- 1. Strength of Materials - I Page 1 CHAPTER 1 SIMPLE STRESSES AND STRAINS Content: Concept of strain and stress Hook’s Law Stress Strain relationship Elastic constants Factor of safety Stress and strain in three dimensions Introduction: In our day to day work we engineers come across different materials like cement, concrete, still etc. which are used to complete our project. For selecting suitable material every one of us is interested to know the strength. The ability of element of the structure to resist its failure under the application of external force is known as ‘strength of material’ and the ability to resist deformation is known as ‘stiffness’ of that material. It has been observed that the material first deforms and then failure takes place. A detailed study of forces and their effects along with suitable protective measures is known as ‘strength of material’. Behavior of material: Whenever force acts on body it undergoes deformation and some resistance to deformation. When external force is removed the resistance force will vanish and body come back to original shape and size. If is only possible if deformation due to external force is within certain limit. This limit is known as ‘Elastic Limit’. Elasticity:
- 2. Strength of Materials - I Page 2 The ability of material to retain its original shape and size perfectly after removal of load is known as ‘Elasticity’. Elastic Limit: Limit within which the body behaves perfectly elastic is known as ‘Elastic Limit’. Stress: Every material is elastic in nature that’s why whenever external force acts on body, body undergoes deformation. As a body undergoes deformation its molecule set up some resistance to the deformation. Resistance per unit deformation is known as stress Stress (σ) = 𝐿𝑜𝑎𝑑 𝐴𝑟𝑒𝑎 = 𝑃 𝐴 Unit: N/mm2 or MPa or GPa Note: Tensile stresses will be treated as positive and compressive stress is treated as negative. Strain: Strain is the ratio of change in dimension to original dimension. Strain is denoted as (e). Strain is a dimensionless quantity. Strain = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 Strain = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ Strain = 𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ− 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ Strain = 𝐿+ 𝛿𝐿−𝐿 𝐿 = 𝛿𝐿 𝐿 ∴ 𝑒 = 𝛿𝐿 𝐿 Hook’s Law: It states that within elastic limit stress is directly proportional to strain.
- 3. Strength of Materials - I Page 3 Mathematically, Stress α strain σ α e σ = E e where E = Young’s Modulus E = 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜎 𝑒 σ = 𝑃 𝐴 and 𝑒 = 𝛿𝐿 𝐿 ∴ 𝐸 = 𝜎 𝑒 𝛿𝐿 𝐿 ∴ 𝛿𝐿 = 𝑃𝐿 𝐴𝐸 Where, P = Force or Load L = length of member A = area of cross section E = modulus of elasticity 𝛿𝐿 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ Stress – strain relationship for ductile Material: By Hook’s law within elastic limit stress is directly proportional to strain. From point O to A stress is directly proportional to strain. Therefore point A is called as proportionality limit. After that, at point B mild still regain its size and shape after removal of load. This point is called as elastic limit. At point C stress remains constant and strain increases and material enters at plastic zone but it can still regain its size and shape after removal of load. Point C is called as upper yield point and
- 4. Strength of Materials - I Page 4 point D is called as lower yield point. As load increases it reaches at point E where he stress becomes highest and it is called ultimate stress. Every material has its own limit. Stress increases can crosses the permissible limit and breaks at point F so point F is called as breaking point. Stress – strain relationship for brittle Material: The initial point OA of the graph is straight line. It indicates that the stress is directly proportional to strain, hence Hook’s law is obeys. The stress corresponding to point A is called elastic limit. If stress corresponding to point E is removed then wire regains its original length completely. Numerical: 1) A bar of length 1.2 m extends through 3 mm under the action of axial pull of 2.5 kN. Compute strain in material if diameter of bar is 20 mm. Find stress in bar. 2) A wooden bar of 8 m long, 100 mm wide and 120 mm thick is subjected to an action pull of 50 kN and stretch is 4 mm. Calculate value of strain for the material a. Stress b. Strain c. Modulus of elasticity 3) A short timber post of rectangular cross section has one post side of section is 2 times the other and is loaded with 10 kN force when it contracts by 0.0525 mm foe 1 m length. Modulus of elasticity is 12 Gpa. Calculate sectional dimensions of post. 4) A load of 10 kN is to be rest with the help of steel wire. Find the minimum diameter of steel wire if stresses is not to exceed 100 MPa. Factor of safety: It is the ratio of maximum stress to the working stress. Factor of safety = maximum stress working stress In case of ductile material failure of material occurs once it reaches plastic deformation hence Factor of safety is based upon yield point stress.
- 5. Strength of Materials - I Page 5 Factor of safety = Yiels stress working stress In case of brittle material Factor of safety depends upon ultimate stress. Factor of safety = ultimate stress working stress Principle of Superposition: According to principle of superposition when number of loads are acting on a body the resulting strain will be sum of stress caused by each individual loads. A = area of cross section which is same throughout E = modulus of elasticity or Young’s Modulus L1 = length of part AB, L2 = length of part BC, L3 = length of part CD δL = δLAB + δLBC + δLCD δLAB = ( 𝑃𝐿 𝐴𝐸 ) 𝐴𝐵 ---------- (Tensile) δLBC = ( 𝑃𝐿 𝐴𝐸 ) 𝐵𝐶 ---------- (Compressive) δLCD = ( 𝑃𝐿 𝐴𝐸 ) 𝐶𝐷 ---------- (Compressive) Numerical: 1) A bar having cross sectional area 1000 mm2 is subjected to axial force as shown in Figure below. Find the total elongation of bar. Take E = 1.02 x 105 N/mm2 . 2) A steel rod ABCD 5 m long and 25 mm in diameter is subjected to forces as shown in Figure below. If value of Young’s modulus for steel material is 200 GPa. Determine its deformation. 3) Find the total elongation of the bar as shown in Figure below. Area of bar 600 mm2 and E = 200 MPa. 4) If a bar of uniform cross section of 50 mm x 50 mm is subjected to the forces as shown in Figure below. Determine total deformation of bar. Take E = 1.02 x 105 N/mm2 .
- 6. Strength of Materials - I Page 6 5) Calculate total elongation of steel bar ABCD for a given Figure having cross sectional area 750 mm2 and subjected to axial forces as shown in Figure below. Take E = 2.10 x 105 N/mm2 . 6) A steel bar ABCD of different section is subjected to an axial force as shown in Figure below. Find the value of P necessary for equilibrium. Take E = 210 kN/mm2 . Determine total elongation of bar. 7) A slender bar ABCD of different section is subjected to an axial force as shown in Figure below. Take E = 210 kN/mm2 . Determine total elongation of bar. 8) A mild steel bar is in three parts each 20 cm long and diameter of each part AB, BC and CD are 3 cm, 1.5 cm and 4.5 cm respectively. It is subjected to an axial pull of 4 ton then find its elongation in three parts of bar. Also find ratio of greatest to least elongation. Take E = 3 x 105 kN/mm2 .
- 7. Strength of Materials - I Page 7 Composite bar subjected to a load: Figure 1 Composite bar subjected to axial load (P) Where, As = cross sectional area of steel Ab = cross sectional area of brass σs = stress in steel σb = stress in brass l = length of member Es = Modulus of Elasticity of steel Eb = Modulus of Elasticity of brass Considering total load shared by steel and brass. P = Ps + Pb ------------- (1)
- 8. Strength of Materials - I Page 8 We know that, Stress = load per unit area to deformation. Load = stress x area ∴ P = σs As + σb Ab By Hook’s Law, Stress α strain σ = E e E = 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜎 𝑒 Strain in steel i.e. es = 𝜎𝑠 𝐸𝑠 and Strain in brass i.e. eb = 𝜎 𝑏 𝐸 𝑏 Brass and steel tubes are joined in such a manner that the system equally extends or contracts as one unit when subjected to tension or compression. Strain in steel = strain in brass es = eb 𝜎𝑠 𝐸𝑠 = 𝜎 𝑏 𝐸 𝑏 ------------- (2) From (1) and (2) we can solve any problem subjected to load carried by composite bar of section. Numerical: Type I: 1) A 250 mm long steel tube of 150 mm internal diameter and 15 mm thick is surrounded closely by brass tube of same length and thickness. The tube carries an axial thrust of 100 kN as shown in Figure 2 below. Estimate load vcarried by each tube and amount of sharing of each tube. Es = 200 x 103 N/ mm2 and Eb = 100 x 103 N/ mm2 .
- 9. Strength of Materials - I Page 9 Figure 2 2) A reinforced concrete column of square section 300 mm sides has four reinforcing bar of 32 mm diameter one in each corner with its centre 56 mm from the edges as shown in Figure 3. Find load carried by column if concrete can be stressed to 5 N/mm2 . What is corresponding stress in steel and reinforcement and what proportion of load is carried by it? Modular ratio of steel to concrete is 18. Figure 3 3) A mild steel bar 20 mm in diameter and 350 mm long is enclosed in a brass tube whose external diameter is 35 mm and internal diameter is 30 mm. The composite bar subjected to axial pull of 50 kN. Es = 200 GPa and Eb = 100 GPa. Find stress in bar and tube, as well as extension of bar. 4) A reinforced concrete column of 230 mm x 530 mm is reinforced with six bar of 20 mm diameter steel bars as shown in Figure below which is subjected to an axial load of 600 kN. Compute stress developed in each material. Consider Es = 200 GPa and Eb = 15 GPa. 5) A steel cylinder is enclosed in copper tube as shown in Figure below. The cylinder and tube are compressed between rigid parallel plates. Find stresses in steel and copper. P = 525 kN, d = 125 mm and D = 250 mm. Es = 200 x 103 N/ mm2 and Eb = 108 x 103 N/ mm2 .
- 10. Strength of Materials - I Page 10 Type II: 1) A steel rod of 32 mm diameter is placed in hollow Aluminum cylinder with internal diameter 40 mm and external diameter 50 mm as shown in Figure below. The steel rod projects 0.15 mm as shown. What maximum load can be apllied to the bearing plate? Es = 200 GPa and Eb = 120 GPa and σs = 165 MPa and σb = 85 MPa 2) A solid steel bar 600 mm long and 65 mm diameter is placed inside an Aluminum tube having 70 mm inside diameter and 90 mm outside diameter. The Aluminum cylinder is 0.2 mm longer than steel rod and axial load of 700 kN is applied to bar and the cylinder so rigid. Find stress developed in Aluminum bar and tube. Es = 210 kN/ mm2 and Eb = 70 kN/ mm2 . 3) Two steel rods and one copper rod each of 30 mm diameter together support a load of 30 kN as shown in Figure. Find stresses in each rod. Es = 205 GN/mm2 and Ec = 110 GN/mm2 . 4) Following Figure shows round steel rod supported and surrounded by co-axial copper tube. The upper end of rod is 0.2 mm below that of the tube and axial load is applied to a rigid plate resting on top of the tube. a. Determine the magnitude of maximum permissible load, if compressive stress in rod is not to exceed 120 MPa and that in the tube not to exceed 20 MPa. b. Find amount by which the tube will be shorten by load if compressive stress in tube is same as that in rod. Es = 210 GPa and Ec = 105 GPa. Compound bar subjected to change in temperature L1= Length of bar AB L2 = length of bar BC A1 = Cross sectional area of bar AB A2 = Cross sectional area of bar BC E1 = Young’s modulus of AB E2 = Young’s modulus of BC t = Temperature in 0 C α1 = Coefficient of linear expansion for AB α2 = Coefficient of linear expansion for BC Free expansion due to change in temperature in a bar can be given by, δL = L α t
- 11. Strength of Materials - I Page 11 Free expansion in bar AB = δL1 = L1 α1 t Free expansion in bar BC = δL2 = L2 α1 t Total free expansion = δL = δL1 + δL2 = L1 α1 t + L2 α1 t ------------- (1) If compound bar is subjected to axial force ‘P’, then each segment will be subjected to same force ‘P’. Here suppose P1 be the axial force in segment AB and P2 is the axial force in segment BC P1 = P2 = P If total expansion is prevented then bar is subjected to compressive forces and compressive force will be induced in bar. δL = 𝑃𝐿 𝐴𝐸 as part AB and BC are subjected to compressive force ‘P’ total change in length is given by δL = 𝑃1𝐿1 𝐴1𝐸1 + 𝑃2𝐿2 𝐴2𝐸2 ------------- (2) by equating 1 and 2 we can find force and stresses in bar. Numerical: 1) Find thermal stresses when temperature is raised by 200 C. Assume L1 = L2 = L, A1 = 2 A2. Take α = 11.7 x 10-6 per 0 C and E = 200 GPa. 2) A compound strut consists of brass portion AB of diameter 80 mm and steel portion BC of diameter 45 mm as shown in Figure below. Support A and C are rigid. If temperature is raised through 150 C. then find force exerted on support B. Take αs = 11.2 x 10-6 per 0 C and αb = 20 x 10-6 per 0 C. Es = 210 kN/mm2 and Eb = 85 kN/mm2. 3) A rod of total length 700 mm is fixed at its ends which is made up of two materials i.e. steel and brass which are rigidly connected to each other. The brass bar is 40 mm in diameter and 400 mm in length while the steel rod is 15 mm in diameter and 300 mm in length. If temperature of bar is raised through 120 0C. Find force exerted on the support. Take αs = 11.6 x 10-6 per 0 C and αb = 20.1 x 10-6 per 0 C. Es = 210 GPa and Eb = 81 GPa.
- 12. Strength of Materials - I Page 12 THERMAL STRESSES AND STRAINS A change in temperature of an object tends to produce change in its dimensions. It is a property of material to expand or contract as per change in temperature. If these natural tendencies are allowed then there is no strain and no stress is developed in the member. If deformation of body is prevented, some stresses are induced in body. Such stresses are called as thermal stresses and corresponding strain are called as thermal strain. Thermal strains: If temperature of an elastic body is changed and corresponding change in length is totally and partially then strain set up in member is called as thermal strain. Thermal Strain = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐿𝑒𝑛𝑔𝑡ℎ 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐿𝑒𝑛𝑔𝑡ℎ Thermal strains are usually reversible in nature since that member retains its original shape when it is brought to original temperature. Thermal stresses: If a body allowed to expand or contract freely with rise or fall in temperature then no stresses are induced in body. If these expansion or contraction is prevented then internal resisting forces are developed. Stresses caused by these internal processes are called as thermal stresses. Stress developed is in accordance with change in length prevented. According to law of linear expansion, the strain in prismatic bar is proportional to change in temperature. Mathematically, e α t e = α t -------------- (1) Where, α = coefficient of thermal expansion. We know, e = 𝛿𝐿 𝐿 = α t 𝛿𝐿 𝐿 = α t ∴ 𝛿𝐿= α t L -------------- (2) From Hook’s Law 𝜎 = 𝑒𝐸
- 13. Strength of Materials - I Page 13 Where, E = Young’s modulus 𝜎 = α t 𝐸 -------------- (3) If P = internal resistive force P = 𝜎 𝐴 P = α t 𝐸 𝐴 -------------- (4) Numerical: 1) A bar is 2.5 m long at 150 C. Find expansion of rod if temperature raised to 80 0 C. If these expansions are prevented find stresses in material. Take E = 100 GPa and 𝛼 = 0.0000120 C. 2) A copper rod 1.8 m long is at temperature 120 C. a. Find extension of rod when temperature raised to 980 C. Also find stress in rod. b. Find contraction of rod if temperature is lowered by 00 C. Also find stress in rod. Take E = 115 GPa and 𝛼 = 17.2 x 10-60 C. 3) A steel rod 20 mm in diameter, 300 mm long heated to 900C and at same time it is subjected to pull ‘P’ if total expansion of rod is 0.5 mm. What should be magnitude of force ‘P’? Take 𝛼 = 12 x 10-60 C and E = 210 GN/m2 . 4) A steel rod AB of diameter 30 mm and length 700 mm is held in between two supports at end A and B temperature of rod is raised uniformly from 260 C to 640 C. Assuming rod is to stress free at 260 C. Find a. Thermal stress and strain if supports do not yield. b. Thermal stress and strain if support yield by 0.21 mm and also find axial force in support. Take 𝛼 = 12.6 x 10-60 C and E = 200 GN/m2 . 5) Following Figure shows rod AB of length 700 mm. When a temperature of rod is 250C the gap of BC is 0.5 mm. Determine a. Thermal stress at 450 C. b. Temperature at which gap will just closed. c. Stress and strain in rod when t = 980 C Take 𝛼 = 16 x 10-60 C and E = 100 GPa.
- 14. Strength of Materials - I Page 14 Types of loaded bar: 1) Prismatic bar: Bar having uniform cross section throughout length is called as Prismatic bar. 2) Non-Prismatic bar: Bar having different cross section throughout length is called as Non- Prismatic bar. 3) Composite Bar: If cross section of bar made up of more than one material such bar is known as composite bar. 4) Compound Bar: Bar is having different material along its length is known as compound bar. Compound bar may be Prismatic or Non-Prismatic Concept of bar with different cross sections: Figure 1 Where, P = Tensile load or force acting on bar or rod L1 = length of part AB, L2 = length of part BC, L3 = length of part CD A1 = cross section of part AB, A2 = cross section of part BC, A3 = cross section of part CD E = Young’s modulus or Modulus of Elasticity I) Evaluation of stress: For part AB, σ1 = 𝑃 𝐴1
- 15. Strength of Materials - I Page 15 For part BC, σ2 = 𝑃 𝐴2 For part CD, σ3 = 𝑃 𝐴3 II) Evaluation of strain: For part AB, e1 = 𝜎1 𝐸 For part BC, e2 = 𝜎2 𝐸 For part CD, e3 = 𝜎3 𝐸 III) Evaluation of deformation: For part AB, δL1 = e1 x L1 For part BC, δL2 = e2 x L2 For part CD, δL3 = e3 x L3 Hence total deformation = δL = δL1 + δL2 + δL3 Numerical: 1) Following Figure 2 shows a bar consisting of three lengths. Find stresses in three parts and total extension if axial pull of 50 kN is applied. Take E = 2.1 x 105 N/mm2 . Figure 2 2) Find decrease in length of steel bar which is loaded as shown in Figure below. Take E = 2.1 x 105 N/mm2 .
- 16. Strength of Materials - I Page 16 Figure 3 3) A steel bar 600 mm long is 16 mm diameter for 200 mm of its length, 28 mm in diameter for 160 mm of its length and 20 mm in diameter for remaining 240 mm as shown in Figure 4 below. It is subjected to an axial pull of 30 kN. Calculate extension of bar. Take E = 2.1 x 105 N/mm2 . Figure 4
- 17. Strength of Materials - I Page 17 ELASTIC CONSTANTS Longitudinal or Linear strain: When a body subjected to axial load there is an axial deformation in the length of body. The ratio of axial deformation to original length of body is known as longitudinal or linear strain. Consider, P = Tensile load acting on body L = length of body δL =Deformation in length of body in direction of ‘P’ Linear strain = 𝛿𝐿 𝐿 Lateral strain: Strain at right angle to direction of applied load is known as lateral strain. Consider a rectangular bar of length L, breath ‘b’ and depth‘d’ is subjected to an axial force ‘P’ as shown in Figure below. Length of bar will increase while breath and depth will decrease. Consider, δL = increase in length δd = Decrease in depth Longitudinal strain = 𝛿𝐿 𝐿 Lateral strain = 𝛿𝑑 𝑑 Lateral strain = 𝛿𝑏 𝑏 Note: 1) Lateral strain compressive in nature then Lateral strain tensile in nature and vise a versa.
- 18. Strength of Materials - I Page 18 2) Every linear strain in direction of load is composed of lateral strain of opposite kind in all direction perpendicular to load. Poisson’s Ratio (μ) or( 𝟏 𝒎 ): When material is stressed within elastic limit ratio of lateral stain to longitudinal strain is constant which is known as Poisson’s Ratio. μ = lateral stain longitudinal strain Numerical: 1) Determine change in length, breadth and thickness of steel bar which is 4 m long, 30 mm wide, and 25 mm thick and subjected to an axial pull of 35 kN in direction of length. Take E = 2 x 105 N/mm2 and μ = 0.8. 2) Determine the value of Young’s modulus and poison’s ratio of metallic bar of length 40 cm, breadth 3.5 cm and depth 3.5 cm. When it is subjected to axial compressive load of 350 kN. Decrease in length is given as 0.075 cm and increase in breath is 0.003 cm. 3) A steel bar 2 m long, 20 mm wide and 10 mm thick is subjected to pull of 20 kN in direction of length. Find change in length, breath and thickness if E = 2 x 105 N/mm2 and μ = 0.3. Volumetric Strain: Change in length = L + δL Change in breadth = b + δb Change in depth = d + δd Original Volume = L x b x d Final Volume = (L + δL) (b + δb) (d + δd) δV = Original Volume – Final Volume δV = Lbd – [(L + δL) (b + δb) (d + δd)]
- 19. Strength of Materials - I Page 19 δV = (Lb δd + δL bd + δd Lb) δV = linear strain – 2μ linear strain δV = linear strain (1 -2μ) ev = 𝛿𝐿 𝐿 (1 -2μ) Q) Find the Volumetric Strain circular rod subjected to axial force ‘P’ along its length. Original Volume = 𝜋 4 L d2 Final Volume = 𝜋 4 L (d + δd) 2 - 𝜋 4 L d2 δV = Original Volume – Final Volume δV = 𝜋 4 L d2 – [ 𝜋 4 L (d + δd)2 − 𝜋 4 L d2 ] δV = linear strain – 2μ linear strain δV = linear strain (1 -2μ) ev = 𝛿𝐿 𝐿 (1 -2μ) ev = Strain in length – 2 (strain in diameter) Note:
- 20. Strength of Materials - I Page 20 Member subjected to loading in X and Y direction by means of passion’s ratio we can explain Hook’s Law of uniaxial loading to element subjected to state of biaxial loading. Strain in X direction Due to σx = 𝜎 𝑥 𝐸 (T) Due to σy = μ 𝜎 𝑦 𝐸 (C) Total strain in X direction ex = 𝜎 𝑥 𝐸 - μ 𝜎 𝑦 𝐸 ------------- (1) Strain in Y direction Due to σx = μ 𝜎 𝑥 𝐸 (C) Due to σy = 𝜎 𝑦 𝐸 (T) Total strain in Y direction ey = 𝜎 𝑦 𝐸 - μ 𝜎 𝑥 𝐸 ------------- (2) Element subjected to state of Tri – axial loading: Strain in x direction Due to σx = 𝜎 𝑥 𝐸 (T) Due to σy = μ 𝜎 𝑦 𝐸 (C) Due to σz = μ 𝜎 𝑧 𝐸 (C) Total strain in X direction ex = 𝜎 𝑥 𝐸 - μ 𝜎 𝑦 𝐸 - μ 𝜎 𝑧 𝐸 ------------- (1) Strain in Y direction Due to σx = μ 𝜎 𝑥 𝐸 (C) Due to σy = 𝜎 𝑦 𝐸 (T) Due to σz = μ 𝜎 𝑧 𝐸 (C)
- 21. Strength of Materials - I Page 21 Total strain in Y direction ey = -μ 𝜎 𝑥 𝐸 + 𝜎 𝑦 𝐸 - μ 𝜎 𝑧 𝐸 ------------- (2) Strain in Z direction Due to σx = μ 𝜎 𝑥 𝐸 (C) Due to σy = μ 𝜎 𝑦 𝐸 (C) Due to σz = 𝜎 𝑧 𝐸 (T) Total strain in Z direction ez = -μ 𝜎 𝑥 𝐸 - μ 𝜎 𝑦 𝐸 + 𝜎 𝑧 𝐸 ------------- (3) by adding 1, 2 and 3 we get (ex + ey + ez) = 𝜎 𝑥 𝐸 - μ 𝜎 𝑦 𝐸 - μ 𝜎 𝑧 𝐸 -μ 𝜎 𝑥 𝐸 + 𝜎 𝑦 𝐸 - μ 𝜎 𝑧 𝐸 -μ 𝜎 𝑥 𝐸 - μ 𝜎 𝑦 𝐸 + 𝜎 𝑧 𝐸 (ex + ey + ez) = 𝜎 𝑥 𝐸 + 𝜎 𝑦 𝐸 + 𝜎 𝑧 𝐸 (1 -2μ) ev = Linear strain (1 -2μ) Numerical: 1) A metallic bar 300 mm x 15 mm x 50 mm subjected to a force of 6 kN tensile, 8 kN tensile and 15 kN tensile along X, Y and Z respectively. Determine change in volume of block. Take E = 2 x 105 N/mm2 and μ = 1 4 . 2) A metallic bar 250 mm x 100 mm x 60 mm is loaded as shown in Figure below. Find change in volume of bar. Take E = 2 x 105 N/mm2 and μ =0.3. Also find the change that should be made in 3 MN load in order that there should be no change in volume of bar. 3) A steel rod 5 m long, 35 mm in diameter is subjected to an axial tensile load of 60 kN. Determine change in length, diameter and volume of rod. Take E = 2 x 105 N/mm2 and μ = 1 3 . Bulk modulus: When body is subjected to mutually perpendicular like and direct stresses the ratio of direct stress to the corresponding volumetric strain is found to be constant for given material. When the deformations are within the certain limit the ratio is known as bulk modulus and it is denoted as k. K = Direct stress volumetric strain
- 22. Strength of Materials - I Page 22 K = 𝜎 𝑒 𝑣 K = 𝜎 𝛿𝑉 𝑉 Relation between Bulk modulus and Modulus of Elasticity: Consider a cube ABCDEFGH subjected to three mutually perpendicular equal and direct stresses as shown in Figure below. Strain in AB = 𝛿𝐿 𝐿 Strain in AB due to σ on AEHD and BFGC = 𝜎 𝐸 Strain in AB due to σ on DHGH and ABFE = -μ 𝜎 𝐸 Strain in AB due to σ on ABCD and EFGH = -μ 𝜎 𝐸 Total strain in cube = 𝜎 𝐸 -μ 𝜎 𝐸 -μ 𝜎 𝐸 ∴ 𝛿𝐿 𝐿 = 𝜎 𝐸 -μ 𝜎 𝐸 -μ 𝜎 𝐸 𝛿𝐿 𝐿 = 𝜎 𝐸 (1 -2μ) V = L3 δL = change in length δV = change in volume
- 23. Strength of Materials - I Page 23 δV = 3L2 δL eV = 𝛿𝑉 𝑉 = 3L2 δL 𝐿3 = 3 𝛿𝐿 𝐿 = 3 𝜎 𝐸 (1 -2μ) k = 𝜎 𝑒 𝑣 k = 𝜎 3 𝜎 𝐸 (1 −2μ) ∴ E = 3 k (1 -2μ) and μ = 3k − E 6𝜇 Numerical: 1) Calculate bulk modulus if E = 1.4 x 105 N/mm2 and passion’s ratio 1 3 . 2) A bar of 40 mm diameter is subjected to a piull of 80 kN. The measured extension of gauge length of 400 mm is 0.2 mm and change in diameter 0.005 mm. Calculate a. Young’s modulus b. Passion’s ratio c. Bulk modulus Shear stress: Stress induced in a body when subjected to two equal and opposite forces whi9ch are acting tangentially across the resisting section is known as ‘Shear stress’ and is denoted by (τ) Τ = 𝑆ℎ𝑒𝑎𝑟 𝐹𝑜𝑟𝑐𝑒 𝑅𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝐴𝑟𝑒𝑎 Shear strain is nothing but distortion per unit length Φ = 𝐴𝐴′ 𝐴𝑑 Φ = 𝛿𝐿 ℎ Shear modulus: It is a ratio of shear stress to shear strain when the material is loaded within elastic limit is known as shear rigidity or shear modulus. It is denoted as C or G or N. G = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
- 24. Strength of Materials - I Page 24 G = 𝜏 𝜑 Principle of complementary shear stress: It states that, “a set of shear stress across a plane is always accompanied by set of balancing shear stresses i.e. of same intensity across the plane and normal to end”.
- 25. Strength of Materials - I Page 25 CHAPTER 2 COMBINED DIRECT AND BENDING STRESSES CONTENT: 1) Concept of direct, bending and combined stresses 2) Analysis of short column subjected to eccentric load 3) Concept of kernel or core of the section 4) Analysis of section subjected to eccentricity at both axes 5) Analysis of chimney subjected to wind pressure 6) Analysis of dams 7) Analysis of retaining walls Theory questions: Introduction: Column: 1) Short column 2) Long column Eccentric load is a load whose line of action does not coincide with axis of the section. The eccentricity of load may be about one of the axis or about both the axes. M = Pex MX = Pex MY = PeY CONCEPT OF DIRECT, BENDING AND COMBINED STRESSES Consider a column subjected to load ‘P’ at an eccentricity ‘e’ from centroidal axis. Hence ‘e’ is known as eccentricity of the load as shown in Figure A.
- 26. Strength of Materials - I Page 26 To understand the concept of Direct and Bending stresses, apply along the axis of the column two equal and opposite forces ‘P’. Due to application of these forces resultant effect is zero and there is no effect on column. With the help of principle of superposition we will split the forces as shown in Figure C and D. In figure C load ‘P’ is acting axially, therefore stresses induced is given by, σ = 𝑃 𝐴 which is known as ‘Direct Stress’. In figure D two equal and opposite forces are acting at a distance ‘e’ and will create couple and couple will induce bending stress in cross section. Hence an eccentric load will produce a direct as well as bending stress. As these stresses are normal to cross section, hence these stresses may be algebraically added into a single resultant stress as shown in Figure A. ANALYSIS OF SHORT COLUMN SUBJECTED TO ECCENTRIC LOAD 1) Direct stress = (σd) = 𝑃 𝐴 2) Bending stress = (σb) = The bending stress (σb) due to moment at any point of the column section at a distance ‘y’ from neutral axis Y – Y is given by, 𝑀 𝐼 = 𝜎 𝑏 ±𝑦 σb = ± 𝑀 𝐼 𝑦 σb = ± 𝑀 ( 𝐼 𝑦 ) σb = ± 𝑀 𝑧 3) Combined stresses or Resultant stresses: σR = σd + σb σR = 𝑃 𝐴 ± 𝑀 𝑧 σMAX = 𝑃 𝐴 + 𝑀 𝑧 and σMIN = 𝑃 𝐴 − 𝑀 𝑧
- 27. Strength of Materials - I Page 27 A= bd M = P e Z = 𝐼 𝑦 I = Iyy = 𝑑𝑏3 12 Ymax = 𝑦 2 = 𝑏 2 Z = 𝑑𝑏3 12 2 𝑏 = 𝑑𝑏2 6 σb = 𝑀 𝑧 = 6 𝑃𝑒 𝑑𝑏2 σb = 6 𝑃𝑒 𝑑𝑏2 σb = ± 6 𝑃𝑒 𝐴 𝑏 σR = 𝑃 𝐴 ± 6 𝑃𝑒 𝐴 𝑏 σR = 𝑃 𝐴 (1 ± 6 𝑒 𝑏 ) (The resultant stress along width of the column will vary by a straight line law) 𝜎MAX = 𝑃 𝐴 (1 + 6 𝑒 𝑏 ) --------------- (1) 𝜎MIN = 𝑃 𝐴 (1 − 6 𝑒 𝑏 ) --------------- (2) as shown in Figure in above Note: Compressive stresses are considered as positive (+ve) and tensile stresses are considered as negative (-ve) Numerical: 1) A rectangular section of width 200 mm and thickness or depth 175 mm carries a point load of 300 kN as shown in Figure below. Determine Max and Min stresses on section and draw stress distribution pattern. 2) In above problem if min stress on section is zero then find out min eccentricity of point load and find corresponding resultant stresses. 3) If load is acting at an eccentricity of 50 mm from centroidal axis of column, find out resultant stress and stress distribution pattern.
- 28. Strength of Materials - I Page 28 CONCEPT OF KERNEL OR CORE OF THE SECTION The cement concrete columns are weak in tension and strong in compression hence the load must be applied in this column in such a way that there should be no tensile stress induced in cross section. But when eccentric load is acting on column it will produce direct as well as bending stress. The resultant stress at any point is algebraic sum of these two stresses. 1) RECTANGULAR CROSS SECTION: Consider a solid rectangular cross section of width ‘b’ and thickness or depth ‘d’as shown in following Figure. Due to eccentric load stresses are induced in column. For no tensile stresses in column direct stress (σd) should be greater than or equal to bending stress (σb). σb ≤ σd 𝑀 𝑧 ≤ 𝑃 𝐴 𝑃 𝑒 𝑧 ≤ 𝑃 𝐴 e ≤ 𝑍 𝐴 Z = 𝐼 𝑦𝑦 𝑋 𝐼 𝑦𝑦 = 𝑑𝑏3 12 and x = 𝑏 2 Z = 𝑑𝑏3 12 𝑏 2 = 𝑑𝑏2 6 Z = 𝑑𝑏2 6 ∴ e ≤ 𝑍 𝐴 ≤ 𝑑 6 CORE OR KERNAL OF SECTION: The area within which the resultant load may be applies so as to avoid tensile stress is known as Core or Kernal of section.
- 29. Strength of Materials - I Page 29 2) HOLLOW RECTANGULAR CROSS SECTION: e ≤ 𝑍 𝐴 Area (A) = BD – bd z = 𝐼 𝑦 I = 𝐵𝐷3 12 − 𝑏𝑑3 12 y = 𝐷 2 z = 𝐵𝐷3− 𝑏𝑑3 6 𝐷 e ≤ 𝐵𝐷3− 𝑏𝑑3 6𝐷 (𝐵𝐷−𝑏𝑑) 3) CIRCULAR CROSS SECTION: σb ≤ σd
- 30. Strength of Materials - I Page 30 𝑀 𝑧 ≤ 𝑃 𝐴 𝑃 𝑒 𝑧 ≤ 𝑃 𝐴 e ≤ 𝑍 𝐴 z = 𝜋𝐷3 32 e ≤ 𝐷 8 4) HOLLOW CIRCULAR CROSS SECTION: ANALYSIS OF SECTION SUBJECTED TO ECCENTRICITY AT BOTH AXES Numerical: 1) A column 750 mm X 300 mm is subjected to an eccentric load of 120 kN as shown in figure below. Find what is the Max and Min intensities of stresses in column. 2) A masonry pipe 3 m X 4 m supports a vertical load of 8 kN as shown in Figure below: a. Find stresses developed at each corner. b. What additional load should be placed at centre so that there is no tensile stress in section. c. What is stress at centre with additional load at centre.
- 31. Strength of Materials - I Page 31 CHAPTER 3 SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR STATICALLY DETERMINATE BEAM Content: Introduction Types of Beams a. Definition of Shear force & Bending Moment b. Types of loading c. Types of support d. Statically determinate & indeterminate structure e. Shear force Diagram & Bending Moment Diagram Sign convention for shear force & Bending Moment Important points for drawing shear force & Bending Moment Diagrams S.F.D & B.M.D. for cantilever beams subjected to a) Point load b) UDL c) UYL + Point loads d) UVL e) Couple f) & Common findings SFD & BMD for simply supported Beams subjected to a) Point load b) UDL c) UVL d) Trapezoidal e) Couple f) Inclined point load g) & Common findings Relation between Load, S.F. & B.M.
- 32. Strength of Materials - I Page 32 Shear force: The algebraic sum of vertical forces at any section of a beam to the right or left of the section is known as shear force. Shear force diagram: A S.F.D. is one which shows the variation of the S.F. along the length of the beam. Bending moment: The algebraic sum of Moments of all the forces acting to the right or left of the section is known as bending Moment. Bending moment diagram: A B.M.D. is one which shows the variation of the B.M. along the length of the beam. Sign conversions for shear force & bending moment: 1) Shear Force: a) The shear force at a section will be considered positive when the resultant of the forces to the left to the section is upwards or to the right of the section is downwards. b) Similarly, the S.F. at a section will be considered negative if the resultant of the forces to the left of the section is downwards, or to the right of the section is upwards. 2) Bending Moment: a) The bending Moment will be considered positive when the Moment of the forces & Reactions on the left portion is clockwise & on the right portion is anticlockwise. b) The Bending Moment will be considered negative when the moment of the Forces & Reactions on the left portion is anti-clockwise & on the right portion clockwise. 3) Important Points for driving S.F.D. & B.M.D. a. Always consider left or right port of the section. b. Add Forces (including reaction) normal to the beam on one of the portion. Note:- a) If right port of the section is chosen a force on the right port acting downwards is positive. While a force on the right portion acting Upwards is negative. b) IF left port of section is chosen, a Force on the Left port acting upwards is positive while a Force on the left port acting downwards is Negative.
- 33. Strength of Materials - I Page 33 c) The positive values of S.F. & B.M. are plotted above the base line, & negative values below the base line. d) The S.F.D. Will increase or decrease suddenly i.e.by a vertical straight line at a section where there is a vertical point load. e) The S.F. between any two vertical loads will be constant & hence the S.F.D. Between two vertical loads will be horizontal. f) The bending Moment at the two supports of a simply supported beam & at the free end of a cantilever will be zero. Derivation: Consider a simply supported beam subjected to uniformly distributed load having cross sectional dimensions b and d. Consider two sections AB and CD at a distance ‘dx’ apart as shown in Figure below.
- 34. Strength of Materials - I Page 34 We know, f = 𝑀 𝐼 y Due to moment (M) on face AB, the stress induced is given by, 𝑓𝐴𝐵 = 𝑀 𝐼 y And due to Moment (M +dM) on face CD 𝑓𝐶𝐷 = (𝑀+𝑑𝑀) 𝐼 y Now, Force acting on face AB is given by, FAB = 𝑓𝐴𝐵A = ( 𝑀 𝐼 y)A Similarly force acting on face CD is given by, FCD = 𝑓𝐶𝐷A = ( (𝑀+𝑑𝑀) 𝐼 y)A Unbalanced force at a distance ‘y’ from N.A. is given by = FCD - FAB = ( 𝑑𝑀 𝐼 y)A Now, Total unbalanced force (F) = ∫ ( 𝑑𝑀 𝐼 y) A = 𝑑𝑀 𝐼 ∫(y)A
- 35. Strength of Materials - I Page 35 F = 𝑑𝑀 𝐼 𝐴𝑦̅ ------------------ (1) Now shear stress (q) = 𝐹 𝐴 F = q A F = q b (dx) ------------------ (2) Equating (1) and (2) q b (dx) = 𝑑𝑀 𝐼 𝐴𝑦̅ q = 𝑑𝑀 𝑑𝑥 𝐴𝑦̅ 𝐼𝑏 (Shear stress) q = 𝐹 𝐴𝑦̅ 𝐼𝑏
- 36. Strength of Materials - I Page 36 ASSIGNMENT 2 SHEAR FORCE AND BENDING MOMENT DIAGRAM CANTILEVER BEAMS: 1) Draw SFD and BMD for following cantilever beam as shown in Figure 1 below. Figure 1 2) Draw SFD and BMD for following cantilever beam as shown in Figure 2 below. Figure 2 3) Draw SFD and BMD for following cantilever beam as shown in Figure 3 below. SIMPLY SUPPORTED BEAMS: 4) Draw SFD and BMD for beam as shown below. 5) Draw SFD and BMD for beam as shown below.
- 37. Strength of Materials - I Page 37 6) Draw SFD and BMD for beam as shown below. 7) A beam of 20 m span, hinged at its both ends as shown in Figure below. Determine the reaction at the ends and Draw SFD and BMD. 8) Draw SFD and BMD for beam as shown below. 9) A beam AB of span 6m is simply supported its ends and carries an UDL of 2500 N/m over a length of 3 m starting from its left end. The beam is also subjected to a clockwise couple of 4000 Nm at the middle of its length. Find the max. Values of SF and BM and plot SFD and BMD. OVERHANG BVEAMS: 10) Draw SFD and BMD for beam as shown below.
- 38. Strength of Materials - I Page 38 11) Draw SFD and BMD for beam as shown below. 12) Draw SFD and BMD for beam as shown below. 13) Draw SFD and BMD for beam as shown below. 14) A 4 m long beam is shown in figure. It carries a load of 12 kN applied through a bracket and also a UDL for 8 m length from reight end. Beam over hanges at right end. Draw SFD and BMD. (12)
- 39. Strength of Materials - I Page 39 15) Sketch SFD and BMD for the beam as shown in Figure below. 16) Draw SFD and BMD for the beam shown in figure below giving all important values of max SF and max BM 17) Draw SFD and BMD for beam as shown below.
- 40. Strength of Materials - I Page 40 CHAPTER 4 BENDING STRESSES IN BEAMS Content: Bending stress Simple bending or pure bending Assumptions made in theory of pure bending Classic flexural equation Definition of section modulus Application of bending equation Moment of resistance THEORY QUESTIONS BENDING STRESSES: The stress induced in the beam to resist the bending moment is known as bending stresses. 1) Whenever a beam is subjected to external transverse loading, the beam will bend. As the beam bends resistance to the action sets up. 2) When every cross section is set up full resistance to bending moment acting on it the process of bending will stop. 3) If bending moment is applied the resistance induced internally is bending stress. 4) The external force is shear force, and then the internal resistance is shear stress. 5) The bending stress is in the form of tension or compressive stresses in the cross section. 6) These stresses are normal stresses. The portion of cross section where bending are tensile are known as tensile zone and where compressive are known as compressive zone. PURE BENDING: Consider a cantilever beam AB, fixed at A and free at B. At any cross section of beam, there is no shear force. Bending moment at any cross is constant and i.e. M. This is the case of pure bending.
- 41. Strength of Materials - I Page 41 ASSUMPTIONS MADE IN THEORY OF PURE BENDING 1) Material is homogenous and isotropic. 2) Beam is straight before bending. 3) Beam has uniform cross section throughout its length. 4) Transverse section of beam which is plane before bending remains plane after bending. 5) The material is elastic, Hook’s law is applicable. 6) Modulus of elasticity remains same in tension and in compression. 7) The effect of shear is totally neglected. Therefore the analysis is made for pure bending. 8) The beam is composed of infinite number of layers along its length. Each layer is free to expand or contract independently of layer above or below to it. 9) The beam is initially straight and all longitudinal fibers bends in circular arc having common centre of curvature. CLASSIC FLEXURAL EQUATION The flexural stresses will be maximum at extreme fibers i.e. maximum compressive stress will be at extreme fiber and maximum tensile stress will be at extreme tensile fiber. fc = 𝑀 𝐼 yc ft = 𝑀 𝐼 yt a) The cross section is subjected to sagging bending moment for finding moment of resistance. Moment of resistance is capacity of cross section to resist the bending moment. b) The moment of resistance cannot be changed for a cross section and is always same for the cross section. c) The applied moment should be less than the moment of resistance. d) If permissible stress in tension and compression are different, than the moment carrying capacity in tension and compression is to be found separately by considering respectively extreme fiber stresses and smallest one is accounted as moment capacity of cross section or moment of resistance of cross section.
- 42. Strength of Materials - I Page 42 SECTION MODULUS: Z = 𝐼 𝑦 𝑚𝑎𝑥 It is the ratio of moment of inertia about neutral axis to the distance of extreme fiber from neutral axis. Section modulus for different cross sections: 1) Rectangular cross sections: I = 𝐵𝐷3 12 ymax = 𝐷 2 z = 𝐼 ymax z = 𝐵𝐷3 12 𝐷 2 = 𝐵𝐷2 6 2) Hollow rectangular cross sections: I = 𝐵𝐷3 12 - 𝑏𝑑3 12 ymax = 𝐷 2 z = 𝐼 ymax z = 𝐵𝐷3 12 − 𝑏𝑑3 12 𝐷 2 3) Circular section: I = 𝜋𝑑4 64 ymax = 𝑑 2
- 43. Strength of Materials - I Page 43 z = 𝐼 ymax z = 𝐼 ymax z = 𝜋𝑑4 64 𝑑 2 = 𝜋𝑑3 32 4) Hollow circular section: I = 𝜋𝐷4 64 − 𝜋𝑑4 64 ymax = 𝐷 2 z = 𝐼 ymax z = 𝜋𝐷4 64 − 𝜋𝑑4 64 𝐷 2 = 𝜋 32 𝐷 (𝐷4 - 𝑑4 ) 5) Triangular section: I = 𝑏ℎ3 36 ymax = 2ℎ 3 z = 𝐼 ymax z = 𝑏ℎ3 36 2ℎ 3 = 𝑏ℎ2 24
- 44. Strength of Materials - I Page 44 Numerical: 1) A beam having E = 2.9 105 N/mm2 is bent with radius of curvature 35 m under the effect of bending moment of 6000 Nm. Calculate the moment of inertia of cross section. 2) A steel is bent into a circular arc of radius 10 m. If the plate section be 140 mm wide and 25 mm thick. Find the maximum stress induced and the bending moment which can produce this stress. E = 2 x 105 N/mm2 . 3) A cantilever beam is subjected to a force and a couple as shown in figure below. Determine maximum tensile and compressive stress developed in beam and also draw the bending stress distribution. 4) A cantilever beam of span 600 mm carries UDL of 6 kN/m from support to mid span. It also carries a point load of 10 kN at the tip of cantilever. If the beam section is 50 mm wide and 200 mm deep throughout the length. Determine maximum bending stress. Neglect the self weight of beam. 5) Two 50 mm x 150 mm rectangular timber section are glued together to form a T section as shown in Figure below. If a sagging bending moment of 6 kNm is applied to this beam at the horizontal axis. Find the stresses at the extreme fiber. 6) A cantilever beam of length ‘l’ and cross sectional area of side ‘a’ is subjected to transverse load of w per unit length. Find maximum bending stress in beam for the section as shown in Figure below. Find the maximum bending stress of beam if cross section is placed as shown in Figure B. Find relation between two stresses. 7) A timber beam of rectangular cross section of length 6 m is simply supported carries UDL of 9kNm and point load of 20 kN at 4 m from left. If the depth is 2 times the width of beam and bending stress is 6 N/mm2 find the suitable dimensions. 8) A rolled steel joist of I section has dimensions as shown in Figure below. This beam carries an UDL of 50 kNm run over a span of 12 m, calculate the maximum stress produced due to bending. 9) The cross section used in the figure is used as simply supported beam of span 5 m carrying UDL of 15 kN/m. Find the stresses at a section which is at a distance of 1.5 m from any support.
- 45. Strength of Materials - I Page 45 Numerical based on Moment of Resistance: 1) A grove in the form of triangle is cut symmetrically from a beam section as shown in figure below. If the stress in bending is not to exceed 30 MPa. Find the safe UDL which the beam can carry on a simply supported span of 4m. 2) Determine moment of resistance of a rectangular beam of cross section 600 mm x 300 mm size of timber. If permissible bending stress is 2 N/mm2 . 3) Determine moment of resistance of a T section having flange 75 mm x 15 mm and web 10 mm x 130 mm. Take bending stress 160 N/mm2 .
- 46. Strength of Materials - I Page 46 CHAPTER 5 SHEAR STRESSES IN BEAMS Content: Definition of Torsion and shaft Derivation of Torsion Formula Concept of Torsional Rigidity Power transmitted by shaft Numricals Sr. No. Section Area Centroidal Distance Ixx IYY 𝑋̅ 𝑦̅ 1 bd 𝑏 2 𝑑 2 𝑏𝑑3 12 𝑑𝑏3 12 Theory Questions: 1) Draw shear stress distribution diagrams for following sections and show important values: a. Rectangle section b. Triangular section c. Circular section d. I-Section e. T-section 2) Sketch the general shear stress profile for following Rectangle section and T section. (04) 3) Give the equation of shear stress with details of each term in it. (03) 4) The shear force acting on a beam at a section is F. The section of the beam is triangular having base B and of an altitude h. The beam is placed with its base horizontal. Find maximum shear stress and shear stress at NA. (14)
- 47. Strength of Materials - I Page 47 Sr. No. Cross section Shear stress Distribution qmax qavg qat y 1. Rectangular Section 2. Circular Section 3. Triangular Section
- 48. Strength of Materials - I Page 48 Shear stress (q): The stress produced due to shear force Figure1 q = 𝑭 𝑨 The stress in any given section at a distance ‘y’ from axis is given by q = 𝑭𝑨𝒚̅ 𝒃 𝑰 Where, F = shear force at the given section A𝑦̅ = moment of the area of the section about the level under consideration b = width of the beam at the level under consideration I = M.I. of the beam section about N. A. q = Intensity of shear stress at the level under consideration
- 49. Strength of Materials - I Page 49 Table 1: Sr No Cross Section Shear stress Distribution Diagram qmax qavg qy 1 2 3
- 50. Strength of Materials - I Page 50 SHEAR STRESS DISTRIBUTION FOR DIFFERENT CROSS SECTIONS: 1) RECTANGULAR SECTION: (Shear stress) q = 𝐹 𝐴𝑦̅ 𝐼𝑏 q = 𝐹 ( 𝑑 2 − 𝑦) 𝑏 1 2 (𝑦+ 𝑑 2 ) 𝐼𝑏 q = 𝐹 2 ( 𝑑 2 − 𝑦) (𝑦 + 𝑑 2 ) q = 𝐹 2I ( 𝑑2 4 − 𝑦2 ) At neutral axis (y = 0), q = 𝐹 2I ( 𝑑2 4 ) = 𝐹𝑑2 8I q = 𝐹𝑑2 8I = 12 𝐹𝑑2 8𝑏 𝑑3 = 1.5 𝐹 𝑏𝑑 q = 1.5 𝐹 𝑏𝑑 qmax = 1.5 qavg At top, (y = 𝑑 2 ), q = 𝐹 2I ( 𝑑2 4 − 𝑑2 4 ) = 0 Numerical: 1) A rectangular section 230 mm and 580 mm is subjected to shear force of 35 kN. Determine the average shear stress and Maximum shear stress. 2) The average shear stress (qavg) at a section of simply supported beam of 150 mm X 380 mm is 0.45 N/mm2 . Dteremine:
- 51. Strength of Materials - I Page 51 a. Shear stress in section. b. Maximum shear stress. c. Shear stress at a point on a section located at 35 mm above NA and 75 mm below NA 3) A wooden beam 100 mm wide and 150 mm deep is simply supported over a span of 4 m. If shear force at a section of beam is 4500 N, Find the shear stress at a distance of 25 mm above NA. 4) A rectangular section 100 mm wide and 250 mm deep is subjected to a Maximum shear force of 50 kN. a. average shear stress (qavg) b. Maximum shear stress. c. Shear stress at a point on a section located of 25 mm above NA (Dec 2010) (10) 5) A beam of square section is used as a beam with one diagonal horizontal. The beam is subjected to shear force (F) at a section. Find the maximum shear stress in the cross-section of the beam and draw the shear stress distribution diagram for the beam. 2) TRIANGULAR SECTION: The shear force acting on a beam at a section is F. The section of the beam is triangular having base B and of an altitude h. The beam is placed with its base horizontal. Find maximum shear stress and shear stress at NA. (14) (Shear stress) q = 𝐹 𝐴𝑦̅ 𝐼𝑏 A𝑦̅ = moment of the area of the section about the level under consideration A𝑦̅ = A (∆ 𝐶𝐸𝐹) x Distance of CG of triangle CEF about NA A𝑦̅ = 1 2 𝑏𝑥 ℎ x 2 3 (h-x)
- 52. Strength of Materials - I Page 52 A𝑦̅ = 1 3 𝑏𝑥2 ℎ (h-x) I = M I of the whole triangle section CAB about NA b = Actual width at the level EF b = 𝑥𝑏 ℎ q = F 1 3 𝑏𝑥2 ℎ (h−x) I 𝑥𝑏 ℎ q = 1 3 𝐹𝑥 (ℎ−𝑥) 𝐼 q = 𝐹 3𝐼 (xh – x2 ) q (x = 0) = 0 q (x = h) = 0 q (NA) (x = 2ℎ 3 ) = 𝐹 3𝐼 ( 2ℎ2 3 – 4ℎ2 9 ) = 𝐹 3𝐼 2ℎ2 9 = 2 27 𝐹ℎ2 𝐼 = 8 3 𝐹 𝑏ℎ MAXIMUM SHEAR STRESS: To get maximum shear stress, differentiating following equation with respect to ‘x’ ane equating to zero. q = 𝐹 3𝐼 (xh – x2 ) 𝑑 𝑑𝑥 ( 𝐹 3𝐼 (xh – x2 )) = 0 𝐹 3𝐼 (h – 2x) = 0 ∴ h = 2x ∴ x = ℎ 2 q = 𝐹 3𝐼 ( ℎ2 2 – ℎ2 4 ) q = 𝐹ℎ2 12 𝐼 qmax = 3𝐹 𝑏ℎ
- 53. Strength of Materials - I Page 53 Note: The shear stress is not maximum at the NA, but it is maximum at a depth of ℎ 2 from the top. In all other cases the shear stress will be maximum at the NA. Numerical: 1) A beam of cross section of an isosceles triangle is subjected to a shear force of 30 kN at a section where base width 150 mm and height 450 mm. Determine: a. Horizontal shear stress at NA b. The distance from the top of the beam where shear stress is maximum and c. Value of maximum shear stress. The distance from the top of the beam where shear stress is maximum
- 54. Strength of Materials - I Page 54 CHAPTER 6 COMPOSITE BEAMS (Flitched Beams / Bending stresses in composite Beams) Content: Introduction Flitched Beam Modular Ratio ( m ) Concept of Equivalent cross section Moment of Resistant to Flitched Beams Theory Questions: 1) Explain the term: Equivalent Section (03) 2) Define the term: Flitched Beam (02) 3) Explain the term: Moment of Resistance(03) Introduction: A section made up of two or more material joined together in such a way that they behave like a single piece & each material cross section bends to the same radius of curvature. Such beams are used when a beam of one material required is of large cross section which does not suite for the space available. Hence one material is reinforced with another material of higher strength in order to reduce the cross section of the beam. This is known as Composite Section. Definition: A beam made up of two or more different materials assumed to be rigidly connected together & behaving like a single piece is known as composite beam or wooden Flitched Beam.
- 55. Strength of Materials - I Page 55 The total moment of resistance of composite section will be equal to the sum of moment of Resistance of Individual cross sections. Other examples of Composite beams are shown below. Consider a composite beam as shown in fig above. Let at a distance y from N.A the stresses in steel & wood are 𝑓𝑠 & 𝑓𝑤 We know, 𝑒𝑠 = 𝑒 𝑤 (Note: for composite sections at common surface strain is same) 𝑓𝑠 𝐸𝑠 = 𝑓 𝑤 𝐸 𝑤 (e = 𝑓 𝐸 ) 𝑓𝑠= 𝐸𝑠 𝐸 𝑤 x 𝑓𝑤 𝑓𝑠=m x 𝑓𝑤 (m = 𝐸𝑠 𝐸 𝑤 =Modular ratio)
- 56. Strength of Materials - I Page 56 Using 𝑀 𝐼 = 𝑓 𝑦 M. R = 𝑓 𝑦 x I M.R = 𝑓 𝑦 x I 𝑀𝑠 = 𝑓𝑠 𝑦 x I 𝑠 & 𝑀 𝑤 = 𝑓 𝑤 𝑦 x I 𝑤 M.R = 𝑀𝑠 + 𝑀 𝑤 M.R = 𝑓𝑠 𝑦 x I 𝑠 + 𝑓 𝑤 𝑦 x I 𝑤 (of whole section) M.R = 𝑚𝑓 𝑤 𝑦 x I 𝑠 + 𝑓 𝑤 𝑦 x I 𝑤 M.R = 𝑓 𝑤 𝑦 (mI 𝑠 + I 𝑤 ) Where (mI 𝑠 + I 𝑤) = ( Equivalent Moment Of Inertia of the cross section ) 𝐼𝑒 = mI 𝑠 + I 𝑤 M = 𝑓 𝑤 𝑦 x 𝐼𝑒 (Equivalent section made up of wooden material) Note: 1) Equivalent section is produced by using I = mI 𝑠 + I 𝑤 2) The composite section can be converted into equivalent section by multiplying the dimensions of material of steel in the direction parallel to N.A. by m. Numerical: 1) A flitched beam consists of a wooden joist 150mm wide & 300 mm deep strengthened by a steel plate 12 mm thick &300 mm deep on either side of the joist is 7 N /mm2 ,Find the corresponding max stress attained in steel. Find the moment of resistance of the complete section. 𝐸𝑠 = 2 x 105 N/mm2 & 𝐸 𝑤 = 1 x 104 N/mm2
- 57. Strength of Materials - I Page 57 2) A timber beam 75 mm wide and 200 mm deep has 2 steel plates 8 mm thick and 160 mm deep symmetrically attached to it on either side to form composite beam section. If the maximum bending stress in timber is 8 MN/m2 find corresponding stress in steel and moment of resistance. Es = 20 Ew. 3) A flitched beam consists of a wooden joist 100 mm wide & 200 mm deep, strengthened by two steel plates 10 mm thick & 200 mm deep as shown in fig below. If the max stress in wooden joist is 7 N/mm2 . Find the corresponding Max stress attained in steel. Also find the moment of resistance of composite section. Es=2 x 105 N/mm2 and Ew = 1 x 104 N/mm2 .
- 58. Strength of Materials - I Page 58 CHAPTER 7 TORSION Content: - Definition of Torsion and shaft - Derivation of Torsion Formula - Concept of Torsional Rigidity - Power transmitted by shaft - Numricals Theory Questions: Definition of Torsion: A torsional moment may be defined as resultant moment about polar or longitudinal axis to the right or left of the section is known as torsional moment or twisting moment. When a tangential force is applied to a shaft at the circumference of its transverse cross section, the shaft is said to be subjected to twisting moment. When prism bar is subjected to equal and opposite couple acting on two parallel planes at right angles to the axis of bar, the cross section twists relative to each other and is said to be subjected to torsion. Definition of Shaft: A member subjected to torsion is known as shaft. A circular shaft is used for power transmission is an example of bar subjected to torsion. Assumptions made in Torsion Theory: 1) Circular cross section before twisting remains Circular after twisting. 2) The shaft is straight and uniform throughout the section. 3) The shaft material is homogeneous and isotropic. 4) Plane section remains plane after twisting. 5) Radius of cross section remains plane after twisting. 6) The twisting moment applied in planes which are perpendicular to the axis of shaft. 7) The torsional stresses are well below the proportional limit of the material. DERIVATION OF TORSIONAL FORMULA: Consider a shaft of length ‘L’ and radius ‘R’ with one end fixed and subjected to torque ‘T’ as shown in Figure
- 59. Strength of Materials - I Page 59 Consider the deformation of line AB on surface of shaft. The cross section is distorted by angle θ and line AB is distorted by angle φ. Shear strain = 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑑𝑖𝑠𝑡𝑜𝑟𝑡𝑖𝑜𝑛 𝐿𝑒𝑛𝑔ℎ𝑡 Φ = 𝐵𝐵′ 𝐿 = 𝑅θ 𝐿 ---------------- (1) We know the material obeys Hooks Law τ α φ τ = Gφ φ = τ 𝐺 ---------------- (2) From (1) and (2) 𝑅θ 𝐿 = τ 𝐺 τ = 𝑅θG 𝐿 𝜏 𝑅 = θG 𝐿 ---------------- (I) Consider point ‘C’ at a distance ‘r’ from the centriodal axis due to twisting moment, the point ‘C’ will be distorted by the same angle θ and we can write the same shear stress induce at point ‘C’.
- 60. Strength of Materials - I Page 60 φ = CC′ 𝐿 = 𝑟θ 𝐿 𝜏 𝑟 𝐺 = θr 𝐿 𝜏 𝑟= Gθr 𝐿 As the G, θ and L are constant G = Torsional Rigidity Therefore torsion is directly proportional to the distance of fibre from centroidal axis. Further it shoes that τ is zero at CA and maximum at extreme fibre of the shaft dF = Force on Elementary strip = Stress x Area
- 61. Strength of Materials - I Page 61 = 𝜏 𝑟 x δa Moment of force about polar axis dT = dF x r = 𝜏 𝑟 x δa x r = 𝜏 𝑚𝑎𝑥 𝑅 r x δa x r dT = 𝜏 𝑚𝑎𝑥 𝑅 δa x 𝑟2 T = ∫ 𝑑𝑇 = ∫ 𝜏 𝑚𝑎𝑥 𝑅 δa 𝑥 𝑟2 = 𝜏 𝑚𝑎𝑥 𝑅 ∫ δa 𝑥 𝑟2 ∫ δa 𝑥 𝑟2 = J ----- (where J is Polar MI) T = 𝜏 𝑚𝑎𝑥 𝑅 J 𝑇 𝐽 = 𝜏 𝑚𝑎𝑥 𝑅 Gθ 𝐿 = 𝜏 𝑚𝑎𝑥 𝑅 Gθ 𝐿 = 𝑇 𝐽 = 𝜏 𝑚𝑎𝑥 𝑅 ---------------- (II) Numerical: 1) A shaft running at 3 Hz has to transmit 120 kW. The shaft must not be stressed beyond 60 N/mm2 and must not twist more than 10 in length of 2 m. Select suitable diameter. Take G = 80 GPa. 2) A solid shaft 100 mm Dia. Transmit 100 kW at 80 rpm. Calculate max. shear stress and the angle of twist if G = 80 GPA and L 9 m. 3) A hollow shaft 250 mm external dia. And 160 mm internal dia. Rotates at 25 Hz. What power can be transmitted if thr permissible T = 80 N/mm2. Also calculate the angle of twist in 10 m length of shaft. Consider G = 80 GPa.
- 62. Strength of Materials - I Page 62 CHAPTER 9 (a) THIN CYLINDRICAL SHELLS Content: Introduction circumferential stress Longitudinal stress circumferential strain Longitudinal strain Volumetric strain Wire wound thin cylinders Thin cylinder when Thickness of shell ≤ 𝑑 20 , where ‘d’ is internal diameter INTRODUCTION: In day to day life we come across cylindrical tanks containing fluids like boiler, pipes, and steel pipes tec. The cylinder which is having metal thickness very small as compared to its diameter is known as thin cylindrical shell. For thin cylindrical shell, the thickness should be less than or equal to 1 20 of its internal diameter. Whenever the cylindrical shell is empty it is subjected t atmospheric pressure from inside as well as outside. Hence, the resultant pressure is zero. When the shell is subjected to internal pressure, its wall is subjected tensile stresses. The failure of thin cylindrical shells may be in two ways. It may split in two through or into two cylinders. When the thin cylindrical shell is subjected to internal pressure the walls are subjected to two types of stress i.e. 1) Circumferential (Hoop) stress 2) Longitudinal stress
- 63. Strength of Materials - I Page 63 1) Circumferential (Hoop) stress (fc): The stress which acts in tangential directional to the circumference of cylinder is known as fc. Consider a thin cylindrical shell subjected to an internal pressure ‘P’ having internal diameter ‘d’ and length ‘L’ and thickness ‘t’. Bursting Force = Pressure x area = P x dL -------- (1) Resisting Force = stress x area = fc x 2 x t x L -------- (2) For the pipe to be in equilibrium (1) = (2) fc = 𝐏𝐝 𝟐𝐭
- 64. Strength of Materials - I Page 64 2) Longitudinal stress (fc): Bursting Force = Pressure x area = P x d2 x 𝜋 4 -------- (1) Resisting Force = stress x area = ft x 𝜋 d t-------- (2) Circumferential (Hoop) stress (fc) equilibrium (1) = (2) fL = fc = 𝐏𝐝 𝟒𝐭 3) Circumferential strain (ec): ec = 𝑓𝑐 𝐸 – 𝜇 𝑓 𝐿 𝐸 ec = 𝑓𝑐 𝐸 – 𝜇 𝑓𝑐 𝐸 ec = 𝑓𝑐 𝐸 (1– 𝜇 ) 𝛿 𝑑 = ec x d Volumetric strain: V = 4 3 𝜋 r3 = 4 3 𝜋 ( 𝑑 2 ) 3
- 65. Strength of Materials - I Page 65 V = 𝜋𝑑3 6 𝛿𝑣 = 𝜋 3𝑑2 6 𝛿𝑑 ev = 𝛿𝑣 𝑣 = 3δl d = 3 ec ev = 3 𝑓𝑐 𝐸 (1– 𝜇 ) 𝛿 𝑑 = ec x d CYNDRICAL SHELLS Numerical: 1) A cylindrical shell 1 m long, 180 mm internal diameter, thickness of metal 80 mm is filled with fluid at atm pressure. If an additional 2000 mm3 of fluid is pumped into the cylinder, find the pressure exerted by the fluid on the wall of the cylinder and hoop stress in the section. Take E = 2 x 105 and μ = 0.3. 2) A cylindrical shell 1500 mm diameter, thickness of 12 mm, 4 m long is subjected to an internal pressure, 2 N/mm2 . Find the hoop stress and longitudinal stress induced in cell. Further determine change in diameter of shell. Take E = 2 x 105 and μ = 0.3. SPHERICAL SHELLS Bursting Force = Pressure x projected area = P x d2 x 𝜋 4 -------- (1) Resisting Force = stress x area = fc x 𝜋 d t-------- (2) For the pipe to be in equilibrium (1) = (2)
- 66. Strength of Materials - I Page 66 fc = Pd 4t ft = 1 2 fc Numerical: 1) A thin cylindrical shell 450 mm internal diameter, thickness 4 mm is full of an incompressible fluid at atm pressure. Find the intensity of radial ‘P’ exerted on wall of shell if 2 x 103 mm3 of fluid is pumped in calculate hoop stress and 𝛿𝑑.Take E = 2 x 105 and μ = 0.25.
- 67. Strength of Materials - I Page 67 CHAPTER 9 STRAIN ENERGY Content: Strain energy Strain energy due to: Axial Load 1) Gradually applied load 2) Suddenly applied load 3) Impact load Strain energy due to 1) Bending Moment 2) Shear force 3) Torsional moment Theory questions: 1) Define the terms : (03) a. Proof Resilliance b. Modulus of Resilliance: 2) What do you mean by Proof Resilliance? (03) 3) Obtain an expression for strain energy stored in a body when the load is applied with impact. (13) 4) Explain the terms: (04) a. Gradually applied load b. Suddenly applied load c. Impact load 5) Obtain an expression for strain energy stored in a body when the load is applied gradually. (04) 6) Give the expression for strain energy due to axial force. (03) 7) Derive the equation of strain energy due to bending. (03) Strain energy: Whenever a body is strained, the energy absorbed by the body due to straining effect is known as Strain energy When the member is loaded, whether gradually or Suddenly or Impact, the bat deforms and work is done. The material behaves like a perfect spring and oscillates about its mean position. If the elastic limit is not exceeded this work stored in the member is Strain energy.
- 68. Strength of Materials - I Page 68 Resilliance: The total Strain energy stored in the body is known as Resilliance. Whenever the applied force is removed from the strained body, the body is capable of doing the work hence Resilliance can be defined as the capacity of trained body for doing work on the removal of applied load. Proof Resilliance: It is defined as the maximum strain energy which can be stored by the body which can be stored by the body without undergoing permanent deformation. The energy stored in the body will be maximum when the body is stressed upto elastic limit. Modulus of Resilliance: It is defined as the proof Resilliance per unit volume. Strain Energy stored due to Axial Force: When a member is subjected to axial load, it undergoes axial deformation. Also the resistance is set up in the member gradually within the limit of proportionality the relation between resistance set up and deformation is always linear. Strain energy ‘u’ = Area of shaded diagram = 1 2 δl R = 1 2 e l σ A = 1 2 e σ A l = 1 2 e σ V Strain energy ‘u’ = 1 2 Stress x Strain x Volume = 1 2 σ 𝜎 𝐸 V
- 69. Strength of Materials - I Page 69 = 1 2 𝜎2 𝐸 V a) Gradually applied load: W = External Work done = 1 2 P δl SE = Internal work done = 1 2 R δl But W = SE ‘u’ 1 2 P δl = 1 2 R δl P δl = R δl P = σ A σ = 𝑃 𝐴 b) Suddenly applied load: W = External Work done = 1 2 P δl SE = Internal work done = 1 2 R δl But W = SE ‘u’ P δl = 1 2 R δl 2P = σ A σ = 2𝑃 𝐴 Stress in this case is twice the stress induced in the body by gradually applied load.
- 70. Strength of Materials - I Page 70 c) Impact Load: W = P (h + δl) u = 1 2 R δl u = 𝜎2 2𝐸 AL But W = SE ‘u’ P (h + δl) = 𝜎2 2𝐸 AL 𝜎2 = 2𝐸 𝐴𝐿 (h + δl) 𝜎2 = 2𝐸𝑃ℎ 𝐴𝐿 + 2𝐸𝑃 𝐴𝐿 𝜎 𝐴 𝜎2 - 2𝐸𝑃ℎ 𝐴𝐿 - 2𝐸𝑃 𝐴𝐿 𝜎 𝐴 = 0 By quadratic equation, σ = 2𝑃 2𝐴 ± √ 4𝑃2 4𝐴2 + 8 4 𝑃ℎ𝐸 𝐴𝐿 σ = 𝑃 𝐴 ± 𝑃 𝐴 √1 + 2𝐴ℎ𝐸 𝑃𝐿 σ = 𝑃 𝐴 ±[1 + √1 + 2𝐴ℎ𝐸 𝑃𝐿 ] Strain energy due to 1) Bending Moment u = ∫ 𝑀2 2𝐸𝐼 𝑑𝑥 𝐿 0 2) Shear force u = 3 20 𝑊2 𝐿 𝐺𝑏𝑑 3) Torsional moment a) Solid shaft: U = 𝜏2 4𝐺 volume b) Hollow shaft U = 𝜏2 4𝐺 [ 𝐷2+ 𝐷2 𝐷2 ]volume
- 71. Strength of Materials - I Page 71 Numerical: 8) A beam of cross section 100 mm x 200 mm and span 6 m is subjected to an UDL of intensity50 kN/m over the whole span. Find the strain energy stored due to bending if Modulus of Elasticity is 2 x 105 N/mm2 . (10) May 2014 9) A bar 100 cm in length is subjected to an axial pull, such that the maximum stress is equal to 150 MN/m2 . Its area of cross section is 2 cm2 over a length of 5 cm and for the middle 5 cm length it is only 1 cm2 . If E = 200 GN/m2 , Calculate the strain energy stored in bar. (10) 10) A bar 100 cm in length is subjected to an axial pull, such that the maximum stress is equal to 150 MN/m2 . Its area of cross section is 2 cm2 over a length of 5 cm and for the middle 5 cm length it is only 1 cm2. If E = 200 GN/m2 , Calculate the strain energy stored in bar. (10) 11) A Beam 4 m long is simply supported at the ends and carries an UDL of 6 kN/m length. Determine the SE stored in the beam. Take E = 200 GN/m2 and I = 1440 cm4 . 12) For a beam of span ‘L’ is simply supported at the ends and carries an UDL of ‘w’ per unit length. Determine the SE stored in the beam due to bending. (06) 13) A 300 mm long stepped bar ‘A’ has a diameter 20 mm for a length of 100 mm and dia of 40 mm for remaining length. Another bar ‘B’ made of same material gas dia of 30 mm throughout the entire length of 300 mm. If the permissible stresses for the material are same, compare the values of max SE stored in them. (10) 14) Calculate the resilience and resilience per unit volume of a bar 300 mm long, 50 mm wide and 40 mm thick if a load of 100 kN was gradually applied to it. Take E = 200 GPa. (07) 15) Find the strain energy stored in the beam as shown in Fig. below. 16) A beam 4 m long simply supported at end and carries a UDL of 6 N/m. Determine strain energy in beam. Take E = 200 GN/m2 and I = 1440 cm4 .
- 72. Strength of Materials - I Page 72 ASSUMPTIONS OF LINEAR ELASTICITY In order to evaluate the stresses, strains and displacements in an elasticity problem, one needs to derive a series of basic equations and boundary conditions. Therefore, some basic assumptions have to be made about the properties of the body considered to arrive at possible solutions. The following are the assumptions in classical elasticity. 1. The Body is Continuous Here the whole volume of the body is considered to be filled with continuous matter, without any void. Only under this assumption, can the physical quantities in the body, such as stresses, strains and displacements, be continuously distributed and thereby expressed by continuous functions of coordinates in space. However, these assumptions will not lead to significant errors so long as the dimensions of the body are very large in comparison with those of the particles and with the distances between neighbouring particles. 2. The Body is Perfectly Elastic The body is considered to wholly obey Hooke's law of elasticity, which shows the linear relations between the stress components and strain components. Under this assumption, the elastic constants will be independent of the magnitudes of stress and strain components. 3. The Body is Homogenous In this case, the elastic properties are the same throughout the body. Thus, the elastic constants will be independent of the location in the body. Under this assumption, one can analyse an elementary volume isolated from the body and then apply the results of analysis to the entire body. 4. The Body is Isotropic Here, the elastic properties in a body are the same in all directions. Hence, the elastic constants will be independent of the orientation of coordinate axes. 5. The Displacements and Strains are Small The displacement components of all points of the body during deformation are very small in comparison with its original dimensions and the strain components and the rotations of all line elements are much smaller than unity. Hence, when
- 73. Strength of Materials - I Page 73 formulating the equilibrium equations relevant to the deformed state, the lengths and angles of the body before deformation are used. In addition, when geometrical equations involving strains and displacements are formulated, the squares and products of the small quantities are neglected. Therefore, these two measures are necessary to linearize the algebraic and differential equations in elasticity for their easier solution.