl = 130 0.8 103 π AE AB 4 252 200 Stress and Strain at a point = +1.059 mm 130 kn A 25 mm 0.8 m B 130 kn δl BC = Pl = 70 1.6 103 π AE BC 4 302 200 = 0.792 mm B 30 mm C 70kN 70 kn δl CD = Pl = 50 0.4 103 π = +0.318 mm AE CD 4 202 200 Total elongation, δl = δl AB + δl BC + δl CD δl = 1.059 0.792 + 0.318 δl =0.585 mm. 1.6 m 20 mm C D 50 kn 50 kn 0.4 m 16

Problem : Stress and Strain at a point A steel bar 3.5 m long is acted upon by forces as shown in Figure. Determine the value of P and the total elongation of the bar. Take E=210 kn/mm 2 40 mm 30 mm 30 mm 45 kn P 30 kn 55 kn 1.0 m 1.5 m 1.0 m 17

Extension of tapering bar of circular cross section dx P d d D P x l Consider an elemental length dx of the bar at a distance x from the smaller end. Let the diameter of the bar be d at distance x from the smaller end. d D d = d + L = d + kx x 18

Extension of tapering bar of circular cross section dx P d d D P x l Cross sectional area at a distance x from the smaller end A = π 4 d 2 = π 4 d + kx 2 Stress intensity on the section, f = P A = P π = 4P 4 d+kx 2 π d+kx 2 19

Extension of tapering bar of circular cross section dx P d d D P x l Strain =e = f E = 4P πe d+kx 2 Extension of the elemental length dx, δl = e dx = 4P πe d+kx 2 dx 20

Extension of tapering bar of circular cross section dx P d d D P Extension of the whole bar, = 4P πe x l 0 l dx d + kx 2 = 4P πe 1 k = 4Pl πe(d d) 1 d + kx 1 l 0 d + kl 1 d 21

Extension of tapering bar of circular cross section dx P d d D P x l 4Pl = πe(d d) = 4Pl πedd d + 1 D d l l 1 d 22

Extension of tapering bar of rectangular cross section x L t P b b x B P Consider an elemental length dx of the bar at a distance x from the smaller end. Let the width of the bar be b x at distance x from the smaller end. B b b x = b + L = b + kx x 23

xtension of tapering bar of rectangular cross section (contd..) x L t P b b x B P Cross sectional area at a distance x from the smaller end A = b x t= b + kx t Extension of the elemental length dx, δl = e dx P dx = b+kx t E 24

xtension of tapering bar of rectangular cross section (contd..) L t x P b b x B P Extension of the whole bar, = P te = P te dx b + kx 0 l 1 k log e b + kx 0 = P 1 te k log e b + kl log e b L 25

xtension of tapering bar of rectangular cross section (contd..) L t x P b b x B P = P te 1 B b L log e b + B b L b L = PL te B b log e B b 26

Extension due to self weight (a) Bar of uniform section: Weight per unit volume of the bar = w Considering a small section dx, Weight of the bar below the small section, w x = wax Deformation δ of the elemental strip will be given by δ = w x dx AE dx L = wax dx AE = wx dx E 27

Extension due to self weight (a) Bar of uniform section (contd ): Deformation δ of the elemental strip will be given by δ = wxdx E Deformation of the whole bar is given by = w E 0 L x dx = w L E x2 2 0 dx L = wl2 2E 28

Extension due to self weight (B) Bar of tapering section: Weight per unit volume of the bar = w Considering a small section dx, area of the section is A x. Weight of the bar below the section, w x = 1 3 wa xx Deformation δ of the elemental strip will be given by δ = w x dx A x E 1 = 3 wa xx dx A x E = 1 wx dx 3 E A dx x d L 29 B

Extension due to self weight (B) Bar of tapering section (contd ): A B Deformation δ of the elemental strip will be given by δ = 1 wxdx 3 E Deformation of the whole bar is given by = 1 3 w E 0 L x dx = w L 3E x2 2 0 dx x d L = wl2 6E 30

Problem 1: Stress and Strain at a point A steel plate of 20 mm thickness tapers uniformly from 100 mm to 50 mm in a length of 400 mm. What is the elongation of the plate, if an axial force of 80 kn acts on it. Take E = 200 10 3 N/mm 2. 31

Solution: Stress and Strain at a point A steel plate of 20 mm thickness tapers uniformly from 100 mm to 50 mm in a length of 400 mm. What is the elongation of the plate, if an axial force of 80 kn acts on it. Take E = 200 10 3 N/mm 2. t = 20 mm B= 100 mm b=50 mm L= 400 mm =? P=80 kn E=200 10 3 N/mm 2 PL = te B b log B e b 80 10 = 3 400 log 20 200 10 3 (100 50) e Elongation, = 0.111 mm. 100 50 32

Problem 2: Stress and Strain at a point A round bar of steel tapers uniformly from a diameter of 2.5 cm to 3.5 cm in length of 50 cm. If an axial force of 60000 N applied at each end, determine the elongation of the bar. Take E= 205 kn/mm 2. 33

Solution: Stress and Strain at a point A round bar of steel tapers uniformly from a diameter of 2.5 cm to 3.5 cm in length of 50 cm. If an axial force of 60000 N applied at each end, determine the elongation of the bar. Take E= 205 kn/mm 2. E= 205 kn/mm 2 P= 60 kn d= 25 mm D=35 mm L=500 mm = 4Pl πedd 4 60 500 = π 205 25 35 = 0.213 mm 34

Composite sections: Stress and Strain at a point 2 Tube P 1 Bar 2 Tube P When a bar consists of two different materials, it is said to be composite. Since there are two unknowns, two equations will be required. The conditions of equilibrium will provide one equation for the stresses in the individual sections. The other equation can be obtained from the consideration of the deformation of the whole structure. 35

Composite sections (contd ): 2 Tube P 1 Bar 2 Tube P Solid bar enclosed in the hollow tube and subjected to a compressive force P through rigid collars as shown in Fig. From the equilibrium equation, P 1 + P 2 = P (1) Since the whole assembly is composite, the deformation of the bar is equal to deformation of tube. Therefore, 1 = 2 P 1 l A 1 E 1 = P 2l A 2 E 2 (2) 36

Composite sections (contd ): 2 Tube P 1 Bar 2 Tube P P 1 + P 2 = P (1) P 1 l A 1 E 1 = P 2l A 2 E 2 (2) By solving (1) & (2), one can find P 1 and P 2 37

Problem: Stress and Strain at a point A copper rod of 30 mm diameter is surrounded tightly by a cast iron tube of 60 mm external diameter, the ends being firmly fastened together when this compound bar is subjected to 12 kn compressive load, what will be the share of load carried by copper rod and cast iron tube? How much will be the shortening of the bar in 1m length. Assume E CI = 100 kn/mm 2, E C = 90 kn/mm 2 38

Solution: Stress and Strain at a point A copper rod of 30 mm diameter is surrounded tightly by a cast iron tube of 60 mm external diameter, the ends being firmly fastened together when this compound bar is subjected to 12 kn compressive load. Cast Iron P=12 kn Copper 30 mm Cast iron 60 mm P=12 kn E CI = 100 kn/mm 2 E C = 90 kn/mm 2 P c =?, P CI =? =? 39

Solution: Stress and Strain at a point P=12 kn Copper P c + P CI = P Cast Iron Cast iron P c + P CI = 12 (1) P C l A C E C = P C l π 4 302 90 = P CIl A CI E CI 30 mm P CI l π 4 (602 30 2 ) 100 60 mm P=12 kn P C = 0.3 P CI (2) 40

Solution: Stress and Strain at a point Cast Iron P=12 kn Copper 30 mm Cast iron 60 mm P=12 kn P c + P CI = 12 (1) P C = 0.3 P CI (2) By solving (1) and (2) equations, P CI = 9.23 kn P C = 2.77 kn Contraction of the bar, = P Cl = 2.77 1000 π A C E C 4 302 90 = 0.044 mm 41

Problem 3: Stress and Strain at a point A steel cylinder 500 mm outside diameter and 10 mm thick is placed inside a brass cylinder whose outside diameter is 520 mm and metal thickness 10 mm. The whole assembly is subjected to a compressive force of 40 kn. Determine the load carried by the inner and outer cylinders when the steel cylinder is shorter by 0.005 mm than the brass cylinder. Find also the deformation of brass cylinder. Height of the brass cylinder is 400 mm. E S = 2 10 5 Mpa, E b = 1 10 5 MPa. 42

400 mm Solution: Stress and Strain at a point A steel cylinder 500 mm outside diameter and 10 mm thick is placed inside a brass cylinder whose outside diameter is 520 mm and metal thickness 10 mm. 40 kn 0.005 mm st The whole assembly is subjected to a compressive force of 40 kn. Determine the load carried by the inner and outer cylinders when the steel cylinder is shorter by 0.005 mm than the brass cylinder. Find also the deformation of brass cylinder. Height of the brass cylinder is 400 mm. E S = 2 10 5 Mpa, E b = 1 10 5 MPa. Brass 500 mm 520 mm Brass 43

400 mm Solution: Stress and Strain at a point Load taken by the brass cylinder, P b. P b L b A b E b = 0.005 P b = 20027.65 N 40 kn 0.005 mm st Remaining Load=40000-20027.65=19972 N P b L = P sl A b E b A s E s Brass 500 mm 520 mm Brass P b = 0.52 P s (1) P b + P s = 19972 (2) 44

400 mm Solution: P b = 0.52 P s (1) Stress and Strain at a point P b + P s = 19972 (2) By solving (1) and (2), P b = 6832.52 N P s = 13139.47 N Brass Deformation of brass cylinder, b = P bl b = 6832.52 400 = 1.706 A b E b 16022.123 1 10 5 10 3 mm 40 kn 0.005 mm st Brass 500 mm 520 mm Total deformation of brass cylinder=0.005+0.001706=0.006706 mm 45

Temperature stresses: Stress and Strain at a point When a material undergoes change in the temperature, its length is varied and if the material is free to do so, no stresses are developed. If the material is constrained so that no change in length is permissible, stresses are developed in the material. These developed stresses are temperature stresses and may be tensile or compressive depending upon whether contraction is checked or extension is checked. 46

Temperature stresses (contd ): Let us consider a bar of length L, snuggly fitting between two walls or supports. If the temperature is increased through t 0 C, the bar will be increased in length by an amount = Lαt, Where α is the coefficient of thermal expansion. Due to external conditions, the bar is not free to expand by the amount and therefore from Hooke s law, = fl, where f is the temperature stress. E f = E Lαt E = = αte L L f = αte (Compressive) 47

Temperature stresses Composite section: If the bar consists of two materials and is subjected to temperature change, opposite kinds of stresses (i.e. tensile and compressive) will be setup in the two materials. Let us take a bar of two materials, i.e. of steel and copper, subjected to an increase in temperature Refer Fig. in which the shaded portion shown the final or common expansion = C = S. Steel (S) Copper (C) s t Final expansion 48

Temperature stresses Composite section: Steel (S) Copper (C) t f s s The dotted lines shows the individual expansions in the steel and copper. Since for copper is more, it will try to expand more than steel and hence, in order to have common expansion, tensile stress will be developed in steel and compressive stress in copper. C t c f Final expansion 49

Temperature stresses Composite section: Steel (S) Copper (C) t f s s From the stress equation, we get tensile force in steel = Compressive force in copper P S = P C = P f s A s = f C A C (1) From the compatibility equation, Extension of steel = Extension in copper S = C = C t c f Final expansion 50

Temperature stresses Composite section: Steel (S) Copper (C) t f s s c f Final expansion From the above Fig., S = S t + S f C = C t C f C t t S + f S = t f C C f S + f C = t t C S f S l + f C l = α E S E C tl α s tl (2) C 51

Temperature stresses Composite section: Steel (S) Copper (C) t f s s c f Final expansion f s A s = f C A C (1) C t f S l E S + f C l E C = α C tl α s tl (2) By solving (1) & (2) one can find f S and f C values. 52

Problem: Stress and Strain at a point A steel tube of 30 mm external diameter and 25 mm internal diameter encloses a copper rod of 20 mm diameter to which it is rigidly connected at each end. If at the temperature of 15 0 C, there is no longitudinal stress, calculate the stresses in the rod and the tube when the temperature is raised to 115 0 C. E S = 200 kn/mm 2, E C = 100 kn/mm 2, α S = 6 10 6 / 0 C, α C = 10 10 6 / 0 C 53

Solution: Stress and Strain at a point A steel tube of 30 mm external diameter and 25 mm internal diameter encloses a copper rod of 20 mm diameter to which it is rigidly connected at each end. Steel tube 30 mm 25 mm Copper rod Steel tube 20 mm A S = π 4 302 25 2 = 215.98 mm 2 A C = π 4 202 = 314.16 mm 2 54

Solution (contd ): Stress and Strain at a point If at the temperature of 15 0 C, there is no longitudinal stress, calculate the stresses in the rod and the tube when the temperature is raised to 115 0 C. E S = 200 kn/mm 2, E C = 100 kn/mm 2, α S = 6 10 6 / 0 C, α C = 10 10 6 / 0 C Due to temperature raise, total tension in steel = total compression in copper f s A s = f C A C f s 215.98 = f C 314.16 f s = 1.455f C (1) t= 115-15=100 55

Solution (contd ): Stress and Strain at a point Actual expansion of steel= Actual expansion of copper f S l + α E s tl = α C tl f C l S E C f S + α E s t = α C t f C S E C 1.455f C 200 10 3 + 6 10 6 100 = 10 10 6 100 f C = 23. 16 N/mm 2 f S = 1.455f C = 1.455 23.16 = 33. 69 N/mm 2 Stress in copper rod=23.16 N/mm 2 Stress in Steel tube= 33.69 N/mm 2 f C 100 10 3 56

Problem: Stress and Strain at a point A composite bar of 900 mm length is made up of two bars, aluminium and steel. The steel bar is of 600 mm length and 200 mm 2 in section and aluminium bar is of 300 mm length and 300 mm 2 in section. The ends of the composite bar is held between rigid supports. Calculate the stresses in steel and aluminium if the temperature is raised by 50 0 C. E st = 2.1 10 5 MPa; E Al = 0.7 10 5 MPa; α st = 11.7 10 6 / 0 C; α Al = 23.4 10 6 / 0 C. 57

Solution: Stress and Strain at a point A composite bar of 900 mm length is made up of two bars, aluminium and steel. The steel bar is of 600 mm length and 200 mm 2 in section and aluminium bar is of 300 mm length and 300 mm 2 in section. The ends of the composite bar is held between rigid supports. A S =200 mm 2 A a =300 mm 2 Steel bar Aluminium bar 600 mm 300 mm 58

Solution (contd ): Stress and Strain at a point The temperature is raised by 50 0 C. E st = 2.1 10 5 MPa; E Al = 0.7 10 5 MPa; α st = 11.7 10 6 / 0 C; α Al = 23.4 10 6 / 0 C. f st =?, f Al =? A S =200 mm 2 Steel bar A a =300 mm 2 Aluminium bar 600 mm 300 mm Since the free expansion is checked, either fully or partly, compressive stresses will be setup in both steel as well as aluminium bars. 59

Solution (contd ): Stress and Strain at a point A S =200 mm 2 A a =300 mm 2 Steel bar Aluminium bar 600 mm 300 mm Since the force in both aluminium and steel bars is the same, A S f S = A a f a 200f S = 300f a f S = 1.5f a (1) 60

Solution (contd ): Stress and Strain at a point A S =200 mm 2 A a =300 mm 2 Steel bar Aluminium bar 600 mm 300 mm Contraction due to compressive stresses in both the materials=free expansion due to temperature rise. f s L s E s + f al a E a = L s α s t + L a α a t (2) f s 600 2.1 10 5 + f a 300 0.7 10 5 = (600 11.7 10 6 50) + (300 23.4 10 6 50) 61

Solution (contd ): Stress and Strain at a point A S =200 mm 2 A a =300 mm 2 Steel bar Aluminium bar 600 mm 300 mm f s 600 2.1 10 5 + f a 300 0.7 10 5 = (600 11.7 10 6 50) + (300 23.4 10 6 50) Substituting f S = 1.5f a (from (1)) in the above equation, 1.5f a 600 2.1 10 5 + f a 300 0.7 10 5 = 0.702 f a = 81.9 N/mm 2 f S = 1.5f a =1.5 81.9 = 122.85 N/mm 2 62

Problem: Stress and Strain at a point A steel rod of 20 mm diameter passes centrally through a tight fitting copper tube of external diameter 40 mm. The tube is closed with the help of rigid washers of negligible thickness and nuts threaded on the rod. The nuts are tightened till the compressive load on the tube is 50 kn as shown in Fig. Determine the stresses in the rod and tube, when the temperature of the assembly falls, by 50 K. Take E for steel and copper as 200 GPa and 100 GPa respectively. Take coefficient of expansion for steel and copper as 12 10 6 /K and 18 10 6 /K respectively. Steel rod Copper tube 50 kn 50 kn 20 mm Copper tube 40 mm 63

Problem: Stress and Strain at a point A composite bar shown in Fig. is rigidly fixed at the ends. An axial pull of P=15 kn is applied at B at 10 0 C. Find the stresses in each material at 80 0 kn C. E st = 210, E mm 2 Al = 70 kn/mm 2 ; α st = 11 10 6 / 0 C; α Al = 24 10 6 / 0 C. 30 mm 40 mm Aluminium bar 15 kn Steel bar 100 mm 200 mm 64

Elastic Constants: E, C,K & μ Poisson s ratio - Lateral Strain: Any direct stress produces a strain in its own direction and an opposite strain in every direction at right angles to this. The ratio of the lateral strain to the longitudinal strain is constant for a given material and is known as poisson s ratio. Poisson s rati, 1 m μ = Lateral strain Longitudinal strain = Constant 1 m 1 m Lateral strain = 1 m Longitudinal strain μ = 0.25 to 0.42 For most of the metals μ = 0.45 to 0.50 For rubber 65

Elastic Constants: E, C,K & μ Single direct stress along the longitudinal axis: P P For rectangular bar: V=Lbd δv = δl b d + L δb d + L b δd δv V = δl L + δb b + δd d Volumetric strain, e V = δv = f 1 f 1 f V E m E m E = f 1 2 E m 66

Elastic Constants: E, C,K & μ Single direct stress along the longitudinal axis: P P For Circular bar: V= π 4 d2 L δv = π 4 d2 δl + 2 π d δd L 4 δv V = δl L + 2 δd d Volumetric strain, e V = δv = f 2 f V E m E = f 1 2 E m 67

Elastic Constants: E, C,K & μ Stress f 1 along the axis and f 2 and f 3 perpendicular to it: f 2 f 3 f 1 f 1 f 3 f 2 Longitudinal strain, e 1 = f 1 f 2 f 3 E me me Similarly, e 2 = f 2 f 1 f 3 E me me e 3 = f 3 f 1 f 2 E me me If some of the stresses are opposite sign, necessary change in the algebraic signs of the above expressions will have to be made. 68

Elastic Constants: E, C,K & μ Stress f 1 along the axis and f 2 and f 3 perpendicular to it: f 2 f 3 f 1 f 1 f 3 f 2 Upper limit of Poisson s ratio: e 1 + e 2 + e 3 = 1 2 f 1 + f 2 + f 3 m E Sum of the three strains on the left is known as dilatation and for small strains it represents change volume per unit volume. In the case of hydrostatic tension, we have f 1 =f 2 =f 3 =f and resulting strains are e 1 = e 2 = e 3 =e. Hence, Ee = f 1 2 m 69

Elastic Constants: E, C,K & μ Stress f 1 along the axis and f 2 and f 3 perpendicular to it: f 2 f 3 f 1 f 1 f 3 f 2 In the expression Ee = f 1 2 ; E, e and f are positive m numbers and hence 1 2 must also be positive. m This limits the 1 to a maximum value of 0.5. m Maximum value of poisson s ration =0.5. 70

Elastic Constants: E, C,K & μ Modulus of rigidity (C): The ratio of shear stress and shear strain C = q Complementary shear stress: A stress in a given direction cannot exist without a balancing shear stress of equal intensity in a direction at right angles to it. q q q q q q 71

Elastic Constants: E, C,K & μ Bulk Modulus (K): f f f f f f When a body is subjected to three mutually perpendicular like direct stresses of the equal intensity, the ratio of the direct stress f to the volumetric strain is defined as the bulk modulus K of the body K = f e v 72

Relation between elastic constants: Relation between E and K: f f f f f f Let us take a cube of side L be subjected to three mutually perpendicular like stresses of equal intensity f. By the definition of the bulk modulus K = f e v Or e v = δv V = f K (1) 73

Relation between elastic constants: Relation between E and K: f f f f f f The total linear strain of each side, e = f f F E me me δl L = e = f E 1 2 m Now V=L 3, e v = δv V = 3L2 δl L 3 = 3 δl L δv = 3L 2 δl = 3e = 3f E 1 2 m (2) 74

Relation between elastic constants: Relation between E and K: f f e v = f K (1) f f e v = 3f E 1 2 (2) m Equating (1) and (2), f K = 3f E 1 2 m f f E = 3K 1 2 m 75

Relation between elastic constants: Relation between E and C: The linear strain e of the diagonal AC, e = FC AC = CC cos 45 CD sec 45 e = CC 2CD = 1 2 = 1 2 q C (i) A state of simple shear produces tensile and compressive stresses along diagonal planes and the value of such direct stresses is numerically equal to q. On diagonal AC, there is a tensile stress f, and on diagonal BD there is a Compressive stress f. B A B C C 45 45 F D 76

Relation between elastic constants: Relation between E and C (contd ): The linear strain e of the diagonal AC, due to the two mutually perpendicular direct Stresses is given by e = f E f me = f E 1 + 1 m But f = q q B A B q C C 45 45 F q q D e = q E 1 + 1 m Equating (i) and (ii), we get 1 q 2 C = q E ii 1 + 1 m E = 2C 1 + 1 m 77

Relation between elastic constants: Relation between K and C: E = 3K 1 2 m (1) E = 2C 1 + 1 (2) m Equating (1) and (2), 1 3K 2C = m 6K + 2C Relation between E, K and C: By eliminating 1 we get, m E = 9KC 3K + C q B A B q C C 45 45 F q q D 78

Problem: Stress and Strain at a point A solid circular rod of diameter 30 mm and length 2m, when subjected to an axial tensile load, the diameter of the bar is reduced by 0.005 mm. If the young s modulus of the material of the rod is 200 kn/mm 2 and modulus of rigidity is 80 kn/mm 2 determine the following: (i) Axial load applied on the bar. (ii) The normal stress induced in the bar. (iii) The change in the length of the bar and (iv) Bulk modulus of the material of the bar. 79

Solution: Stress and Strain at a point A solid circular rod of diameter 30 mm and length 2m, when subjected to an axial tensile load, the diameter of the bar is reduced by 0.005 mm. d= 30 mm L= 2m=2000 mm P=? δd = 0.005 mm P P If the young s modulus of the material of the rod is 200 kn/mm 2 and modulus of rigidity is 80 kn/mm 2 E= 200 kn/mm 2 C=80 kn/mm 2 f =? δl=? K =? 80

Solution: We know the relation, Stress and Strain at a point E = 2C 1 + 1 m 1 + 1 m = E 2C 1 m = E 2C 1 P P = 2 10 5 2 0.8 10 5 1 = 0.25 1 m = 0.25 81

Solution: We know the relation, 1 lateral strain m = Linear strain Stress and Strain at a point δd d δl L P P 0.25 = 0.005 30 δl 2000 δl = 1.33 mm (iii) The change in length of the bar=1.33 mm 82

Solution: Stress and Strain at a point Young s modulus, E = f = E δl L Direct stress (f) δl Linear strain L f = 2 10 5 1.33 = 133 N/mm2 2000 (ii) The normal stress induced in the bar, f = 133 N/mm 2 P P f = P A P = fa = 133 π 4 302 = 94012.16 N (i) Axial load applied on the bar, P = 94.01 kn 83

Solution: Stress and Strain at a point iv Bulk modulus of the material of the bar, E = 3K E K = 3 1 2 m P 2 10 K = 3 1 (2 0.25) = 1.33 105 N/mm 2 Bulk modulus, K = 1.33 10 5 N/mm 2 1 2 m P 84

Problem 2: Stress and Strain at a point A rod of length 1 m and diameter 20 mm is subjected to a tensile load of 20 kn. The increase in length of the rod is 0.3 mm and decrease in diameter is 0.0018 mm. Calculate poisson s ratio and three moduli. 85

Solution: d = 20 mm L = 1 m = 1000 mm P = 20 kn = 20 10 3 N δl = 0.3 mm δd = 0.0018 mm 1 m =?, E =?, C =?, K =? 1 m = 0.3 E = 2.12 10 5 N/mm 2 C = 0.82 10 5 N/mm 2 K = 1.77 10 5 N/mm 2 86

Thin Cylinders Y f 2 f 2 X f 1 A B dp θ θ dp f 1 X p p p C D f 1 f 1 f 2 f 2 Fig. shows a thin cylindrical shell whose internal diameter is d, the thickness of the shell being t. Let the length of the shell be L. Let the shell be subjected to an internal pressure of P. Y 87

Thin Cylinders (contd ) Let us consider a longitudinal section XX through the axis, dividing the shell into two halves A and B. Now let us consider two elementary strips subtending an angle dθ at the centre at an angle θ on either side of the vertical through the centre. X f 1 f 1 A B dp n θ θ dp n f 1 f 1 X Normal force on each strip dp n = p r dθ L Where r is the radius of the shell. 88

Thin Cylinders (contd ) The resultant of the two normal forces on the two elemental strips, dp = 2prL dθ cos θ acting vertically, i.e. normal to XX. X f 1 A B dp θ θ dp f 1 X Total force normal to XX on one side of XX= Total bursting force= P = π 2 2prL cos θ dθ = 2prL = p d L f 1 f 1 0 = Intensity of radial pressure Projected area. 89

Thin Cylinders (contd ) Let f 1 be the intensity of tensile stress induced in the metal across the section XX. X Resisting force offered by the section XX=f 1 2tL Equating the resisting force to the bursting force f 1 2tL = pd L f 1 f 1 A B dp θ θ dp f 1 f 1 X f 1 = p d 2 t 90

Thin Cylinders (contd ) Just like the section XX, if we had considered any other longitudinal section, the intensity of tensile stress would be found to be the same. X f 1 A B dp θ θ dp f 1 X Hence the direction of the tensile stress is along the circumference of the shell. A stress so induced is called a hoop or circumferential stress. f 1 f 1 91

Thin Cylinders (contd ) Let us now consider a section YY normal to the axis of the shell. Let the section divide the shell into two parts C and D. Force acting on the end of the shell= P = p π 4 d2 In order the shell may not be split up at the section YY, the section will offer a resistance. P P P Y f 2 f 2 C D f 2 Y f 2 92

Thin Cylinders (contd ) Let f 2 be the tensile stress on the section YY. Equating the resistance offered by the section YY to the total force on one end of the shell f 2 πd t = p π 4 d2 P P P Y f 2 f 2 C D f 2 = pd 4t This stress is called the longitudinal stress f 2 Y f 2 93

Thin Cylinders (contd ) Hence at any point in the metal of the shell there are two principal stresses, namely a hoop stress of pd acting circumferentially and a longitudinal stress of pd 4t 2t acting parallel to the axis of the shell. Greatest shear stress, q max = f 1 f 2 2 = pd 2t pd 4t 2 q max = pd 8t = pd 8t 94

Circumferential strain, Stress and Strain at a point Thin Cylinders (contd ) e 1 = f 1 E f 2 me = pd 2tE pd 4tmE = pd 2tE 1 1 2m e 1 = pd 4tE 2 1 m 95

Thin Cylinders (contd ) Longitudinal strain, e 2 = f 2 E f 1 me = pd 4tE pd 2tmE e 2 = pd 1 2tE 2 1 m Circumferential strain = Change in circumference original circumference =πδd πd =e 1 δd d = e 1 Similarly, Longitudinal strain δl L = e 2 96

Thin Cylinders (contd ) Volume of the cylinder, Change in volume, V = π 4 d2 L δv = π 4 d2 δl + π 2d δd L 4 Volumetric strain, π δv V = 4 d2 δl + π 2d δd L 4 π = δl 4 d2 L L + 2 δd d e v = e 2 + 2e 1 δv V = pd 1 2tE 2 1 m + 2pd 2tE 1 1 2m δv V = pd 5 2tE 2 2 m 97

Thin Cylinders (contd ) δv V = pd 2tE 5 2 2 m δv = pd 4tE 5 4 m V 98

Problem: A thin cylindrical shell 3.5 m long, 1.5 m in diameter and 15 mm thick is subjected to a pressure of 1.5 Mpa. Calculate the maximum shear stress and change in dimensions of the shell. E= 2 10 5 N/mm 2 and poisson s ratio is 0.3 also calculate change in volume. 99

Solution: A thin cylindrical shell 3.5 m long, 1.5 m in diameter and 15 mm thick is subjected to a pressure of 1.5 MPa. Calculate the maximum shear stress and change in dimensions of the shell. E= 2 10 5 N/mm 2 and poisson s ratio is 0.3 also calculate change in volume. L=3.5 m=3500 mm, d=1.5 m=1500 mm, t=15 mm, p=1.5 MPa. E=2 10 5 N/mm 2, 1 m = 0.3. q max =?, δl =?, δd =? δv =? 100

Solution (contd ): Circumferential stress, f 1 = pd 2t Stress and Strain at a point = 1.5 1500 2 15 = 75 N/mm 2 Longitudinal stress, f 2 = pd 4t = 1.5 1500 4 15 = 37.5 N/mm 2 Maximum shear stress, q max = f 1 f 2 2 = 75 37. 5 2 = 18. 75 N/mm 2 101

Solution (contd ): Circumferential strain, Stress and Strain at a point e 1 = f 1 E f 2 me = 75 37.5 0.3 2 105 2 10 5 = 31.875 10 5 Longitudinal strain, e 2 = f 2 E f 1 me = 37.5 75 0.3 2 105 2 10 5 = 7.5 10 5 Change in diameter, δd = e 1 d = 31.875 10 5 1500 = 0.478 mm δd = 0. 478 mm 102

Solution (contd ): Stress and Strain at a point Change in length, δl = e 2 L = 7. 5 10 5 3500 = 0. 2625 mm Volumetric strain, Change in volume, e v = 2e 1 + e 2 = (2 31.875 10 5 ) + 7.5 10 5 = 71.25 10 5 e v = 71. 25 10 5 δv = e v V = 71.25 10 5 π 4 15002 3500 = 4.41 10 6 mm 3 δv = 4. 41 10 6 mm 3 103

Problem: A built up cylindrical shell of 300 mm diameter and 3 m long and 6 mm thick is subjected to an internal pressure of 2 MPa. Calculate the change in diameter, change in length and change in volume of cylinder. Efficiency of longitudinal joint is 80% and that of circumferential joint is 50%. Assume E= 200 GPa and poisson s ratio is 0.29. 104

Solution (contd ): Stress and Strain at a point Circumferential stress, f 1 = pd = 2 300 = 62.5 N/mm2 2tη L 2 6 0.8 Longitudinal stress, f 2 = pd = 2 300 4tη c 4 6 0.5 = 50 N/mm2 Circumferential strain, e 1 = f 1 E f 2 me = 62.5 50 0.29 2 105 2 10 5 = 24 10 5 106

Solution (contd ): Longitudinal strain, Stress and Strain at a point e 2 = f 2 E f 1 me = 50 62.5 0.29 2 105 2 10 5 = 15.94 10 5 Change in diameter, δd = e 1 d = 24 10 5 300 = 0.072 mm δd = 0. 072 mm Change in length, δl = e 2 L = 15. 94 10 5 3000 = 0. 4782 mm 107

Solution (contd ): Stress and Strain at a point Volumetric strain, Change in volume, e v = 2e 1 + e 2 = (2 24 10 5 ) + 15.94 10 5 = 63.94 10 5 e v = 63. 94 10 5 δv = e v V = 63.94 10 5 π 4 3002 3000 = 135.59 10 3 mm 3 δv = 135. 59 10 3 mm 3 108

Strain energy: When a member is deformed under the action of external loading, the member is said to have stored energy which is called the strain energy or the resilience of the member. The strain energy stored by the member is equal to the amount of work done by the external forces to produce deformation. 109

Strain energy due to axial force A member subjected to an external load W. Let the extension of the member be δ. Since the load is applied gradually, the magnitude of the load is increased gradually from zero to the value W and the member also has gradually extended. External work done, W e = Avg.load displacement = 0+W 2 δ = 1 2 Wδ. Let the energy stored by the member be W i. We have W e =W i, Let the tension in the member be S. For the equilibrium of the member, S = W. w δ A w

Strain energy due to axial force (contd ) Tensile stress, f = S A, Tensile strain, e = f E = S AE, Where E is the young s modulus of the material. Change in length, δ= e l = Sl AE Strain energy stored = work done = 1 2 Wδ U A = 1 2 S Sl AE = s2 l 2AE U A = s2 l 2AE Strain energy stored per unit volume = s2 l 2AE w Al = f2 2E δ A w

Stresses due to various types of loads: (i) Gradually applied load: Let a load of magnitude W be applied axially on a member of length L and uniform cross sectional area A. Let δl be the extension of the rod. Let f be the stress intensity in the rod. Strain energy stored by the member, U = f2 2E Volume of the rod = f2 2E AL Work done by the external load= Average load extension = 1 2 W δl 112

Stresses due to various types of loads: (i) Gradually applied load: By equating the strain energy stored by the member to the work done by the loading We get, f 2 2E AL = 1 2 W δl But δl = fl E f2 2E AL = 1 2 W fl E f = W A 113

Stresses due to various types of loads: (ii) Suddenly applied load: Let the load W be suddenly applied. Let the extension of the member be δl. In this case the magnitude of load is constant as W throughout the process of extension. Let f be the maximum stress induced. Equating the strain energy stored by the member to the work done by the loading, But δl = fl E f 2 2E f2 2E AL = W δl AL = W fl E f = 2W A 114

Stresses due to various types of loads: (iii) Impact load: In this case the load W is dropped from a height h before it commences to stretch the bar. L Fig. shows a vertical bar when upper end is fixed at top and a collar is provided at the lower end. h The load W drops by a height h on the collar and will thus extend the member by δl. δl Let f be the maximum intensity of stress produced in the bar. 115

Stresses due to various types of loads: (iii) Impact load (contd ): Extension of the bar, δl = fl E Equating the bar potential energy to the strain energy by the rod W h + δl W f 2 2E h + fl E AL fwl E = f2 2E AL = f2 2E AL = Wh 116

Stresses due to various types of loads: (iii) Impact load (contd ): f 2 2Wf A = 2WhE AL Adding W2 A2 to both sides of this equation, we get f 2 2W W2 f + A A 2 = W2 A 2 + 2WhE AL f W A 2 = W2 A 2 + 2WhE AL 117

Stresses due to various types of loads: (iii) Impact load (contd ): f W A 2 = W2 A 2 + 2WhE AL f W A = W2 A 2 + 2WhE AL f = W A + W2 A 2 + 2E W A h L f = W A 1 + 1 + 2E A h W L 118

Problem: Stress and Strain at a point A bar 12 mm diameter gets stretched by 2 mm under a gradually applied load of 10 kn. What stress would be produced in the same bar by a weight of 12 kn which falls vertically a distance of 4 cm on to a rigid collar attached at its end. The bar is initially unstressed. E= 200 GPa. 119

solution: Stress and Strain at a point The stress due to gradually applied load, f= 88.42 N/mm 2 L=4523.86 mm The stress due to impact load, f = W A 1 + 1 + 2E A W h L f = 727.81 N/mm 2 120

Problem: Stress and Strain at a point An unknown weight falls through 10 mm on to a collar rigidly attached to the lower end of a vertical bar 3m long and 600 mm 2 in section. If the maximum instantaneous extension is 2 mm, what is the corresponding stress and the value of unknown weight. E= 2 10 5 MPa. 121

Solution: Stress and Strain at a point f = 133.33 N/mm 2 W = 6666.67 N 122