Strength of Materials Notes PDF

Strength of material 1. Mechanical properties of material a) Elasticity: It is the property by virtue of which a material deformed under the load is enabled to return to its original dimension when load is removed. b) Homogeneity and isotropy: A material is hom...

Strength of material 1. Mechanical properties of material a) Elasticity: It is the property by virtue of which a material deformed under the load is enabled to return to its original dimension when load is removed. b) Homogeneity and isotropy: A material is homogeneous if it has same composition throughout the body. For such a material, the elastic properties are same at each and every point in the body. If the properties of material are not dependent on direction then it is isotropic. Anisotropic material has different property in different direction. c) Plasticity: Plasticity is the converse of elasticity. A material in plastic state is permanently deformed by the application of load, and it has no tendency to recover. The characteristic of the material by which it undergoes inelastic strain beyond those at the elastic limit is known as plasticity. d) Ductility Ductility is the ability of a material to be permanently deformed without breaking when a force is applied. There are two common measures of ductility. More ductile material can easily be drawn into wires. Ductile materials are weakest in shear. A material is called ductile if post elastic strain is more than 5%. ¨ Platinum is most ductile material and ¨ Gold is most malleable material. Malleability is the property to deform under compressive load into thin sheets. ¨ Lead is highly malleable but has low Axis of Ductile Brittle ductility. failure (mild steel) (cast iron) The ductile to brittle transition temperature is Tension test 45º from axis 90º from axis the temperature at which the failure mode of a Torsion test 90º from axis 45º from axis material changes from ductile to brittle fracture. e) Brittleness Brittleness implies lack of ductility. In a brittle material, failure takes place under load without significant deformation. Brittle materials fail suddenly without indication of yielding, post elastic strain is less than 5%. ¨ Tungsten is brittle at room temperature and ductile at higher temperature. Example: Grey cast iron, high carbon steel, concrete, stone, glass, ceramic materials. f) Toughness It is the property of a material which enables it to absorb energy without fracture. This property is very desirable in components subjected to cyclic or shock loading. It is measured in terms of energy required per unit volume of material to cause rapture under action of gradually increasing tensile load. ¨ Modulus of toughness is the entire area under stress strain diagram. The Modulus of Resilience, the area contained under the elastic portion of a stress-strain curve, is the elastic energy that a material absorbs during loading and subsequently releases when the load is removed. (Proof Resilience is under Load Deflection curve) The energy absorbed by a material prior to fracture is known as tensile toughness and is sometimes measured as the area under the true stress–strain curve (also known as the work of fracture) g) Hardness It is the ability of a material to resist indentation or surface abrasion. (scratch test and indentation test) 1  h) Stress strain diagram We can determine the nominal or engineering stress by dividing the applied load P by the specimen’s original cross- sectional area. This calculation assumes that the stress is constant over the cross section and throughout the gauge length The nominal or engineering strain is found directly from the strain gauge reading, or by dividing the change in the specimen’s gauge length, by the specimen’s original gauge length. i) Fatigue It is the lowering of strength or failure of a material due to repetitive stress which may be above or below the yield strength. It is a common phenomenon in load-bearing components in cars and airplanes, turbine blades, springs, crankshafts and other machinery, biomedical implants, and consumer products, such as shoes, that are subjected constantly to repetitive stresses in the form of tension, compression, bending, vibration, thermal expansion and contraction, or other stresses. These stresses are often below the yield strength of the material; however, when the stress occurs a sufficient number of times, it causes failure by fatigue. The endurance limit, which is the stress below which there is a 50% probability that failure by fatigue will never occur, is our preferred design criteria. Most materials are notch sensitive, with the fatigue properties particularly sensitive to flaws at the surface. ¨ Design or manufacturing defects concentrate stresses and reduce the endurance limit, fatigue strength, or fatigue life. ¨ Sometimes highly polished surfaces are prepared in order to minimize the likelihood of a fatigue failure. ¨ Shot peening is a process that is used very effectively to enhance fatigue life of materials. Small metal spheres are shot at the component. This leads to a residual compressive stress at the surface similar to tempering of inorganic glasses. j) Creep Time dependent permanent deformation under a constant load or constant stress and at high temperatures is known as creep. 2. Simple stress and strains a) Simple Stress When a body is acted upon by external force or load, internal resisting force is setup. Such a body is then said to be in a state of stress, where stress is the resistance offered by the body to deformation. Normal stress: force acting perpendicular to area per unit area. Shear stress: force acting tangential to area per unit area. ¨ 𝜏"# o 𝑖 represents direction of plane o 𝑗 represents direction of force ¨ Sign convention o +𝑣𝑒 force on +𝑣𝑒 plane is +𝑣𝑒 shear o −𝑣𝑒 force on −𝑣𝑒 plane is +𝑣𝑒 shear 2  ¨ 𝜏+, = 𝜏,+ b) Strain Normal strain ¨ We define the normal strain as the change in length of a line per unit length. It is denoted by 𝜖 ¨ Hence, when 𝜖 is positive the initial line will elongate, whereas if 𝜖 is negative the line contracts. Shear strain ¨ If we select two line segments that are originally perpendicular to one another, then the change in angle that occurs between these two line segments is referred to as shear strain. o This angle is denoted by 𝛾 (gamma) and is always measured in radians (rad), which are dimensionless. ¨ Sign convention 3 o Φ+, + 𝑣𝑒 if 𝜃 = 2 4 − Φ+, 5 < 909 3 o Φ+, − 𝑣𝑒 if 𝜃 = 2 − Φ+, 5 > 909 4 Φ;< = 𝛾+= + 𝛾=+ c) State of stress and strain 𝜎++ 𝜏+, 𝜏+= Stress tensor >𝜏,+ 𝜎,, 𝜏,= @ 𝜏=+ 𝜏=, 𝜎== Φ+, Φ+= ⎡ 𝜖++ ⎤ ⎢ 2 2 ⎥ Strain tensor ⎢Φ,+ 𝜖,, Φ,= ⎥ ⎢ 2 2 ⎥ ⎢ Φ=+ Φ=, ⎥ ⎣ 2 𝜖== ⎦ 2 d) Axial deformation 𝑃𝐿 𝛿= 𝐴𝐸 Elongation of rectangular bar due to self- 1 𝑊𝐿 𝛾𝐿4 𝛿= = weight 2 𝐴𝐸 2𝐸 1 𝑊𝐿 Elongation of conical bar due to self- 𝛿= 2 𝐴𝐸 weight Where, 𝐴 is area of base 4𝑃𝐿 𝑃𝐿 𝛿= = 𝜋 𝐸𝜋𝐷R 𝐷4 2 𝐷R 𝐷4 5 𝐸 4 Elongation of frustum due to load P Ø Solving this with average diameter will cause and error of 11.11% Ø And the deformation calculated will be is less than actual value. UV+ Deformation of bar of Uniform strength 𝐴+ = 𝐴𝑒 W 3. Elastic Constants a) Elastic constants Elastic constants are those factors which determine the deformations produced by a given stress system acting on a material. These factors are constant within the limits for which Hooke’s laws are obeyed. b) Longitudinal strain: Modulus of elasticity When an axial strain is applied along the longitudinal axis of a bar, the length of the bar will be increased. This change in the length per unit length of bar is termed as longitudinal strain. The 3  ratio of stress to strain, within elastic limits, is called the modulus of elasticity. (Young’s Modulus) 𝜎 𝐸= 𝜖 c) Lateral strain: Poisson’s ratio When axial force is applied along the longitudinal axis of a bar, the length of the bar will increase, but at the same time its lateral dimensions will be decreased. Thus, any direct stress produces a strain in its own direction and an opposite kind of strain in every direction at right angles to this. This ratio of the lateral strain is called the longitudinal strain is constant for a given material. This constant is known as Poisson’s ratio. 𝜖YZ[ 𝜇=− 𝜖Y9\V 0 ≤ 𝜇 ≤ 0.5 for cork to purely elastic rubber 1 1 ≤𝜇≤ for metals (generally, is Poisson’s ratio increases ductility increases) 4 3 −1.5 ≤ 𝜇 ≤ 0.5 rheotropic material (human tissue) Stress strain relations for triaxial loading 1 𝜖+ = [𝜎+ − 𝜇(𝜎, + 𝜎= )] 𝐸 1 𝜖, = d𝜎, − 𝜇(𝜎= + 𝜎+ )e 𝐸 1 𝜖= = [𝜎= − 𝜇(𝜎+ + 𝜎, )] 𝐸 d) Volumetric Strain Due to single direct stress ∆𝑉 = 𝜖h 𝑉 = 𝜖+ + 𝜖, + 𝜖= 𝜎+ + 𝜎, + 𝜎= = (1 − 2𝜇) 𝐸 ¨ It is Volumetric strain due to mutually perpendicular stress system If all stresses are equal (hydrostatic loading) 3𝜎 𝜖h = (1 − 2𝜇) 𝐸 ¨ For 𝜇 = 0.5 there is no change in volume, therefore fluid becomes incompressible. Volumetric strain in circular thin pipe 𝜖h = 𝜖i + 2𝜖j Volumetric strain in sphere 𝜖h = 3𝜖j Volumetric strain for uniaxial loading 𝜖h = 𝜖+ (1 − 2𝜇) e) Shear modulus or Modulus of Rigidity 𝜏 = 𝐺𝛾 Complimentary shear stress ¨ A shear stress in a given direction cannot exist without a balancing shear stress of equal intensity in a direction at right angles to it. State of simple shear (pure shear) 𝜎 = 𝜏 (tensile) 𝜎 = −𝜏 (compressive) Linear strain of diagonal due to shear 𝜙 𝜏 𝜖= = 2 2𝐺 Relation between E and G 4  𝐸 𝐺= 2(1 + 𝜇) f) Bulk Modulus When a body is subjected to three mutually perpendicular like stresses of equal intensity, (𝜎) the ratio of direct stress to corresponding volumetric strain is defined as the bulk modulus. 𝜎 𝐾= 𝜖h Relation between E and K 𝐸 𝐾= 3(1 − 2𝜇) g) Relation between E G K and 𝜇 9𝐺𝐾 3𝐾 − 2𝐺 𝐸= 𝜇= 𝐺 + 3𝐾 6𝐾 + 2𝐺 h) Independent elastic constants Homogeneous and isotropic = 2 Orthotropic = 9 Anisotropic = 21 4. Basic Analysis a) Axial deflection in composite member Two materials are connected in parallel having equal deformation 𝑃𝐿 𝐴R 𝐸R + 𝐴4 𝐸4 Δ= 𝐸|} = 𝐴R 𝐸R + 𝐴4 𝐸4 𝐴 ~9[ZY b) Temperature stresses ∆𝐿 = 𝐿𝛼∆𝑇 Δ𝐿 = 𝛼Δ𝑇 = 𝜖 ~ 𝐿 𝜎~ = 𝐸𝜖 ~ Example problems ¨ Cube restrained in xy plane but free to move in z direction ‚~ o 𝜎= Rƒ„ R†„ o ∆𝐿= = 𝐿𝛼𝑇 … ‡ Rƒ„ ¨ Cube restrained in all direction ‚~ o 𝜎 = Rƒ4„ ¨ Temperature variation in Beam (top bottom) iˆ o Upward deflection 𝛿 = ‰Š (from basic equation of cords in circle) ‹ R , o From Š = , ⇒ Š = , ⇒ 𝜖 = Š , o For temperature strain 𝛼Δ𝑇 = Š Combining two equations o i. Upward deflection due to temperature change in beam of length L and depth 𝐿4 𝜖 𝛿= 8𝑦 4𝛿𝑡 𝜖= 4 𝐿 c) Rod rotating about one end U‘ ˆ i’ 𝛿= “ 5  5. Principal stress a) Sign convention Tensile stress is considered positive and compressive stress is considered negative. Angle 𝜃 is considered positive when it is in the anticlockwise direction. A shear stress acting on a positive face of an element in positive direction and acting on negative face in negative direction is considered positive. ¨ If positive face negative direction and negative face positive direction is negative shear. b) Stress transformation equations 𝜎+ + 𝜎, 𝜎+ − 𝜎, 𝜎+ ” = +2 5 cos 2𝜃 + 𝜏+, sin 2𝜃 2 2 𝜎+ + 𝜎, 𝜎+ − 𝜎, 𝜎, ” = −2 5 cos 2𝜃 − 𝜏+, sin 2𝜃 2 2 𝜎+ − 𝜎, 𝜏+ ” , ” = −2 5 sin 2𝜃 + 𝜏+, cos 2𝜃 2 c) Principal stress 𝜎+ + 𝜎, 𝜎+ − 𝜎, 4 𝜎 = – — + ˜2 4 5 + 𝜏+, 2 2 𝜎+ + 𝜎, 𝜎+ − 𝜎, 4 𝜎 = – — − ˜2 4 5 + 𝜏+, 2 2 Angle of principal plane 2𝜏+, tan 2𝜃š = 𝜎+ − 𝜎, d) Maximum in plane shear stress 𝜎+ − 𝜎, 4 𝜏›Z+ = ˜2 4 5 + 𝜏+, 2 𝜎+ − 𝜎, Angle of maximum shear tan 2𝜃œ = − 2𝜏+, 6. Principal strain a) Strain transformation 𝜖+ + 𝜖, 𝜖+ − 𝜖, 𝛾+, 𝜖+ ” = +2 5 cos 2𝜃 + sin 2𝜃 2 2 2 𝜖+ + 𝜖, 𝜖+ − 𝜖, 𝛾+, 𝜖, ” = −2 5 cos 2𝜃 − sin 2𝜃 2 2 2 𝛾+ ” , ” 𝜖+ − 𝜖, 𝛾+, = −2 5 sin 2𝜃 + cos 2𝜃 2 2 2 b) Stress value from strain 𝐸 𝜎 = (𝜖 + 𝜇𝜖 ) 1 − 𝜇4 𝐸 𝜎 = (𝜇𝜖 + 𝜖 ) 1 − 𝜇4 6  Ø For plane stress condition µ 𝜖 = − (𝜖 + 𝜖 ) 1−µ c) Strain rosettes The stress in a bar subjected to uniaxial stress can be determined experimentally by attaching a strain gauge oriented in the direction of stress. When the material is strained, the length of the strain gauge wire also changes, resulting in the change of its resistance which can be measured on a wheat stone Bridge. Since each gauge measures the normal strain in only one direction, it is often necessary to use three strain gauges in combination. From three such measurements, it is possible to compute the strains in any direction on the surface. The group of three gauges arranged in a particular pattern is called a strain rosette. 45˚ strain rosette or rectangular strain rosette 𝜖9 + 𝜖¡¢ 𝜖9 − 𝜖¡¢ 𝜙+, 𝜖žŸ˚ = +2 5 cos 2 × 45˚ + sin 2 × 45˚ 2 2 2 ¨ Calculate 𝜙+, and then 𝜖 and 𝜖 7. Strain energy, Resilience and Impact Loading a) Strain energy It is defined as the energy absorbed by the bar during the loading process. It is equal to work done by the load, provided that no energy is added or subtracted in the form of heat. ∆ 𝑈 = 𝑊 = ¥ 𝑃𝜕∆ ¢ 1 = × stress × strain × volume 2 𝑃4 𝐿 = 2𝐴𝐸 Δ𝑈 1 𝑢= = 𝜎𝜖 ΔV 2 1 𝜎4 = 2𝐸 b) Modulus of Resilience The strain energy per unit volume of the material is known as resilience. Resilience is also known as strain energy density. It represents the ability of the material to absorb energy within elastic limit. When the stress 𝜎 is equal to proof stress (𝑓) at the elastic limit, the corresponding resilience is known as Proof Resilience. The proof resilience is also known as Modulus of Resilience. 1 𝑓4 𝑢« = 2𝐸 c) Modulus of Toughness This quantity represents the entire area under the stress–strain diagram, and therefore it indicates the strain-energy density of the material just before it fractures. d) Strain energy of various sections 𝑃4 𝐿 𝑃4 𝐿R 𝐿4 Bar with varying section 𝑈=¬ = – + + ⋯— 2𝐴𝐸 2𝐸 𝐴R 𝐴4 i Non-prismatic bar with 𝑃+4 𝑈=¥ 𝑑𝑥 varying axial force ¢ 2𝐸𝐴+ 𝛾 4 𝐴𝐿“ Hanging bar under own weight 𝑈= 6𝐸 7  𝛾 4 𝐴𝐿“ 𝑃4 𝐿 𝛾𝑃𝐿4 𝑈= + + 6𝐸 2𝐴𝐸 2𝐸 Freely hanging bar with axial This shows that the strain energy of an elastic body due to more load (P) than one load cannot be found merely by adding the strain energies obtained from individual loads. e) Sudden loading If the load P is applied suddenly, the value of load is P throughout the deformation. Therefore, the work done on the bar = 𝑃∆ However, the resistance setup in the body is zero when the deformation is zero, and is equal to R when the deformation is ∆. Equating the two 1 𝑃∆= 𝑅∆ 2 1 𝑃∆= 𝜎𝐴∆ 2 2𝑃 𝜎= 𝐴 Thus, maximum stress set up in the body is equal to twice that in the case of gradual loading. f) Impact loading Work done by falling weight P from a point h 𝜎 4 𝐴𝐿 𝑃(ℎ + ∆) = 2𝐸 𝜎𝐿 𝜎 4 𝐴𝐿 𝑃 –ℎ + — = 𝐸 2𝐸 𝐴𝐿 𝑃𝐿 𝜎 4 – — − 𝜎 – — − 𝑃ℎ = 0 2𝐸 𝐸 Finding the root of the equation we get 𝑃 2𝐴𝐸ℎ 𝜎"›šZ²[ = ³1 + ˜1 + ´ 𝐴 𝑃𝐿 2ℎ 𝜎"› = 𝜎œ[Z[ ³1 + ˜1 + ´ ∆œ[Z[ Deformation ∆ ∆= ∆œ[ + µ∆4œ[Z[ + 2ℎ∆œ[Z[ When ℎ → 0 ¨ 𝜎" ≥ 2𝜎œ ¨ ∆" ≥ 2∆œ g) Strain energy in terms of principal stresses Two-dimensional case 𝑢 = 𝑢R + 𝑢4 1 𝑢 = 𝜎R 𝜖R + ⋯ 2 1 𝑢= 𝜎 (𝜎 − 𝜇𝜎4 ) + ⋯ 2𝐸 R R 1 𝑢= (𝜎 4 + 𝜎44 − 2𝜇𝜎R 𝜎4 ) 2𝐸 R Three-dimension 1 𝑢= 2𝜎 4 + 𝜎44 + 𝜎“4 − 2𝜇(𝜎R 𝜎4 + 𝜎4 𝜎“ + 𝜎“ 𝜎R )5 2𝐸 R Hydrostatic pressure 8  𝑝4 𝑈= 𝑉 2𝐾 8. Theories of failure a) Maximum Principal stress theory: RANKINE’S THEORY This is the simplest and oldest theory of failure. 𝜎R ≤ 𝜎, Limitations ¨ This theory disregards the effect of other principal stresses and effect of shearing stresses. ¨ When the element is in pure shear (𝜎R = 𝜎“ = 𝜏›Z+ ), this theory predicts that failure will occur when the magnitude of 𝜏›Z+ reaches a value 𝑓, but experiments show that failure occurs much earlier, when 𝜏›Z+ reaches 𝟎. 𝟓𝟕𝒇𝒚. ¨ Material in tension test piece slips along 45˚ to the axis of the test piece, where normal stress is neither maximum nor minimum, but shear stress is maximum. ¨ It has been experimentally shown that even a material weak in compression sustains very high hydrostatic pressure without fracture. For brittle materials which do not fail by brittle fracture, this theory is considered to be reasonably satisfactory. b) Maximum principal strain theory: SAINT VENANT’S THEORY This theory is satisfactory for brittle materials but overestimates the elastic strength of ductile materials. 1 𝜎, [𝜎R − 𝜇(𝜎4 + 𝜎“ )] ≤ 𝐸 𝐸 Limitations ¾ W ¨ When 𝜎R = 𝜎4 and 𝜎“ = 0, the failure value of 𝜎R = Rƒ„ , i.e. 𝜎R > 𝑓,. o Theory predicts that tensile stresses will be higher than yield stress in uniaxial- tension, which contradicts experimental results. ¨ If the material is subjected to Hydrostatic pressure p, this theory predicts the failure value of W ¾ 𝑝 = Rƒ4„. Value predicted is much lower than experimental value. W¾ ¨ In case of pure shear, 𝜇 = 0.3 it predicts failure at 𝜏›Z+ = = 𝟎. 𝟕𝟕𝒇𝒚 which is better R.“ than principle stress theory but still maintains a good gap with experimental value. c) Maximum Shear Stress theory Guest’s theory (Coulombs theory), TRESCA’S THEORY Ductile materials yield along 45˚ slip plane which are planes of maximum shear stress, hence this theory is fairly well justified for ductile materials. It is also considered most conservative theory. 1 𝑓, 𝜏›Z+ = (𝜎R − 𝜎“ ) ≤ 2 2 Limitations ¨ This theory does not give accurate results for the state of stress of pure shear in which the maximum amount of shear is developed (torsion test). 𝜏›Z+ = 𝟎. 𝟓𝝈𝒚 whereas the results give 0.57 value which is about 15% inaccurate. ¨ This theory is not applicable to Hydrostatic Pressure, since it will predict the shear stress to be almost zero, meaning thereby the material will never fail, which is physically impossible. ¨ The results differ from experimental results for materials for which elastic limit stresses in tension and compression differ by large amount. This is so for brittle materials like cast iron. d) Maximum strain energy theory: HAIGH’S THEORY 9  According to this theory, a body under complex stresses fails when the total strain energy on the body is equal to the strain energy at elastic limit in simple tension. Strain energy stored per unit volume 1 𝑢= {𝜎 4 + 𝜎44 + 𝜎“4 − 2𝜇(𝜎R 𝜎4 + 𝜎4 𝜎“ + 𝜎“ 𝜎R )} 2𝐸 R 4 𝑓, 𝜎R4 + 𝜎44 + 𝜎“4 − 2𝜇(𝜎R 𝜎4 + 𝜎4 𝜎“ + 𝜎“ 𝜎R ) = Â Æ 𝐹𝑂𝑆 This theory has considerable experimental support from tests on ductile materials, especially thick cylinders. Limitations ¨ Theory does not apply to brittle materials for which elastic limit stresses in tension and in compression are quite different. ¨ It cannot explain hydrostatic loading; this theory has largely been replaced by the maximum-distortion energy theory. e) Maximum shear strain energy theory (Distortion energy) VON MISES HENKY THEORY This theory differs from the maximum strain energy theory in that the portion of the strain energy, producing volume change, is considered ineffective in causing failure by yielding. ¨ The portion of strain energy producing change in shape of the element is assumed completely responsible for the failure of material by yielding. The energy of distortion can be obtained by subtracting the energy of volume change from the total energy. Shear strain energy 1+µ 𝑢œ = [(𝜎R − 𝜎4 )4 + (𝜎4 − 𝜎“ )4 + (𝜎“ − 𝜎R )4 ] 6𝐸 4 1 𝑓, [(𝜎R − 𝜎4 )4 + (𝜎4 − 𝜎“ )4 + (𝜎“ − 𝜎R )4 ] ≥ Â Æ 2 𝐹𝑂𝑆 ¨ For plane stress when 𝜎“ = 0 4 𝑓, 𝜎R4 − 𝜎R 𝜎4 + 𝜎44 ≥Â Æ 𝐹𝑂𝑆 This theory is in good agreement with experimental results with various combination of principle stresses for ductile materials. ¨ For example: in pure shear, the value of maximum shear stress at yielding given by this theory is 0.577𝑓, which coincides with the average experimental value. Limitations ¨ The theory does not apply to brittle materials for which the elastic limit stress in tension and compression is quite different. ¨ Does not holds good for hydrostatic pressure. f) Octahedral shear stress theory This theory is equivalent to the maximum distortion energy theory. However, the distortion energy theory is based on energy concept while octahedral shear stress theory is based on stress conception and hence is preferred by engineers. 1 R √2 [(𝜎R − 𝜎4 )4 + (𝜎4 − 𝜎“ )4 + (𝜎“ − 𝜎R )4 ]4 ≥ 𝑓 3 3 , 9. Centroid and MI of Plane area a) Centroid The centre of gravity of a body is that point through which the resultant of system of parallel forces formed by the weights of all the particles of the body passes, for all positions of the body. 10  The plane areas, like triangle, rectangle, quadrilateral, circle etc., have only areas and no mass. The centre of area of such figures is known as centroid. ∫ 𝑥𝑑𝐴 𝑥̅ = ∫ 𝑑𝐴 Centroid of composite area 𝑥R 𝐴R + 𝑥4 𝐴4 + ⋯ 𝑥̅ = 𝐴R + 𝐴4 + ⋯ b) Moment of Inertia The axial (or equatorial) moment of inertia of a plane area (or of a cross-section) is the geometrical characteristic of the area. It is defined by integrals: 𝐼++ = ¥ 𝑦 4 ∙ 𝑑𝐴 𝐼,, = ¥ 𝑥 4 ∙ 𝑑𝐴 ¨ Since the area is multiplied by the square of the distance, the moment of inertia is also called second moment of area. Parallel axis theorem 𝐼++ = 𝐼²² + 𝐴ℎ4 ¨ ℎ - is distance between xx and cc axis Polar Moment of inertia ¨ The moment of inertia about an axis perpendicular to the plane and passing through the intersection of the other two axes x-x and y-y contained by the plane is equal to the sum of moments of inertia about x-x and y-y 𝐼š (𝐼== ) = 𝐼++ + 𝐼,, Moment of inertia of composite section 𝐼ÍÍ = ¥ 𝑦R4 ∙ 𝑑𝐴R + ¥ 𝑦44 ∙ 𝑑𝐴4 + ¥ 𝑦“4 ∙ 𝑑𝐴“ Radius of gyration ¨ Radius of gyration of an area can be considered to be a distance from the axis at which the entire area could be concentrated and still have the same moment of inertia. 𝐼++ 𝑟+ = ˜ 𝐴 1 1 𝜋𝑑4 𝐴 = 𝑏ℎ 𝐴= 𝑏ℎ (𝑎 + 𝑏)ℎ 2 2 8 𝑏 𝑎+𝑏 2𝑎 + 𝑏 ℎ 4𝑟 𝑥̅ = 𝑥̅ = 𝑦Ñ = × 𝑦Ñ = 2 2 𝑎+𝑏 3 3𝜋 𝑏ℎ“ 𝑏ℎ“ 𝑏ℎ“ 𝑏ℎ“ 𝜋𝑑ž 𝐼+ = ,𝐼 = ” 𝐼+ = ,𝐼 =” 𝐼+ = ” 12 + 3 36 + 12 128 11  𝜋𝑑4 2𝑏ℎ 𝑏ℎ 4 3 3 𝜋𝑑 ž 3𝑏 3𝑏 𝐼+ = 𝑥̅ = 𝑥̅ = 64 8 4 𝜋𝑑ž 𝑛+1 𝑏 𝑛+1 𝐼= = =– — =– —𝑏 32 𝑛+2 2 𝑛+2 c) Product of inertia The product of inertia of a plane is defined by the following integral. 𝐼+, = ∫ 𝑥𝑦𝑑𝐴 In the above expression, each elementary area 𝑑𝐴 is multiplied by the product of the coordinates of the area; hence product of inertia may be positive, negative or zero. If one axis is the axis of symmetry, the product of inertia of the area with respect to the pair of axes is zero. Parallel axis theorem 𝐼+, = 𝐼ÔÕ + 𝐴𝑑+ 𝑑, d) Moment of inertia about Inclined axes In mechanical or structural design, it is sometimes necessary to calculate the moments and product of inertia 𝐼+ ” 𝐼, ” and 𝐼+ ” , ” for an area with respect to a set of inclined 𝑥 Ö and 𝑦′ axes when the values for 𝜃, 𝐼+ , 𝐼, and 𝐼+, are known. 𝐼+ + 𝐼, 𝐼+ − 𝐼, 𝐼+ ” = +– — cos 2𝜃 − 𝐼+, sin 2𝜃 2 2 𝐼+ + 𝐼, 𝐼+ − 𝐼, 𝐼, ” = −– — cos 2𝜃 − 𝐼+, sin 2𝜃 2 2 𝐼+ − 𝐼, 𝐼+ ” , ” = − – — sin 2𝜃 + 𝐼+, cos 2𝜃 2 Just like stress transformation we can use Mohr’s circle to find the maximum and minimum moment of inertia. 10. Shear Force and Bending Moment a) Definitions Shear force is algebraic sum of all the transverse forces either moving left to right or right to left and it always act along the plane of cross-section. Bending moment at any section of beam is algebraic summation of moment given by transverse loading and concentrated moment either moving left to right or right to left. Sign convention: ¨ A clockwise shear is taken as positive and an anticlockwise shear is taken as negative. ¨ A bending moment causing concavity upwards is taken as positive and is called sagging BM. Similarly, a BM causing convexity upwards is taken as negative and is called hogging BM. b) Constructing SF and BM diagram from Load Δ𝑉 = ∫ 𝑤(𝑥)𝑑𝑥 12  𝑑𝑉 = −𝑤(𝑥) 𝑑𝑥 ¨ Thus, the rate of change of shear force (slope of SFD) is equal to −𝑤 Δ𝑀 = ∫ 𝑉𝑑𝑥 𝑑𝑀 =𝑉 𝑑𝑥 ¨ Thus, the rate of change of bending moment 𝑀 along the length of the beam is equal to the shear force. ÚÛ Form BM to be maximum = 0. Ú+ ÚÛ But = 𝑉 hence BM is maximum where SF is zero or Ú+ changes sign. c) Loading diagram and SF diagram from BM diagram Properties of SF diagram ¨ The SF diagram consists of a rectangle if the beam is loaded with point load. ¨ The SF diagram consists of inclined line for the portion on which UDL acts. ¨ The SF diagram consists of parabolic curve for portion over which triangular or trapezoidal load distribution act ¨ Point where sign of SFD changes, there BM will be maximum or minimum. Properties of BM diagram ¨ The BM diagram consists of inclined lines for the beam loaded with point loads. ¨ The BM diagram consists of parabolic curve for the portion on which UDL acts. ¨ The BM diagram consists of cubic curve for portion over which triangular or trapezoidal load distribution act ¨ Point where sign of BM changes in point of contra-flexure (slope of beam changes its direction, also called inflection point) d) Examples: 13  11. Bending Stress a) Introduction Locus of all cross-sectional point where bending is zero is Neutral Axis. In elastic analysis neutral axis is centroidal axis, which is center of area. (in plastic analysis, NA is equal area axis) Symmetrical bending: when plane of bending and plane of couple coincides. Unsymmetrical bending: when the plane of bending and plane of couple do not coincide. Assumptions ¨ Material is homogeneous and isotropic; elastic and hook’s law are valid. ¨ E in tension and compression is same. ¨ Plane section remains plane after bending. (shear force cause warping, which is neglected) ¨ Before and after bending cross-section remains vertical. b) Simple Bending The strain of a fiber is proportional to its distance from the neutral axis. 𝑦 𝜖= 𝑅 𝑀 𝜎 𝐸 = = 𝐼 𝑦 𝑅 R Û The curvature Š = of the section is directly proportional to the bending moment and is inversely proportional to the flexural rigidity (𝐸𝐼) of the section. c) Design criterion The beam section should section should have sufficient strength to resist the external bending moment. To ensure strength of a beam, it is necessary that the greatest tensile and greatest compressive stress at the critical section should not exceed their corresponding allowable stresses for the material. 14  𝑍= is known as section modulus, which represent the ,ÝÞß strength of the section. Moment of Resistance: it is the maximum moment that a section can resist without any point in its cross-section having stress greater than permissible bending stress. If area of I section, Circular section, Rectangular section is same, then 𝑍 > 𝑍«|²[ > 𝑍²"« ¨ More the area concentrated away from the NA better is the section. Ratio of section modulus of square when the side is kept horizontal (𝑍R ) to that of horizontal diagonal (𝑍4 ) 𝑍R = √2 𝑍4 ¨ Moment capacity decreases by 29.27% Width and depth of strongest beam cut out from circular log of wood of diameter D. 𝐷 𝑏= = 0.577𝐷 √3 2 𝑑 = ˜ 𝐷 = 0.814𝐷 3 d) Flitched Beam Beams that are made of more than one material are called composite beams. They can be analyzed by using plane section remains plain after bending. Thus, the strain distribution along the depth remains linear. However, such structures are statically indeterminate, and the position of neutral axis is not centroid of the section. The criterion of strain compatibility has to be used, i.e. strain in the two materials, at a given vertical distance from the NA, has to be the same. If both the materials are rigidly joined together, they will behave like a unit piece and the bending will take place about the combined axis. On the other hand, if both the materials have been simply placed one above the other, they will bend about their respective geometric axes. Symmetrical section ¨ Strain are same at a level 𝜖R = 𝜖4 𝜎R 𝜎4 = 𝐸R 𝐸4 𝐸4 𝜎4 = 𝜎R 𝐸R 𝜎4 = 𝑚𝜎R o 𝑚 = ˆ is called modular ratio á ¨ If it is required to find stresses induced in the section corresponding to a given bending moment, Radius of curvature is same 𝑅R = 𝑅4 𝐸R 𝐼R 𝐸4 𝐼4 = 𝑀R 𝑀4 𝑀R 𝐸R 𝐼R = 𝑀4 𝐸4 𝐼4 ¨ Moment of resistance 15  o 𝑀 = 𝜎R 𝑍R + 𝜎4 𝑍4 o Find the governing stress, i.e. the stress value which is reached first and the corresponding stress in the other material. Unsymmetrical section ¨ The first step is to find the position of new axis of bending. This is best done by drawing the equivalent section. ¨ Find Neutral axis of transformed section. ¨ Find the MI for the transformed section about cg. ¨ Find the governing stress, which is not being exceeded above the allowable stress. ¨ Then find moment of resistance. e) Strain energy due to bending Total strain energy 1 𝑈= ¥ 𝑀+4 𝑑𝑥 2𝐸𝐼 𝑊 4 𝐿“ Simply supported beam with central load 96𝐸𝐼 𝑤 4 𝐿Ÿ Simple supported beam with UDL 240𝐸𝐼 𝑤 4 𝐿Ÿ Cantilever with UDL 40𝐸𝐼 Simply supported beam with constant M 𝑀4 𝐿 throughout 2𝐸𝐼 12. Shearing stress a) Distribution of Shear stress 𝑉𝑄 𝜏= 𝐼𝑏 𝑉𝐴𝑦Ñ = 𝐼𝐵 Assumptions: ¨ For all value of 𝑦, 𝜏 is uniform across the width of the cross-section, irrespective of its shape. ÚÛ ¨ 𝑉= Ú+ is derived from the assumption that bending moment varies linearly across the section and is zero at the centroid. ¨ The material is homogenous and isotropic and the value of E is same for tension and compression. b) Relation between average and maximum shear Section 𝜏›|Z\ 𝜏›Z+ 𝑉 𝜏›Z+ 3 𝜏̅ = = 𝑏𝑑 𝜏ZhV 2 Occurs at y from NA 𝑑 𝜏›Z+ occurs at NA 𝑦=± 𝑦=0 2√3 16  𝑉 𝜏›Z+ 4 𝜏̅ = 𝜋 = 𝑑4 𝜏ZhV 3 4 at 𝑑 at NA 𝑦=± 𝑦=0 4 To derive this, use r and y 𝑉 2𝑉 𝜏ZhV = = 𝜏›Z+ 3 1 𝑏ℎ = 2 𝑏ℎ 𝜏ZhV 2 at from bottom 4 8𝑉 𝑥 = ℎ/2 𝜏@æÍ = 𝜏ZhV = 3 3𝑏ℎ derivation integrate from top 9 𝜏 8 ZhV 𝑉 At from top 𝜏ZhV = 𝐵4 3 at NA and other two points 𝑥= 𝑏 8 R è or 𝑏 = from NA ‰ ž√4 𝑉 4 𝑅4 + 𝑅𝑟 + 𝑟 4 𝜏ZhV = = 𝜋(𝑅4 − 𝑟 4 ) 3 𝑅4 + 𝑟 4 éÝÞß Shape factor is ration of éÞêë c) Qualitative shear distribution over some sections d) Shear center The shear center is the point through which a force can be applied which will cause a beam to bend and yet not twist. The shear center will always lie on an axis of symmetry of the cross section. The location of the shear center is only a function of the geometry of the cross section, and does not depend upon the applied loading. Shear center of sections consisting of two intersecting narrow rectangles always lies at the intersection of center line of two rectangles. 17  ž« Shear center of semi-circular arc strip of radius R will be at a distance of from the center of “3 the arc. ìíˆ îˆ For a given channel section (§PYQ) shear center žï e) Shear flow Shear flow is a measure of the force per unit length along the axis of a beam. This value is found from the shear formula and is used to determine the shear force developed in fasteners and glue that holds the various segments of a composite beam together. ñò The shear flow formula 𝑞 = can be used to determine the distribution of the shear flow throughout a thin-walled member provided the shear V acts along an axis of symmetry or principal centroidal axis of inertia for the cross section. If a member is made from segments having thin walls, only the shear flow parallel to the walls of the member is important. The shear flow varies linearly along segments that are perpendicular to the direction of the shear V. The shear flow varies parabolically along segments that are inclined or parallel to the direction of the shear V. On the cross section, the shear “flows” along the segments so that it results in the vertical shear force V and yet satisfies horizontal force equilibrium. 13. Deflection a) Introduction Curvature equation óˆ ¾ R – — ó߈ ¨ Š =− ’ ó¾ ˆ ˆ ôR†2 5 õ óß Differential equation for small deflection 𝑑4 𝑦 𝐸𝐼 = −𝑀 𝑑𝑥 4 𝑑𝜃 𝑀 =− 𝑑𝑥 𝐸𝐼 This equation can be integrated to find the angle of rotation 𝜃 or the deflection 𝑦 provided the bending moment is known. Sign convention ¨ x is positive when measured towards the right. ¨ y is positive when measured downwards ¨ 𝜃 is positive when the rotation is clockwise from the x-axis ¨ 𝑀 is positive when sagging. b) Double integration method This method is useful when one uniform 𝑀(𝑥) equation in the whole span of the beam. For concentrated point load or moment load, equation varies which increases computations. 𝑑4 𝑦 𝐸𝐼 4 = −𝑀 𝑑𝑥 𝐸𝐼𝜃 = − ¥ 𝑀 𝐸𝐼𝑦 = − ö 𝑀 Steps ¨ Integrate once to get slope, apply boundary condition to find the value of constant of integration. ¨ Integrate second time to find deflection profile and apply BC to find constant. 18 c) Macaulay’s method. When the beam is subjected to several point loads, or a combination of point loads and UDL for some portion, the procedure of double integration becomes highly cumbersome. However, the complication can be avoided by the use of discontinuity functions for different segments of the beam. Two kinds of discontinuity functions are used ¨ Macaulay’s function ¨ Singularity function For concentrated load it is preferred, not for non-prismatic beam. d) Super position method This method is suitable when more than one type of loading act and slope and deflection of concerned point is known by standard results, then final slope and deflection is obtained by superposition. e) Moment Area method (Mohr’s method) The method is especially suited when the deflection and angle of rotation at only one point of the beam is required, because it is possible to find these quantities directly without first finding the complete equation for the deflection curve. The deflection of a point is denoted as 𝑦 ÍÔ where, ¨ lower suffix denotes the deflection of the point under consideration and ¨ the upper suffix mean the point with respect to which the deflection is being calculated. Deviation of a point 𝑑 ÍÔ from a tangent A Deflectance of a point 𝐷 ÍÔ 𝐷ÔÍ = 𝑦ÔÍ + 𝑑²Í Mohr’s first theorem: ¨ The angle 𝜃èÍ between tangents to the deflection curve at two points A and B is equal to the negative of area of 𝑀/𝐸𝐼 diagram between points. Í 𝑀 Í 𝜃èÍ = −¥ 𝑑𝑥 = − ¬ 𝐴Û/ 𝐸𝐼 è è Mohr’s second theorem: ¨ The deviation of B form tangent at A is equal to negative of first moment with respect to B, Û of diagram area between A and B. Í 𝑀𝑥 Ö Í 𝑑èÍ = −¥ 𝑑𝑥 = − ¬ 𝐴Û/ ∙ 𝑥̅ 𝐸𝐼 è è o 𝑥̅ - has its origin at point where deviation is being calculated 1 1 1 𝐴 = 𝑏ℎ 𝐴 = 𝑏ℎ 𝐴 = 𝑏ℎ 𝐴 = (𝑎 + 𝑏)ℎ 2 2 2 𝑏 𝐿 (2𝑎 + 𝑏) 𝑥̅ = 𝑏 1 2 𝑥̅ = 𝑥̅ = (𝑏 + 𝑐) 𝑥̅ = 3 3 3 (𝑎 + 𝑏) 19  1 1 2 𝑛 𝐴 = 𝑏ℎ 𝐴= 𝑏ℎ 𝐴 = 𝑏ℎ 𝐴= 𝑏ℎ 3 𝑛+1 3 𝑛+1 3 𝑛+1 3 𝑛+1 𝑏 𝑥̅ = 𝑏 𝑥̅ = – —𝑏 𝑥̅ = 𝑏 𝑥̅ = – — 4 𝑛+2 8 𝑛+2 2 f) Conjugate beam method This approach is good for beams having variable flexural rigidity. A conjugate beam is an imaginary secondary beam method, which when loaded with 𝑀/𝐸𝐼 diagram of the real beam, yield directly the slope and deflection of the real beam in the form of shear force and bending moment of the conjugate beam. Theorem 1: The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam. 𝜃«|ZY = 𝑉²9\# Theorem 2: The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam. 𝛿«|ZY = 𝑀²9\# Change in support Sign convention ¨ Loading on the conjugate beam is equal to 𝑀/𝐸𝐼. If it is positive it acts down wards, and negative acts upwards. ¨ Slope of the real beam will be clockwise if the SF is positive and anticlockwise if the SF is negative. ¨ Deflection of the real beam will be downward, if the BM at the corresponding point of the conjugate beam is positive and upward if the BM is negative. g) Strain Energy method Strain energy stored in a beam 20  i 𝑀+4 𝑑𝑥 𝑈=¥ 2𝐸𝐼 ¢ ¨ Equate to find deflection 1 𝑃∙Δ=𝑈 2 ¨ The only deflection we can find is the deflection under the load, and not the deflection elsewhere, by this direct method. Similarly, when a beam is subjected to couple, the R rotation can be found by equating it to 4 𝜇𝜃. Here also the rotation is found at the point where the couple is applied. ¨ However, we can find the deflection at any other point of the beam by two other methods: o Unit load method o Fictious load method or Castigliano’s method Unit load ¨ Deflection at any point i 𝑀 ∙ 𝑚 ∙ 𝑑𝑥 Δ=¥ 𝐸𝐼 ¢ ¨ 𝑚 is moment due to unit load. Castigliano’s first theorem ¨ States that deflection caused by an external force is equal to the partial derivative of the strain energy with respect to that force. 𝜕𝑈 𝜕𝑈 = 𝛿R = 𝜙R 𝜕𝑊R 𝜕𝑀R 𝑀 𝜕𝑀 𝛿R = ¥ – — 𝑑𝑥 𝐸𝐼 𝜕𝑊R 𝑀 𝜕𝑀 𝜙R = ¥ – — 𝑑𝑥 𝐸𝐼 𝜕𝑀R ¨ In the above expression, 𝜕𝑀/𝜕𝑊R is evaluated by differentiating inside the integral sign before integrating. ¨ If no load is acting at a point where the deflection is desired, a fictious load W is applied at the point in the direction, the deflection is required, then after differentiating, but before integrating, the fictious load is set to zero. This method is also known as fictious load method. If deflection is happening in opposite direction of applied load then negative sign comes. 14. Fixed beam a) Statically indeterminate beams When the number of unknown reaction components exceeds the static conditions of equilibrium, the beam is said to be statically indeterminate. To determine the reaction components at the supports of such beams we will have to consider the equations of deformations or displacements, in addition to the equation of statics. b) Area moment method Area of free and fixed bending moment diagrams are equal, as the slope is zero at both ends. Also, moment of area is equal as the deflection is also zero at the ends. c) Method of superposition The method of super-position is an important tool in analyzing statically indeterminate structures. The method essentially consists of identifying the statical redundants from the indeterminate structure, and removing the redundant reactions, thus making the structure statically determinate. Such a structure is known as the released structure or the basic determinate structure. 21  The principle of superposition is then applied, according to which the final displacements due to both the actual loads and the redundant reactions must be equal to the sum of those displacements as calculated separately. 15. Column a) Buckling failure: Euler’s theory The determination of critical force 𝑃²« is of utmost importance, since it represents the ultimate capacity of an ideal structure. 𝑛4 𝜋 4 𝐸𝐼 𝑃²« = 𝐿4| 𝑛 = 1 single bow condition 𝑛 = 2 double bow condition Critical stress 𝜋4𝐸 𝜎²« = 4 𝜆 As the critical stress is inversely proportional to slenderness ratio, for smaller and intermediate columns, stress is more than crushing strength of material, which is not possible. Hence, Euler theory gives higher value for buckling load (over estimates). b) Assumptions of Euler’s theory Axis of the column is perfectly straight when unloaded. The line of thrust coincides exactly with the unstrained axis of the column. Flexural rigidity 𝐸𝐼 is uniform. Material is isotropic and homogeneous. The buckling value of 𝑃 = 𝑃²« is assumed to obtain all degrees of flexure. c) Rankine theory It takes into account, the failure by both crushing and buckling so it is generally used for intermediate columns. However, it can be used for both short column and long column. Rankin’s load 1 1 1 = + 𝑃« 𝑃² 𝑃| ¨ Where, 𝑃« = Rankin’s load, 𝑃² = crushing load, 𝑃| = Euler’s buckling load ¨ Rearranging the equation, we can get 𝜎² 𝐴/FOS 𝑃« = 𝑙 4 1 + 𝛼 – 𝑟| — d) Effective Length and support conditions Distance between two point of contra-flexure is called effective length. Effective length of column with given end conditions is equal to equivalent column of same material with both ends hinged and having magnitude of critical load equal to that of given column. Support Effective Critical force condition length 𝜋 4 𝐸𝐼 Both hinged 𝐿 𝐿4 𝜋 4 𝐸𝐼 Fixed and free 2𝐿 4𝐿4 4𝜋 4 𝐸𝐼 Both fixed 𝐿/2 𝐿4 2𝜋 4 𝐸𝐼 Fixed and hinge 𝐿/√2 𝐿4 e) Radius of Gyration 22  𝐼 𝑟=˜ 𝐴 Ú 𝑟²"«²Y| = ž f) Slenderness Ratio For a compression member is defined as the ratio of its effective length to the appropriate radius of gyration. 𝐿| 𝜆= 𝑟 g) Combined Direct and Bending stress Eccentric loading 𝑃 Direct compressive stress 𝜎9 = 𝐴 𝑀 𝑃𝑒 Bending stress 𝜎ý = ± 𝑦=± 𝑦 𝐼++ 𝐼++ 𝑃 𝑀 𝜎›Z+ = + 𝐴 𝑍++ Total stress at the extreme fibers 𝑃 𝑀 𝜎›"\ = − 𝐴 𝑍++ Rectangular section: middle third rule ¨ When, 𝜎9 < 𝜎ý due to which the stress changes sign, being partly compressive and partly tensile. In many structures such as masonry structures or concrete structures, development of tensile stress in a section is neither desirable nor permissible as they are weak in tension. ¨ In order that tension may not develop 𝜎9 ≥ 𝜎ý 𝑃 𝑃𝑒 𝑑 𝑃𝑒𝑑 ≥ ≥ 𝐴 𝐼 2 2𝐴𝑘 4 2𝑘 4 𝑒≤ 𝑑 ¨ For a rectangular section 2 𝑑4 𝑑 𝑒≤ ≤ 𝑑 12 6 𝑑 𝑒›Z+ = 6 ¨ The stress will be of the same sign throughout the section, if the load line is within the middle third of the section. This requirement is commonly known as the middle third rule. Circular section: Middle fourth rule ¨ Solid circular section Úˆ o 𝑘 4 = Rÿ 2 𝑑4 𝑒≤ 𝑑 16 𝑑 𝑒›Z+ = 8 o Thus, in order that tension is not developed, the load line must fall within the middle fourth of the section. ¨ Hollow circular section ! $j # ƒÚ # % j ˆ †Ú ˆ 4 o 𝑘 = "# ! ˆ (j ƒÚ ˆ ) = Rÿ # 2 𝐷4 + 𝑑4 𝑒≤ 𝐷 16 23  𝐷4 + 𝑑4 𝑒›Z+ = 8𝐷 h) The core of a section When load is applied in the kern area, then no tension occurs in the column. It is of diamond shape (rhombus), for rectangular section, with diagonals of d/3. Ú For solid circular column middle fourth portion is kern. 2𝑒 = ‰ 5 Kern of I section is rhombus Kern of square is square. i) General question Gravity dam, or retaining wall of elementary triangular shape, & ¨ = √𝐺 where G is specific gravity ý For eccentricity in dam is è ' '| ¨ 𝑒≤ stress can be calculated by ± ÿ Í ( è ‰) ¨ 𝑒> ÿ stress is “è where W is weight of dam and B is width at base. (it is calculated using force balance and resultant of triangular distribution acts at 1/3rd from side) 16. Torsion a) Torque diagram From right hand thumb rule if the direction of torque is in the direction where the thumb points. Moving from left to right, if direction of torque is same as movement, then it is positive, else negative. b) Basic equation 𝑇 𝜏 𝐺𝜃 = = 𝐽 𝑟 𝑙 The above equation states that the intensity of shear stress at any point in the cross-section of a shaft subjected to pure torsion is proportional to its distance from the centre. The variation is linear. 𝐽/𝑟 is known as torsional sectional modulus. 𝐺𝐽 is known as torsional rigidity of the bar. The relative stiffness of the shafts is measured by the inverse ratio of the angles of twist in equal lengths of shafts when subjected to equal torques. Polar moment of inertia 3Ú # ¨ Solid shaft 𝐽 = “4 3 ¨ Hollow circular shaft 𝐽 = “4 (𝐷ž − 𝑑ž ) 16𝑇 R/“ 𝑑=– — π 𝜏,-; c) Strain energy in Torsion The torsional Strain energy of a shaft is equal to the work done in twisting 1 𝑈 = 𝑇𝜃 2 ¨ The above from of torsional strain energy is useful only of 𝑇 and 𝜃 have been previously found. As the shear stress varies linearly along the radius, the value of strain energy stored needs to be integrated to find the expression. 24  ¨ Strain energy stored in solid shaft 4 𝜏./01-23 𝑈= × volume of shaft 4𝐺 ¨ For hollow shaft. 𝜏œ4 𝐷4 + 𝑑4 𝑈= Â Æ × volume of shaft 4𝐺 𝐷4 d) Power Transmitted 2𝜋𝑁 𝑃 = 𝑇𝜔 = 𝑇 60 Questions have been asked on finding diameter or maximum shear if power (kW) is given 𝑃 ∝ 𝜏›Z+ 𝑑“ 4 ¨ Use 𝜋 ≈ 10 Q. for a solid shaft of same weight as hollow shaft, and mean diameter of hollow shaft equal to diameter of solid shaft. ~: ¨ Ratio of torque transmitted = 1.7 ~; ': ¨ Ratio of power transmitted '; = 1.7 ˆ = 0 ˆ Solving we can find the unknown reactions. g) Compound shaft Two shafts of different materials can be joined together in two ways ¨ Shafts in series: both the shaft carry the same torque T and the total angle of twist at the resisting end is the sum of separate angles of twist of two shafts. 𝑇𝐿R 𝑇𝐿4 𝜃 = 𝜃R + 𝜃4 = + 𝑁R 𝐽R 𝑁4 𝐽4 ¨ Shafts in parallel: the torque is applied at the junction of two shafts and the reactive torques 𝑇R and 𝑇4 are at the remote ends. Naturally the angle of twist for both the shafts is the same. 𝜃R = 𝜃4 𝑇R 𝐿R 𝑇4 𝐿4 = 𝑁R 𝐽R 𝑁4 𝐽4 25  h) Torsion in non-circular and thin-walled sections For torsion of rectangular section, the maximum shear stress 𝜏›Z+ is given by 𝑇 𝜏›Z+ = 𝑘R 𝑑𝑏4 For thin walled closed section 𝑇 𝜏= 2𝐴› 𝑡 ¨ Where A is the area enclosed by the median line and t is the thickness. 17. Thin cylinder and sphere a) Circumferential stress / hoop stress Balancing force on projected area and on the skin 𝜎& 2𝑡𝐿 = 𝑝(𝐷𝐿) 𝑝𝐷 𝜎& = 2𝑡 è The variation of hoop stress is hyperbolic 2𝜎& = « ˆ + 𝐴5 in radial direction, maximum at inner diameter and decreasing outward to outer diameter. (the variation is given by Lame’s equation; whose constants are positive) b) Longitudinal stress 𝜋 𝜎Y (𝜋𝐷𝑡) = 𝑝 2 𝐷4 5 4 𝑝𝐷 𝜎Y = 4𝑡 For a thin closed cylinder, combining both longitudinal and hoop stresses we get 𝑝𝑑 𝜖Y = (1 − 2µ) 4𝑡𝐸 𝑝𝑑 𝜖[ = (2 − µ) 4𝑡𝐸 𝑝𝑑 𝜖h = (5 − 4µ) 4𝑡𝐸 c) Thin sphere šÚ 𝜎& = ž[ “šÚ 𝜖h = ž[ (1 − µ) 18. Spring a) Closed coiled Helical spring: Axial pull 𝑅 = mean radius of the coil 𝑛 = number of turns 𝑑 = diameter of wire or rod 𝐿 = length of the spring = 2𝜋𝑅𝑛 𝑃 = axial pull 𝛿 = 𝑅𝜃 axial deflection of load Torque 𝑇 = 𝑃𝑅 Angle of twist and axial deflection 𝑇𝐿 𝑃𝑅 (2𝜋𝑅𝑛) 𝜃= = 𝜋 𝐺𝐽 𝐺 232 𝑑ž 5 64𝑃𝑅4 𝑛 𝜃= 𝐺𝑑ž 64𝑃𝑅“ 𝑛 𝛿=