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Mechanics of solid.pptx

The document discusses the mechanics of solid stress, defining it as force per unit area and detailing various types of stress such as tensile, compressive, and shear stress. It explains concepts of strain and deformation, along with the relationship between stress and strain, including Hooke's law and stress-strain diagrams that categorize materials into ductile and brittle. Additionally, it covers strain energy, thermal stresses, and the principle of superposition in understanding how applied loads affect materials.

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Mechanics of solid
STRESS Stress is defined as force per unit area. 𝜎 = 𝐹 𝐴 Normal stress Compressive and tensile stresses constitute normal stress because the force vector from which they are calculated is normal (perpendicular) to the area on which it acts. Tensile stress Compressive stress Shear stress Shearing Stress is a type of stress that acts coplanar with a cross section of material. Mathematically 𝜏 = 𝐹 𝐴
STRAIN A change in shape due to the application of force is known as deformation. Strain is the amount of deformation experienced by the body in the direction of force applied, divided by the initial dimensions of the body. 𝐿 ∆𝐿 + 𝐿 The deformation that is easiest to observe is the stretching of a bar made of a single material when a tensile force is applied to it. Suppose that a force pulling on a bar of initial length L stretches it by an amount ΔL, as in Fig. This quantity is known as longitudinal strain and is usually denoted ε, that is 𝑆𝑡𝑟𝑎𝑖𝑛 𝜀 = ∆𝐿 𝐿 Shear strain Shear strain can mathematically written as 𝛾 = 𝐹 𝐺 G is called modulas of rigidity Poisson’s ratio The ratio of lateral strain to longitudinal strain is as poisson’s ratio. Mathematically 𝑚 = − 𝜀𝑙𝑎𝑡 𝜀𝑙𝑜𝑛𝑔
RELATION BETWEEN STRESS AND STRAIN Hooke’s law; modulus of elasticity the stress is directly proportional to the strain (within elastic limit) , and we can write 𝜎 ∝ 𝜀 → 𝜎 = 𝐸𝜀 𝐸 = 𝜎 𝜀 = 𝐹𝐿 𝐴∆𝐿 The coefficient E is called the modulus of elasticity of the material involved, or also Young’s modulus. Stress-strain diagram to distinguish some common characteristics among the stress strain diagrams of various groups of materials and to divide materials into two broad categories on the basis of these characteristics, namely, the ductile materials and the brittle materials. • Ductile materials, which comprise structural steel, as well as many alloys of other metals, are characterized by their ability to yield at normal temperatures. • As the specimen is subjected to an increasing load, its length first increases linearly with the load and at a very slow rate. Thus, the initial portion of the stress-strain diagram is a straight line with a steep slope. • after a critical value of the stress has been reached, the specimen undergoes a large deformation with a relatively small increase in the applied load. • The elongation of the specimen after it has started to yield can be 200 times as large as its deformation before yield. • After a certain maximum value of the load has been reached, the diameter of a portion of the specimen begins to decrease. This phenomenon is known as necking.
RELATION BETWEEN STRESS AND STRAIN Stress-strain diagram to distinguish some common characteristics among the stress strain diagrams of various groups of materials and to divide materials into two broad categories on the basis of these characteristics, namely, the ductile materials and the brittle materials. • Brittle materials, which comprise cast iron, glass, and stone, are characterized by the fact that rupture occurs without any noticeable prior change in the rate of elongation. • Thus, for brittle materials, there is no difference between the ultimate strength and the breaking strength. • Also, the strain at the time of rupture is much smaller for brittle than for ductile materials.
STRAIN ENARGY AND IMPACT LOADING Strain energy when a load is applied gradually Strain Energy is defined as the internal work done to deform a body by the action of externally applied forces. Let’s say we have an axially loaded bar on which load is applied gradually which increases from 0 to P and due to this load the length of the bar increases by ΔL. We can write, work done due to force P 𝑑𝑈 = 𝑃∆𝐿 Total work 𝑈 = 0 𝐿 𝑃∆𝐿 In the case of a linear and elastic deformation, the portion of the load- deformation diagram involved can be represented by a straight line of equation 𝑃 = 𝑘𝐿 𝑈 = 0 𝐿 𝑘𝐿∆𝐿 = 𝑘𝐿 𝐿 2 = 𝑃𝐿 2 It can also interpreted as work done = average load × deformation 𝑈 = 𝑃 2 ∆𝐿 = 𝜎. 𝐴. 𝜀. 𝐿 2 = 𝑉. 𝜎. 𝜀 2 = 𝑉. 𝜎. 𝜎 𝐸 . 1 2 → 𝑼 = 𝝈𝟐𝑽 𝟐𝑬 Strain Energy stored in a body when a load is applied suddenly Work done = Strain energy 𝑃 × ∆𝐿 = 𝜎2. 𝐴. 𝐿 2𝐸 𝑃 × 𝜀𝐿 = 𝑃 × 𝜎 𝐸 . 𝐿 = 𝜎2 . 𝐴. 𝐿 2𝐸 → 𝜎 = 2 𝑃 𝐴 Then σ = 𝑃 𝐴 for gradually applied load σ = 2𝑃 𝐴 for suddenly applied load The stress applied by sudden load is twice the stress of gradual load when the load is same. So the strain energy will be 𝑼 = 𝟒𝝈𝟐𝑽 𝟐𝑬 The strain energy stored in a body due to suddenly applied load Compared to when it is applied gradually is four times.
STRAIN ENARGY AND IMPACT LOADING Strain Energy stored in a body when a load is applied with impact We know, Work done = force × distance travelled = 𝑃 ℎ + ∆𝐿 We can write → 𝑃 ℎ + ∆𝐿 = 𝜎2 . 𝐴. 𝐿 2𝐸 𝑃ℎ + 𝑃∆𝐿 = 𝜎2 . 𝐴. 𝐿 2𝐸 𝑃ℎ + 𝑃. 𝜀. 𝐿 = 𝑃ℎ + 𝑃. 𝜎 𝐸 . 𝐿 = 𝜎2. 𝐴. 𝐿 2𝐸 𝜎2 2 − 𝜎𝑃 𝐴 − 𝑃ℎ𝐸 𝐴𝐿 = 0 This is a quadratic equation 𝜎 = 𝑃 𝐴 + 𝑃2 𝐴2 + 4 1 2 𝑃ℎ𝐸 𝐴𝐿 1 = 𝑃 𝐴 + 𝑃2 𝐴2 + 4 1 2 𝑃ℎ𝐸 𝐴𝐿 𝜎 = 𝑃 𝐴 + 𝑃 𝐴 1 + 2 ℎ𝐸𝐴 𝑃𝐿 𝜎 = 𝑃 𝐴 1 + 1 + 2 ℎ𝐸𝐴 𝑃𝐿 When ΔL is very small as compared to h, then 𝑃ℎ = 𝜎2 . 𝐴. 𝐿 2𝐸 𝜎 = 2𝑃ℎ𝐸 𝐴𝐿 h P ∆𝐿
STRAIN Once s is obtained, deformation Δl or strain energy can be calculated: 𝑼 = 𝑺𝟐 𝑽 𝟐𝑬 If Δl is very small as compared to h, then, 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝑃ℎ 𝑆2𝑉 2𝐸 = 𝑃ℎ → 𝑆 = 2𝑃ℎ𝐸 𝑉 = 2𝑃ℎ𝐸 𝐴𝐿 Strain Energy stored in a body due to shear stress We know Work done = Force × Displacement 𝑤 = 𝑃 2 × ∅ℎ = 1 2 𝑞𝐴 ∅ℎ 𝑤 = 1 2 𝑞ℎ𝐿 ∅ℎ 𝑤 = 1 2 . 𝑞. ∅. ℎ. ℎ. 𝐿 𝑤 = 1 2 𝑞∅𝑉 = 1 2 . 𝑞. 𝑞 𝐺 . 𝑉 = 𝑞2 . 𝑉 2𝐺 𝒘 = 𝒒𝟐. 𝑽 𝟐𝑮 The strain energy stored in a body subjected to shear stress 𝑼 = 𝟏 𝟐 . 𝑽𝝉𝟐 𝑮 ∅ ∅ ℎ 𝐿
DEFORMATION A change in shape due to the application of force is known as deformation. Deformation due to axial loading ∆𝐿 = 𝐹𝐿 𝐴𝐸 or ∆𝐿 = 𝑃𝐿 𝐴𝐸 Deformation due to own weight We know, ∆𝐿 = 𝑃𝐿 𝐸𝐴 Specific weight of the bar = 𝑊 𝐴𝐿 Load below small section = 𝑊 𝐴𝐿 × 𝐴𝑥 = 𝑊𝑥 𝐿 Deformation in the small section = 𝑊𝑥𝑑𝑥 𝐸𝐴𝐿 Deformation of the bar ∆𝐿 = 𝑊 𝐸𝐴 0 𝐿 𝑥𝑑𝑥 = 𝑊 𝐸𝐴𝐿 . 𝑥2 2 0 𝐿 = 𝑊 𝐸𝐴𝐿 . 𝐿2 2 ∆𝑳 = 𝟏 𝟐 . 𝑾𝑳 𝑨𝑬 Deformation of uniformly tappering circular bar We know ∆𝐿 = 𝑃𝐿 ( 𝜋 4 )𝑑2𝐸 Change in diameter with respect to length 𝑘 = 𝑑1 − 𝑑2 𝐿 Diameter in the small section = 𝑑1 − 𝑑1 − 𝑑2 𝐿 𝑥 = 𝑑1 − 𝑘𝑥 Deformation in small section = 𝑃𝑑𝑥 𝜋 4 𝑑1 − 𝑘𝑥 2𝐸 L x dx d1 d2 dx L x d
DEFORMATION Deformation of the bar ∆𝐿 = 4𝑃 𝜋𝐸 0 𝐿 𝑑𝑥 𝑑1 − 𝑘𝑥 2 = − 4𝑃 𝜋𝑘𝐸 𝑑1 𝑑2 𝑑𝑧 𝑧2 = 4𝑃 𝜋𝐸𝑘 1 𝑧 𝑑1 𝑑2 = 4𝑃 𝜋𝐸 𝑑1 − 𝑑2 𝐿 × 1 𝑑2 − 1 𝑑1 = 4𝑃 𝜋𝐸 𝑑1 − 𝑑2 𝐿 × 𝑑1 − 𝑑2 𝑑1𝑑2 So the deformation of tapering circular bar ∆𝑷 = 𝟒𝑷𝑳 𝝅𝑬𝒅𝟏𝒅𝟐 Deformation of uniformly tappering flat bar Change in width with respect to length 𝑎 − 𝑏 𝐿 = 𝑘 Width at the small section 𝑎 − 𝑘𝑥 From the definition of deformation at small section = 𝑃𝑑𝑥 𝐸 𝑎 − 𝑘𝑥 𝑡 Deformation for bar ∆𝐿 = 𝑃 𝐸𝑡 0 𝐿 𝑑𝑥 𝑎 − 𝑘𝑥 ∆𝐿 = − 𝑃 𝐸𝑡 𝑎 − 𝑏 𝐿 𝑎 𝑏 𝑑𝑧 𝑧 ∆𝐿 = − 𝑃 𝐸𝑡 𝑎 − 𝑏 𝐿 ln 𝑏 𝑎 = 𝑷𝑳 𝒕𝑬 𝒂 − 𝒃 𝐥𝐧 𝒂 𝒃 a b t
STRESS Stress on composite bar A composite bar may be defined as a bar made up of two or more different materials, joined together, in such a manner that the system extends or contracts as one unit, equally, when subjected to tension or compression. In case of composite bars, the following points should be kept in view: • The extension or contraction of the bar being equal, the strain i.e. deformation per unit length is also equal. ∆𝐿1 = ∆𝐿2 • The total external load on the bar is equal to the sum of the loads carried by different materials. 𝑃 = 𝑃1 + 𝑃2 𝐴1 𝐸1 𝑃1 ∆𝐿1 𝐴2 𝐸2 𝑃2 ∆𝐿2 ∆𝐿1 = ∆𝐿2 𝑃1𝐿 𝐸1𝐴1 = 𝑃2𝐿 𝐸2𝐴2 → 𝑃1 = 𝑃2𝐸1𝐴1 𝐸2𝐴2 From the assumption 𝑃 = 𝑃1 + 𝑃2 𝑃 = 𝑃2𝐸1𝐴1 𝐸2𝐴2 + 𝑃2 = 𝑃2 1 + 𝐸1𝐴1 𝐸2𝐴2 𝑷 = 𝑷𝟐 𝟏 + 𝑬𝟏𝑨𝟏 𝑬𝟐𝑨𝟐 𝑷 = 𝑷𝟏 𝟏 + 𝑬𝟐𝑨𝟐 𝑬𝟏𝑨𝟏 Again 𝑃1𝐿 𝐸1𝐴1 = 𝑃2𝐿 𝐸2𝐴2 We know 𝜎 = 𝑃 𝐴 Putting these value 𝜎1 𝐸1 = 𝜎2 𝐸2 → 𝜎1 = 𝐸1 𝐸2 𝜎2 The ratio E1 / E2 is known as modular ratio of the two materials.
STRESS 2m 2m 1.5m 25 N Diameter 2 cm for all bar 𝐸𝑠 = 2.05 𝐺𝑁 𝑚2 𝐸𝑐 = 1.1 𝐺𝑁 𝑚2 We know, ∆𝐿𝑠 = ∆𝐿𝑐 = ∆𝐿𝑠 𝑃 = 𝑃𝑠 + 𝑃𝑐 + 𝑃𝑠 = 2𝑃𝑠 + 𝑃𝑐 So, ∆𝐿𝑠 = ∆𝐿𝑐 = ∆𝐿𝑠 𝑃𝑠𝐿𝑠 𝐸𝑠𝐴𝑠 = 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 = 𝑃𝑠𝐿𝑠 𝐸𝑠𝐴𝑠 𝑃𝑠𝐿𝑠 𝐸𝑠𝐴𝑠 = 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 𝑃𝑠 = 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 × 𝐸𝑠𝐴𝑠 𝐿𝑠 𝐴𝑐 = 𝜋 4 0.02 2 = 3.1415 × 10−4 𝑚2 𝐴𝑠 = 𝜋 4 0.02 2 = 3.1415 × 10−4 𝑚2 𝐴𝑐 = 𝐴𝑠 S C S We get 𝑃 = 2𝑃𝑠 + 𝑃𝑐 𝑃 = 2 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 × 𝐸𝑠𝐴𝑠 𝐿𝑠 + 𝑃𝑐 25 = 2 𝑃𝑐1.5 1.1 × 109 × 2.05 × 109 2 + 𝑃𝑐 25 = 2.79𝑃𝑐 + 𝑃𝑐 𝑃𝑐 = 6.59 𝑁 Now 𝑃 = 2𝑃𝑠 + 𝑃𝑐 𝑃𝑠 = 25 − 6.59 2 = 9.21 𝑁 Stress 𝑆𝑠 = 𝑃𝑠 𝐴𝑠 = 9.21 3.1415 × 10−4 = 29.317 𝑘𝑁/𝑚2
STRESS Thermal stresses in simple bar The thermal stresses or strains, in a simple bar, may be found out as discussed below: • Calculate the amount of deformation due to change of temperature with the assumption that bar is free to expand or contract. • Calculate the load (or force) required to bring the deformed bar to the original length. • Calculate the stress and strain in the bar caused by this load. ∆𝐿 𝜎 𝐿 Let L = Original length of the body T = Increase in temperature ⍺ = Coefficient of linear expansion We know that increase in length due increase in temperature ∆𝐿 = 𝐿. 𝛼. 𝑡 If the ends of the bar are fixed to rigid supports, so that its expansion is prevented, then compressive strain induced in the bar. 𝜺 = ∆𝑳 𝑳 = 𝑳. 𝜶. 𝒕 𝑳 = 𝜶. 𝒕 Stress 𝝈 = 𝑬. 𝜺 = 𝜶. 𝒕. 𝑬 If the supports yeild by Δ then 𝜺 = ∆𝑳 − ∆ 𝑳 = 𝑳. 𝜶. 𝒕 − ∆ 𝑳 = 𝜶. 𝒕 − ∆ 𝑳 Stress 𝝈 = 𝑬 𝜶. 𝒕 − ∆ 𝑳
STRESS Thermal stresses in bars of circular Tapering bar Let L = Original length of the body T = Increase in temperature ⍺ = Coefficient of linear expansion We know that increase in length due increase in temperature ∆𝐿 = 𝐿. 𝛼. 𝑡 P = Load required to bring the bar to the orginal length We know that decrase in length due to load P ∆𝐿 = 4𝑃𝐿 𝜋. 𝐸. 𝑑1. 𝑑2 𝐿. 𝛼. 𝑡 = 4𝑃𝐿 𝜋. 𝐸. 𝑑1. 𝑑2 𝑃 = 𝛼. 𝑡. 𝜋. 𝐸. 𝑑1. 𝑑2 4 Max stress 𝜎𝑚𝑎𝑥 = 𝑃 𝐴 = 𝛼. 𝑡. 𝜋. 𝐸. 𝑑1. 𝑑2 4 𝜋 4 𝑑2 2 = 𝜶. 𝒕. 𝑬. 𝒅𝟏 𝒅𝟐 If we substitute 𝑑1 = 𝑑2 we get 𝝈 = 𝜶. 𝒕. 𝑬 𝑑1 𝑑2 𝐿 𝐿1 𝐿2 Thermal stresses in bars of varying section We know that as a result of the increase in temperature, the bar will tend to expand. But since it is fixed at its ends, therefore it will cause some compressive stress in the body. Moreover, as the thermal stress is shared equally by both the portions, therefore 𝜎1𝐴1 = 𝜎2𝐴2 Total deformation of the body ∆𝐿 = ∆𝐿1 + ∆𝐿2 = 𝜎1𝐿1 𝐸1 + 𝜎2𝐿2 𝐸2
NUMERICAL 1. A rod 2m long at a temperature of 10℃. Find the expansion of the rod when the temparature rise to 80℃ . If the expansion is prevented. Find the stress induceed in the material. 𝛼 = 0.000012℃; 𝐸 = 1.0 × 105 𝑀𝑁/𝑚2 ∆𝐿 = 𝛼. 𝑡. 𝐿 = 0.000012 80 − 10 2 = 1.68 × 10−3 𝑚 Stress induced 𝜎 = 𝐸. 𝑡. 𝛼 = 1.0 × 1011 × 70 × 0.000012 = 8.4 × 107 𝑁 𝑚2 2. Steel rod of 3cm diameter and 5m long is connected to two bricks. And the rod is maintained at temperature of 95℃. Determine the stress and pull excerted when temperature falls to 30℃. i. The ends do not yield ii. The ends yeild 0.12 cm Stress 𝜎 = 𝛼. 𝐸. 𝑡 = 3 × 10−2 × 2 × 1011 × 95 − 30 × 11 × 10−6 = 4.29 × 106 𝑁 𝑚2 Stress when ends yield 𝜎 = 𝐸. 𝛼. 𝑡 − ∆ 𝐿 = 2 × 1011 × 11 × 10−6 × 65 − 1.2 × 10−3 3 × 10−2 𝜎 = 7.85 × 109 𝑁/𝑚2
PRINCIPLE OF SUPER POSITION The principle of superposition states that when there are numbers of loads are acting together on an elastic material, the resultant strain will be the sum of individual strains caused by each load acting separately.
50 kN P kN 120 kN 40 kN 60 kN 0.5 m 0.5 m 0.5 m 0.5 m 50 kN 50 kN 60 kN 60 kN 20 kN 20 kN 140 kN 140 kN From the above figure 𝑃 = 90 𝑘𝑁 𝐸 = 2 × 105 𝑁 𝑚𝑚2 = 2 × 1011 𝑁 𝑚2 We know ∆𝐿1 = 𝑃1𝐿1 𝐴1𝐸 = 50000 × 0.5 𝜋 4 0.1 2 × 2 × 1011 = 1.6 × 10−5 𝑚 ∆𝐿2 = 𝑃2𝐿2 𝐴2𝐸 = 140000 × 0.5 𝜋 4 0.2 2 × 2 × 1011 = 1.1 × 10−5 𝑚 ∆𝐿3 = 4𝑃𝐿 𝜋𝐸𝑑1𝑑2 = 4 × 0.5 × 20000 𝜋 × 2 × 1011 × 0.2 × 0.4 = 7.96 × 10−7 𝑚 ∆𝐿4 = 𝑃4𝐿4 𝐴4𝐸 = 0.5 × 60000 𝜋 4 0.4 2 × 2 × 1011 = 1.19 × 10−6 𝑚 Now ∆𝐿 = ∆𝐿1 + ∆𝐿2 + ∆𝐿3 + ∆𝐿4 = 1.6 × 10−5 + 1.1 × 10−5 + 7.96 × 10−7 + 1.19 × 10−6 = 2.8989 × 10−5 𝑚 10 cm 20 cm 40 cm 20 cm 40 cm PRINCIPLE OF SUPER POSITION
ELASTICITY Elasticity, ability of a deformed material body to return to its original shape and size when the forces causing the deformation are removed. Hooke's law The modern theory of elasticity generalizes Hooke's law to say that the strain (deformation) of an elastic object or material is proportional to the stress applied to it. Mathematically 𝐸 = 𝐹 𝐴 ∆𝐿 𝐿 = 𝐹𝐿 𝐴∆𝐿 E is called elastic constant or young modulus 𝐿 ∆𝐿 + 𝐿 Relation between K, G and E The relation among different elastic modulas/constant • 𝐾 = 𝑚𝐸 3(𝑚−2) • 𝐺 = 𝑚𝐸 2(1+𝑚) Here • m = Poisson’s ratio • E = Elasticity • G = Rigidity modulus • K = Bulk modulus
THIN CYLINDRICAL AND SPHERICAL SHELLS In engineering field, we daily come across vessels of cylindrical and spherical shapes containing fluids such as tanks, boilers, compressed air receivers etc. Generally, the walls of such vessels are very thin as compared to their diameters. These vessels, when empty, are subjected to atmospheric pressure internally as well as externally. In such a case, the resultant pressure on the walls of the shell is zero. But whenever a vessel is subjected to internal pressure (due to steam, compressed air etc.) its walls are subjected to tensile stresses. In general, if the thickness of the wall of a shell is less than 1/10th to 1/15th of its diameter, it is known as a thin shell. Failure of a cylindrical shell due to an internal pressure if these stresses exceed the permissible limit, the cylinder is likely to fail in any one of the following two ways as shown in Fig. 1. It may split up into two troughs 2. It may split up into two cylinder Stresses in thin cylindrical shells A little consideration will show that the walls of the cylindrical shell will be subjected to the following two types of tensile stresses: 1. Circumferential stress or Hoop stress 2. Longitudinal stress.
THIN CYLINDRICAL AND SPHERICAL SHELLS Circumferential stress Consider a thin cylindrical shell subjected to an internal pressure as shown in Fig. We know that as a result of the internal pressure, the cylinder has a tendency to split up into two troughs as shown in the figure. Let • L = Length of the shell, • d = Diameter of the shell, • t = Thickness of the shell and • p = Intensity of internal pressure. Total pressure along the diameter (say X-X axis) of the shell, 𝑃 = Intensity of the pressure × area = 𝑝. 𝑑𝑥. 𝐿 and circumferential stress in the shell, 𝜎𝑐 = Total pressure Resisting section = 𝑝. 𝑑. 𝐿 2𝑡𝐿 = 𝑝𝑑 2𝑡 Longitudinal stress Consider the same cylindrical shell, subjected to the same internal pressure as shown in Fig. We know that as a result of the internal pressure, the cylinder also has a tendency to split into two pieces as shown in the figure. Total pressure along its length (say Y-Y axis) of the shell 𝑃 = Intensity of the pressure × area = 𝑝. 𝜋 4 . 𝑑2 Longitudinal stress in the shell 𝜎𝑙 = Total pressure Resisting section = 𝑝. 𝜋 4 . 𝑑2 𝜋. 𝑡. 𝑑 = 𝑝𝑑 4𝑡 = 1 2 𝜎𝑐
THIN CYLINDRICAL AND SPHERICAL SHELLS Change in dimension due to internal pressure We know, 𝜎𝑐 = 𝑝𝑑 2𝑡 ; 𝜎𝑙 = 𝑝𝑑 4𝑡 ; poison′ s ratio = 1 m ; ΔL = Change in length; Δd = Change in diameter Change in length ΔL = 𝜀𝐿. 𝐿 → εL = ΔL L = 𝜎𝐿 𝐸 − 1 𝑚 . 𝜎𝑐 𝐸 = 1 𝐸 𝑝𝑑 4𝑡 − 1 𝑚 . 𝑝𝑑 2𝑡 ΔL L = 1 𝐸 . 𝑝𝑑 2𝑡 1 2 − 1 𝑚 𝚫𝐋 = 𝒑𝒅𝑳 𝟐𝒕𝑬 𝟏 𝟐 − 𝟏 𝒎 𝜺𝑳 = 𝒑𝒅 𝟐𝒕𝑬 𝟏 𝟐 − 𝟏 𝒎 Change in diameter Δd = 𝜀𝑑. 𝑑 → εd = Δd d = σc E − 1 𝑚 . 𝜎𝐿 𝐸 = 𝑝𝑑 2𝑡𝐸 − 1 𝑚 . 𝑝𝑑 4𝑡𝐸 Δd d = 𝑝𝑑 2𝑡𝐸 1 − 1 2𝑚 𝚫𝐝 = 𝒑𝒅𝟐 𝟐𝒕𝑬 𝟏 − 𝟏 𝟐𝒎 𝜺𝒅 = 𝒑𝒅 𝟐𝒕𝑬 𝟏 − 𝟏 𝟐𝒎 Change in volume ΔV = 𝑉 + ΔV − V ∆𝑉 = 𝜋 4 𝑑 + Δ𝑑 2 𝐿 + ∆𝐿 − 𝜋 4 𝑑2 . 𝐿 ∆𝑉 = 𝜋 4 𝑑2 ∆𝐿 + 2∆𝑑. 𝑑. 𝐿 ∆𝑉 𝑉 = 𝜋 4 𝑑2 ∆𝐿 + 2∆𝑑. 𝑑. 𝐿 𝜋/4 𝑑2. 𝐿 = ∆𝐿 𝐿 + 2 ∆𝑑 𝑑 = 𝜀𝐿 + 2𝜀𝑐 . 𝑉
NUMERICAL 1. Given that 𝑑 = 800 𝑚𝑚 = 0.8 𝑚; 𝑡 = 10 𝑚𝑚 = 0.01; 𝑃 = 2.5 𝑀𝑃𝑎 = 2.5 × 106 𝑁/𝑚2 𝜎𝑐 = 𝑃𝑑 2𝑡 = 2.5 × 109 × 0.8 2 × 0.01 = 1 × 108 𝑁/𝑚2 𝜎𝑙 = 0.5 × 𝜎𝑙 = 1 × 1011 × 0.5 = 5 × 107 𝑁/𝑚2 2. Given that 𝑑 = 40 𝑚𝑚 = 0.04 𝑚; 𝑡 = 5 𝑚𝑚 = 0.005; 𝜎𝑙 = 30 𝑀𝑃𝑎 = 30 × 109 𝑁/𝑚2 𝜎𝑙 = 𝑃𝑑 2𝑡 → 30 × 106 = 𝑃 × 0.04 2 × 0.005 → 𝑃 = 30 × 106 4 = 7.5 × 106 𝑁/𝑚2 3. Given that 𝑑 = 800 𝑚𝑚 = 0.8 𝑚; 𝑡 = 10 𝑚𝑚 = 0.01; 𝑃 = 2.5 𝑀𝑃𝑎 = 2.5 × 106 𝑁/𝑚2 𝐿 = 4𝑚; 1 𝑚 = 0.25; 𝐸 = 200 × 109 𝑁/𝑚2 Δd = 𝑃𝑑2 2𝐸𝑡 1 − 1 2𝑚 = 2.5 × 106 × 0.8 2 2 × 200 × 109 × 0.01 1 − 0.25 2 Δd = 3.5 × 10−4𝑚 = 0.35 𝑚𝑚 ∆𝑙 = 𝑃𝑑𝑙 2𝑡𝐸 1 2 − 1 𝑚 = 2.5 × 106 × 4 × 0.8 2 × 200 × 109 × 0.01 0.5 − 0.25 ∆𝑙 = 5 × 10−4 = 0.5 𝑚𝑚
THIN CYLINDRICAL AND SPHERICAL SHELLS Thin spherical shell Consider a thin spherical shell subjected to an internal pressure as shown in Fig. We know that as a result of the internal pressure, the shell is likely to be torn away along the centre of the sphere. Let, • d = Diameter of the shell, • t = Thickness of the shell and • p = Intensity of internal pressure. 𝑃 = Intensity of the pressure × area = 𝑝. 𝜋 4 . 𝑑2 Stress in the shell, 𝜎 = Total pressure Resisting section = 𝑝. 𝜋 4 . 𝑑2 𝜋. 𝑡. 𝑑 = 𝑝𝑑 4𝑡
TORTION OF CIRCULAR SHAFT Shear stress in a circular shaft subjected to torsion Consider a circular shaft fixed at one end subjected to a torque at the other end as shown in fig. We know, Shear strain = Deformation per unit length Φ = 𝐴𝐴′ 𝐶𝐴 = 𝐴𝐴′ 𝑙 = 𝑅. 𝜃 𝑙 Also Shear stress = 𝜏; Modulas of rigidity = C 𝐶 = 𝜏 Φ 𝜏 = 𝐶. Φ = C. 𝑅. 𝜃 𝑙 𝜏 𝑅 = 𝐶. 𝜃 𝑙 If 𝜏𝑥 is the intensity of shear stress at a distance 𝑥 from the centre of the shaft 𝜏𝑥 𝑥 = 𝜏 𝑅 = 𝐶. 𝜃 𝑙 Maximum torque transmitted by a circular solid shaft We know, Shear stress at x 𝜏𝑥 𝑥 = 𝜏 𝑅 Shear stress at that section, 𝜏𝑥 = 𝜏 𝑅 . 𝑥 Turning force at that section 𝐹 = Stress × Area 𝐹 = 𝜏 𝑅 . 𝑥 2𝜋𝑥. 𝑑𝑥 𝐹 = 𝜏 𝑅 . 2𝜋𝑥2 . 𝑑𝑥
TORTION OF CIRCULAR SHAFT 𝐹 = 𝜏 𝑅 . 2𝜋𝑥2 . 𝑑𝑥 Turning moment at this section 𝑑𝑇 = 𝜏 𝑅 . 2𝜋𝑥2. 𝑑𝑥 . 𝑥 𝑑𝑇 = 𝜏 𝑅 . 2𝜋𝑥3 . 𝑑𝑥 0 𝑇 𝑑𝑇 = 0 𝑅 𝜏 𝑅 . 2𝜋𝑥3 . 𝑑𝑥 𝑇 = 2𝜋𝜏 𝑅 𝑥4 4 0 𝑅 = 2𝜋𝜏 𝑅 𝑅4 4 𝑇 = 𝜋𝜏𝑅3 2 = 𝜋 16 × 𝐷3 × 𝜏 Polar moment of inertia Reletion between angle of twist and torque 𝑇 = 𝜋 16 × 𝜏 × 𝐷3 → 𝜏 = 𝑇 × 16 𝜋 × 1 𝐷3 Also 𝜏 = 𝐶. 𝑅. 𝜃 𝑙 Equiting the both equation 𝑇 × 16 𝜋 × 1 𝐷3 = 𝐶. 𝑅. 𝜃 𝑙 𝑇 × 16 𝜋 × 1 𝐷3 × 1 𝑅 = 𝐶. 𝜃 𝑙 𝑇32 𝜋𝐷4 = 𝐶𝜃 𝑙 → 𝑇 𝜋 32 𝐷4 = 𝐶𝜃 𝑙 We consider 𝐽 = 𝜋 32 𝐷4 This is called polar moment of inertia. Then we get 𝑻 𝑱 = 𝑪𝜽 𝒍 = 𝝉 𝑹
TORTION OF CIRCULAR SHAFT Maximum torque transmitted by a circular hollow shaft 𝑇 = 𝜋 16 × 𝜏 × 𝐷4 − 𝑑4 𝐷 Polar moment of inertia 𝐽 = 𝜋 16 𝐷4 − 𝑑4 Power transmitted by a shaft We have already discussed that the main purpose of a shaft is to transmit power from one shaft to another. 𝑃 = 2𝜋𝑁𝑇 60 The unit will be watt T = Torque N = r. p. m
BENDING MOMENT AND SHEAR FORCE Shear force Shear force is the force that acts perpendicular to the longitudinal axis of a structural member, such as a beam or a column, and tends to cause the member to slide or shear along its cross section. In other words, it is the force that is exerted on a section of a structure perpendicular to the plane of the section. Shear force is usually denoted by the symbol V and its unit is Newtons (N) or pounds (lb). • When a force acts downwards on the left-hand side of the section, the shear force is positive (+) and acts upwards on the right-hand side of the section. Conversely, when a force acts upwards on the left-hand side of the section, the shear force is negative (-) and acts downwards on the right-hand side of the section. • The shear at any given point of a beam is positive when the external forces (loads and reactions) acting on the beam tend to shear off the beam at that point as indicated in Fig. • In other words, the sign convention of shear force is such that it is positive when it tends to create a clockwise moment on the section, and negative when it tends to create an anticlockwise moment on the section.
BENDING MOMENT AND SHEAR FORCE Bending moment Bending moment, on the other hand, is the moment that tends to cause a structural member to bend or deform around a neutral axis. It is the algebraic sum of the moments about the neutral axis of all the forces acting on one side of the section. Bending moment is usually denoted by the symbol M and its unit is Newton-meters (Nm) or pound-feet (lb-ft). • The sign convention of bending moment is based on the direction of the moment acting on a section of a structure. If the moment is causing clockwise rotation on a section, it is considered negative, and if it is causing counter clockwise rotation, it is considered positive. • The bending moment at any given point of a beam is positive when the external forces acting on the beam tend to bend the beam at that point as indicated in Fig. • For example, if a beam is supported at both ends and a load is applied at the midpoint of the beam, the bending moment will be maximum at the midpoint and will be negative on the right side of the section and positive on the left side of the section, as the moment is causing clockwise rotation on the right side and counter clockwise rotation on the left side.
BENDING MOMENT AND SHEAR FORCE Shear force and bending moment diagram A shear force diagram (SFD) shows the variation of shear force along the length of a structural member. The diagram is plotted with shear force on the y-axis and the length of the member on the x-axis. The diagram shows positive values of shear force as upward or rightward forces and negative values of shear force as downward or leftward forces. A bending moment diagram (BMD) shows the variation of bending moment along the length of a structural member. The diagram is plotted with bending moment on the y-axis and the length of the member on the x-axis. Simply supported beam with a point load at its mid point: 𝐴 𝐵 𝑊 𝐿 𝑅𝐴 = 𝑊 2 𝑅𝐵 = 𝑊 2 Shear force at A and B Since the load is in the mid point, the reaction force will be, 𝑅𝐴 = 𝑅𝐵 = 𝑊 2 Bending at poin C with respect to A and B is same 𝑀𝑐 = 𝑅𝐴 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑊 2 × 𝐿 2 = 𝑊𝐿 4 𝐶 𝐶 𝑊𝐿 4 Simply supported beam with a point load at any point: 𝐴 𝐵 𝑊 𝐿 𝑅𝐴 = 𝑊𝑏 𝐿 𝑅𝐵 = 𝑊𝑎 𝐿 𝐶 𝑏 𝑎 Reaction force at A 𝑅𝐴 × 𝐿 − 𝑊 × 𝑏 = 0; 𝑅𝐴 = 𝑊𝑏 𝐿 Reaction force at B 𝑅𝐵 × 𝐿 − 𝑊 × 𝑎 = 0; 𝑅𝐵 = 𝑊𝑎 𝐿 Bending moment at C 𝑀𝐶 = 𝑅𝐴 × 𝑎 = 𝑊𝑎𝑏 𝐿 ; 𝑀𝐶 = 𝑅𝐵 × 𝑏 = 𝑊𝑎𝑏 𝐿 𝐶 𝑊𝑎𝑏 𝐿
BENDING MOMENT AND SHEAR FORCE Reaction force at A 𝑅𝐴 × 2.5 − 2 × 1.5 − 4 × 1 = 0; 𝑅𝐴 = 7 2.5 = 2.8 Reaction force at B 𝑅𝐵 × 2.5 − 4 × 1.5 − 2 × 1 = 0; 𝑅𝐵 = 3.2 Bending moment at C 𝑀𝐶 = 𝑅𝐴 × 1 = 2.8 × 1 = 2.8 Bending moment at D 𝑀𝐷 = 𝑅𝐴 × 1.5 = 2.8 × 1.5 − 2 × 0.5 = 3.2 Shear force diagram 𝐹𝐴 = +𝑅𝐴 = +2.8; 𝐹𝐶 = 2.8 − 2 = 0.8; 𝐹𝐷 = 0.8 − 4 = −3.2; 𝐹𝐵 = −3.2; Bending moment diagram 𝑀𝐴 = 𝑀𝐵 = 0; 𝑀𝐶 = 2.8; 𝑀𝐷 = 3.2
BENDING MOMENT AND SHEAR FORCE Simply suppoted beam with uniformly distributed load 𝐴 𝐵 𝑤 = 𝑢𝑛𝑖𝑡 𝑙𝑜𝑎𝑑 𝐿 𝑅𝐴 = 𝑤𝐿 2 𝑅𝐵 = 𝑤𝐿 2 𝐶 𝐶 𝑤𝐿2 8 𝑥 Reaction force at A and B 𝑅𝐴 = 𝑅𝐵 = 𝑤𝐿 2 Shear force diagram 𝐹𝐴 = +𝑅𝐴 = + 𝑤𝐿 2 ; 𝐹𝐶 = 𝑤𝐿 2 − 𝑤𝐿 2 = 0; 𝐹𝐵 = − 𝑤𝐿 2 ; 𝐹𝑥 = −𝑅𝐵 + 𝑤𝑥 = 𝑤𝑥 − 𝑤𝐿 2 Bending moment diagram 𝑀𝐴 = 𝑀𝐵 = 0; 𝑀𝐶 = 𝑅𝐴. 𝐿 2 = 𝑤𝐿2 4 − 𝑤𝐿 2 . 𝐿 4 = 𝑤𝐿2 8 ; 𝑀𝑥 = 𝑅𝐵. 𝑥 = 𝑤𝐿𝑥 2 − 𝑤𝑥. 𝑥 2 = 𝑤𝐿𝑥 2 − 𝑤𝑥2 2 𝐴 𝐵 𝑤 = 1000 𝑘𝑔/𝑚 4𝑚 𝐶 𝐸 𝑅𝐴 = 𝑅𝐵 = 1000 × 4 2 = 2000 SFD 𝐹𝐴 = 2000; 𝐹𝑐 = 2000 − 1000 × 2 = 0: 𝐹𝐵 = −2000; BMD 𝑀𝐴 = 𝑀𝐵 = 0 𝑀𝐷 = 2000 × 1 − 1000 × 1 × 1 2 = 1500; 𝑀𝐶 = 2000 × 2 − 1000 × 2 × 2 2 = 2000; 𝐷 1 1 𝐶 𝑤𝐿2 8
BENDING MOMENT AND SHEAR FORCE 𝐴 𝐵 𝑤 = 500 𝑘𝑔/𝑚 4 𝐶 𝐸 𝐷 2 2 1000 𝑁 1000 𝑁 Reaction force at A and B 𝑅𝐴 × 8 = 500 × 4 × 4; 𝑅𝐴 = 4000 8 = 1000 𝑅𝐵 = 1000 𝑘𝑔 SFD 𝐹𝐴 = +1000; 𝐹𝐷 = +1000; 𝐹𝐶 = 1000 − 500 × 2 = 0; 𝐹𝐸 = −500; 𝐹𝐵 = −2000; BMD 𝑀𝐴 = 𝑀𝐵 = 0; 𝑀𝐷 = 1000 × 2 = 2000; 𝑀𝐶 = 1000 × 4 − 500 × 2 × 2 2 = 3000; 𝑀𝐸 = 1000 × 6 − 500 × 4 × 4 2 = 2000; 𝐶 𝐸 𝐷 𝐴 𝐵 𝑤 = 500 𝑘𝑔/𝑚 4 𝐶 𝐸 𝐷 2 2 1000 𝑁 1000 𝑁 Reaction force at B (taking moment at A) 𝑅𝐵 × 4 + 1000 × 2 = 500 × 4 × 2 + 1000 × 6 𝑅𝐵 = 8000 4 = 2000 𝑘𝑔 𝑅𝐴 = 𝑅𝐵 = 2000 𝑘𝑔 SFD 𝐹𝑐 = −1000; 𝐹𝐴 = −1000 + 2000 = 1000 𝐹𝐸 = 1000 − 500 × 2 = 0 𝐹𝐵 = 0 − 500 × 2 + 2000 = 1000 𝐹𝐷 = 1000 − 1000 = 0 BMD 𝑀𝑐 = 0; 𝑀𝐴 = −1000 × 2 = −2000 𝑀𝐸 = −1000 × 4 + 2000 × 2 − 500 × 2 × 1 = −1000 𝑀𝐵 = −1000 × 2 = −2000
Reaction force at B taking moment at A 𝑅𝐵 × 5 = 0.5 × 2 × 4 + 1.5 × 2 × 2 + 0.5 × 1 × 0.5 = 10.25 𝑅𝐵 = 2.05 𝑘𝑁 𝑅𝐴 = 0.5 × 1 + 1.5 × 2 + 0.5 × 2 − 2.05 = 2.45 𝑘𝑁 SFD 𝐹𝐴 = 2.45 𝑘𝑁; 𝐹𝐶 = +2.45 − 0.5 × 1 = 1.95 𝑘𝑁; 𝐹𝐷 = 1.95 − 1.5 × 2 = −1.05 𝑘𝑁; 𝐹𝐵 = −2.05 BMD 𝑀𝐴 = 0; 𝑀𝐶 = 2.45 × 1 − 0.5 × 1 × 0.5 = 2.2 𝑀𝐷 = 2.05 × 2 − 0.5 × 2 × 1 = 3.1 𝑀𝐵 = 0 The maximum bending moment is at M, x distance from c where shear force changes sign. 1.95 𝑥 = 1.05 2 − 𝑥 → 𝑥 = 1.3 𝑚 𝑀𝑀 = 2.45 × 2.3 − 0.5 × 1 × (0.5 + 1.3) − 1.5 × 1.3 × 0.65 = 3.47
Reaction force at A by taking moment at B 𝑅𝐴 × 5 = 10 × 2 × 3 𝑅𝐴 = 12 𝑘𝑁 Reaction force at A 𝑅𝐵 = 10 × 2 − 12 = 8 𝑘𝑁 Reaction force at A by taking moment at B 𝑅𝐴 × 4 = 4 × 2.5 + 2 × 1 × 2 𝑅𝐴 = 3.5 𝑘𝑁 Reaction force at A 𝑅𝐵 = (2 × 1 + 4) − 3.5 = 2.5 𝑘𝑁 Reaction force at A by taking moment at B 𝑅𝐴 × 6 = 4 × 1.5 × 5.25 + 2 × 3 × 1.5 + (5 × 1.5) 𝑅𝐴 = 8 𝑘𝑁 Reaction force at A 𝑅𝐵 = (4 × 1.5 + 5 × 3 + 5) − 8 = 18 𝑘𝑁
Reaction force at A by taking moment at B 𝑅𝐴 × 3 = 4.5 × 3 × 1.5 − 4.5 × 1 × 0.5 𝑅𝐴 = 6 𝑘𝑁 Reaction force at A 𝑅𝐵 = 4.5 × 4 − 6 = 12 𝑘𝑁
Cantilever with point load at free end BENDING MOMENT AND SHEAR FORCE We know that shear force at any section X from a distance x from free end, is equal to the total unbalance vertical force 𝐹𝑥 = 𝑤 𝑀𝑥 = −𝑤. 𝑥 0.5 0.5 0.5 1000 1000 1000 SFD 𝐹𝐵 = 1000; 𝐹𝑐 = 2000; 𝐹𝐷 = 3000 BMD 𝑀𝐵 = 0; 𝑀𝐶 = −1000 × 0.5 = −500 𝑀𝐷 = −1000 × 1 − 1000 × 0.5 = −1500 𝑀𝐴 = −1000 × 1.5 − 1000 × 1 − 1000 × 0.5 = −3000 B C D
COLUMNS AND STRUTS Introduction A structural member, subjected to an axial compressive force, is called a strut. As per definition, a strut may be horizontal, inclined or even vertical. But a vertical strut, used in buildings or frames, is called a column. Failure of a column or struts t has been observed, that when a column or a strut is subjected to some compressive force, then the compressive stress induced, 𝜎 = 𝑃 𝐴 • A little consideration will show, that if the force or load is gradually increased the column will reach a stage, when it will be subjected to the ultimate crushing stress. Beyond this stage, the column will fail by crushing. The load corresponding to the crushing stress, is called crushing load. It has also been experienced that sometimes, a compression member does not fail entirely by crushing, but also by bending i.e., buckling. This happens in the case of long columns. It has also been observed that all the short columns fail due to their crushing. But, • If a long column is subjected to a compressive load, it is subjected to a compressive stress. If the load is gradually increased, column will reach a stage, when it will start buckling. The load, at which the column just buckles is called buckling load, critical load or crippling load and the column is said to have developed an elastic instability. • A little consideration will show that for a long column, the value of buckling load will be less than the crushing load. Moreover, the value of buckling load is low for long columns and relatively high for short columns.
EULER’S COLUMN THEORY Assumption 1. Initially the column is perfectly straight and the load applied is truly axial. 2. The cross-section of the column is uniform throughout its length. 3. The column material is perfectly elastic, homogeneous and isotropic and thus obeys Hooke's law. 4. The length of column is very large as compared to its cross- sectional dimensions. 5. The shortening of column, due to direct compression (being very small) is neglected. 6. The failure of column occurs due to buckling alone. Sign convention Types of end conditions of column 1. Both ends hinged, 2. Both ends fixed, 3. One end is fixed and the other hinged, and 4. One end is fixed and the other free.
COLUMN WITH BOTH ENDS HINGED Consider a long column AB with a length of l is hinged at its both ends. Its carrying a critical load P at B. As a result of loading the column deflect into a curve AX1B as shown in Fig. Now consider a secttion X at a distance x from A. 𝑃 = Load on the column 𝑦 = Deflection of the column at X Moment due to critical load 𝑀 = −𝑃. 𝑦 𝐸𝐼 𝑑2 𝑦 𝑑𝑥2 = −𝑃. 𝑦 𝐸𝐼 𝑑2𝑦 𝑑𝑥2 + 𝑃. 𝑦 = 0 𝑑2𝑦 𝑑𝑥2 + 𝑃. 𝑦 𝐸𝐼 = 0 The general solution for above differential equation, 𝑦 = 𝐶1. cos 𝑥. 𝑃 𝐸𝐼 + 𝐶2 sin 𝑥. 𝑃 𝐸𝐼 We know, when 𝑥 = 0, 𝑦 = 0, then C1 = 0 When 𝑥 = 𝑙, 𝑦 = 0 then sin 𝑙. 𝑃 𝐸𝐼 = 0 sin 𝑙. 𝑃 𝐸𝐼 = sin 0 = sin 𝜋 = sin 2𝜋 Taking the least significant value 𝑙. 𝑃 𝐸𝐼 = 𝜋 → 𝑃 = 𝝅𝟐𝑬𝑰 𝒍𝟐
EULER’S FORMULA AND EQUIVALENT LENGTH In the previous articles, we have derived the relations for the crippling load under various end conditions. Sometimes, all these cases are represented by a general equation called Euler's formula, 𝑃𝐸 = 𝜋2 𝐸𝐼 𝐿𝑒 2 Where 𝐿𝑒 equivalent length. The equivalent length for all the condition are given below. Conditions Relation between 𝒍 and 𝑳𝒆 Formula Both end hinged 𝐿𝑒 = 𝑙 𝑃 = 𝜋2𝐸𝐼 𝑙2 Both end fixed 𝐿𝑒 = 𝑙 2 𝑃 = 4𝜋2𝐸𝐼 𝑙2 One end hinged, other end fixed 𝐿𝑒 = 𝑙 2 𝑃 = 2𝜋2𝐸𝐼 𝑙2 One end fixed, other end free 𝐿𝑒 = 2𝑙 𝑃 = 𝜋2𝐸𝐼 4𝑙2 Slenderness ratio We know that, 𝑃𝐸 = 𝜋2𝐸𝐼 𝐿𝑒 2 Now substituting 𝐼 = 𝐴𝑘2 where 𝐴 is cross sectional area and 𝑘 is the least redius of gyration. 𝑃𝐸 = 𝜋2 𝐸(𝐴𝑘2 ) 𝐿𝑒 2 = 𝑃𝐸 = 𝜋2 𝐴 𝐿𝑒 𝑘 2 Where 𝐿𝑒 𝑘 is called the slenderness ratio.
PRINCIPAL STRESSES AND STRAIN Principal planes Principal planes are the planes within the material such that the resultant stresses across them are completely normal stresses or planes across which the shear stresses is zero. Principal stress Principal stresses are those stresses which are acting on the principal planes. • The plane carrying the maximum normal stress is called the major principal plane and the stress acting on it is called major principal stress. • The plane carrying minimum normal stress is known as minor principal plane and the stress acting on it is called as minor principal stress. Normal stress and shear stress on obli
• আখলাক অর্ থকক? • ভাললা আখলাক এর উধারণ দাও? ৫টি • সৃটির সসবা কক? • সৃটির সসবায় উধাহরন দাও? ৫টি • "রাহমাতুকিল আলামীন" অর্ থকক? • তুকম ককভালব সদশ সক ভাললাবাসলব কলখ? • ক্ষমা কার গুণ? • মহানবী (স) সকান শহলর দাওয়াত কদলত সেলয় লাকিত হলয়কিললন? • ক্ষমার গুরুত্ব কলখ? • সততা কক? • সতামালদর মলধে সলব থ াত্তম সসই বোক্তি,
• আখলাক অর্ থকক?

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Mechanics of solid.pptx

  • 1.
  • 2.
    STRESS Stress is definedas force per unit area. 𝜎 = 𝐹 𝐴 Normal stress Compressive and tensile stresses constitute normal stress because the force vector from which they are calculated is normal (perpendicular) to the area on which it acts. Tensile stress Compressive stress Shear stress Shearing Stress is a type of stress that acts coplanar with a cross section of material. Mathematically 𝜏 = 𝐹 𝐴
  • 3.
    STRAIN A change inshape due to the application of force is known as deformation. Strain is the amount of deformation experienced by the body in the direction of force applied, divided by the initial dimensions of the body. 𝐿 ∆𝐿 + 𝐿 The deformation that is easiest to observe is the stretching of a bar made of a single material when a tensile force is applied to it. Suppose that a force pulling on a bar of initial length L stretches it by an amount ΔL, as in Fig. This quantity is known as longitudinal strain and is usually denoted ε, that is 𝑆𝑡𝑟𝑎𝑖𝑛 𝜀 = ∆𝐿 𝐿 Shear strain Shear strain can mathematically written as 𝛾 = 𝐹 𝐺 G is called modulas of rigidity Poisson’s ratio The ratio of lateral strain to longitudinal strain is as poisson’s ratio. Mathematically 𝑚 = − 𝜀𝑙𝑎𝑡 𝜀𝑙𝑜𝑛𝑔
  • 4.
    RELATION BETWEEN STRESSAND STRAIN Hooke’s law; modulus of elasticity the stress is directly proportional to the strain (within elastic limit) , and we can write 𝜎 ∝ 𝜀 → 𝜎 = 𝐸𝜀 𝐸 = 𝜎 𝜀 = 𝐹𝐿 𝐴∆𝐿 The coefficient E is called the modulus of elasticity of the material involved, or also Young’s modulus. Stress-strain diagram to distinguish some common characteristics among the stress strain diagrams of various groups of materials and to divide materials into two broad categories on the basis of these characteristics, namely, the ductile materials and the brittle materials. • Ductile materials, which comprise structural steel, as well as many alloys of other metals, are characterized by their ability to yield at normal temperatures. • As the specimen is subjected to an increasing load, its length first increases linearly with the load and at a very slow rate. Thus, the initial portion of the stress-strain diagram is a straight line with a steep slope. • after a critical value of the stress has been reached, the specimen undergoes a large deformation with a relatively small increase in the applied load. • The elongation of the specimen after it has started to yield can be 200 times as large as its deformation before yield. • After a certain maximum value of the load has been reached, the diameter of a portion of the specimen begins to decrease. This phenomenon is known as necking.
  • 5.
    RELATION BETWEEN STRESSAND STRAIN Stress-strain diagram to distinguish some common characteristics among the stress strain diagrams of various groups of materials and to divide materials into two broad categories on the basis of these characteristics, namely, the ductile materials and the brittle materials. • Brittle materials, which comprise cast iron, glass, and stone, are characterized by the fact that rupture occurs without any noticeable prior change in the rate of elongation. • Thus, for brittle materials, there is no difference between the ultimate strength and the breaking strength. • Also, the strain at the time of rupture is much smaller for brittle than for ductile materials.
  • 6.
    STRAIN ENARGY ANDIMPACT LOADING Strain energy when a load is applied gradually Strain Energy is defined as the internal work done to deform a body by the action of externally applied forces. Let’s say we have an axially loaded bar on which load is applied gradually which increases from 0 to P and due to this load the length of the bar increases by ΔL. We can write, work done due to force P 𝑑𝑈 = 𝑃∆𝐿 Total work 𝑈 = 0 𝐿 𝑃∆𝐿 In the case of a linear and elastic deformation, the portion of the load- deformation diagram involved can be represented by a straight line of equation 𝑃 = 𝑘𝐿 𝑈 = 0 𝐿 𝑘𝐿∆𝐿 = 𝑘𝐿 𝐿 2 = 𝑃𝐿 2 It can also interpreted as work done = average load × deformation 𝑈 = 𝑃 2 ∆𝐿 = 𝜎. 𝐴. 𝜀. 𝐿 2 = 𝑉. 𝜎. 𝜀 2 = 𝑉. 𝜎. 𝜎 𝐸 . 1 2 → 𝑼 = 𝝈𝟐𝑽 𝟐𝑬 Strain Energy stored in a body when a load is applied suddenly Work done = Strain energy 𝑃 × ∆𝐿 = 𝜎2. 𝐴. 𝐿 2𝐸 𝑃 × 𝜀𝐿 = 𝑃 × 𝜎 𝐸 . 𝐿 = 𝜎2 . 𝐴. 𝐿 2𝐸 → 𝜎 = 2 𝑃 𝐴 Then σ = 𝑃 𝐴 for gradually applied load σ = 2𝑃 𝐴 for suddenly applied load The stress applied by sudden load is twice the stress of gradual load when the load is same. So the strain energy will be 𝑼 = 𝟒𝝈𝟐𝑽 𝟐𝑬 The strain energy stored in a body due to suddenly applied load Compared to when it is applied gradually is four times.
  • 7.
    STRAIN ENARGY ANDIMPACT LOADING Strain Energy stored in a body when a load is applied with impact We know, Work done = force × distance travelled = 𝑃 ℎ + ∆𝐿 We can write → 𝑃 ℎ + ∆𝐿 = 𝜎2 . 𝐴. 𝐿 2𝐸 𝑃ℎ + 𝑃∆𝐿 = 𝜎2 . 𝐴. 𝐿 2𝐸 𝑃ℎ + 𝑃. 𝜀. 𝐿 = 𝑃ℎ + 𝑃. 𝜎 𝐸 . 𝐿 = 𝜎2. 𝐴. 𝐿 2𝐸 𝜎2 2 − 𝜎𝑃 𝐴 − 𝑃ℎ𝐸 𝐴𝐿 = 0 This is a quadratic equation 𝜎 = 𝑃 𝐴 + 𝑃2 𝐴2 + 4 1 2 𝑃ℎ𝐸 𝐴𝐿 1 = 𝑃 𝐴 + 𝑃2 𝐴2 + 4 1 2 𝑃ℎ𝐸 𝐴𝐿 𝜎 = 𝑃 𝐴 + 𝑃 𝐴 1 + 2 ℎ𝐸𝐴 𝑃𝐿 𝜎 = 𝑃 𝐴 1 + 1 + 2 ℎ𝐸𝐴 𝑃𝐿 When ΔL is very small as compared to h, then 𝑃ℎ = 𝜎2 . 𝐴. 𝐿 2𝐸 𝜎 = 2𝑃ℎ𝐸 𝐴𝐿 h P ∆𝐿
  • 8.
    STRAIN Once s isobtained, deformation Δl or strain energy can be calculated: 𝑼 = 𝑺𝟐 𝑽 𝟐𝑬 If Δl is very small as compared to h, then, 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝑃ℎ 𝑆2𝑉 2𝐸 = 𝑃ℎ → 𝑆 = 2𝑃ℎ𝐸 𝑉 = 2𝑃ℎ𝐸 𝐴𝐿 Strain Energy stored in a body due to shear stress We know Work done = Force × Displacement 𝑤 = 𝑃 2 × ∅ℎ = 1 2 𝑞𝐴 ∅ℎ 𝑤 = 1 2 𝑞ℎ𝐿 ∅ℎ 𝑤 = 1 2 . 𝑞. ∅. ℎ. ℎ. 𝐿 𝑤 = 1 2 𝑞∅𝑉 = 1 2 . 𝑞. 𝑞 𝐺 . 𝑉 = 𝑞2 . 𝑉 2𝐺 𝒘 = 𝒒𝟐. 𝑽 𝟐𝑮 The strain energy stored in a body subjected to shear stress 𝑼 = 𝟏 𝟐 . 𝑽𝝉𝟐 𝑮 ∅ ∅ ℎ 𝐿
  • 9.
    DEFORMATION A change inshape due to the application of force is known as deformation. Deformation due to axial loading ∆𝐿 = 𝐹𝐿 𝐴𝐸 or ∆𝐿 = 𝑃𝐿 𝐴𝐸 Deformation due to own weight We know, ∆𝐿 = 𝑃𝐿 𝐸𝐴 Specific weight of the bar = 𝑊 𝐴𝐿 Load below small section = 𝑊 𝐴𝐿 × 𝐴𝑥 = 𝑊𝑥 𝐿 Deformation in the small section = 𝑊𝑥𝑑𝑥 𝐸𝐴𝐿 Deformation of the bar ∆𝐿 = 𝑊 𝐸𝐴 0 𝐿 𝑥𝑑𝑥 = 𝑊 𝐸𝐴𝐿 . 𝑥2 2 0 𝐿 = 𝑊 𝐸𝐴𝐿 . 𝐿2 2 ∆𝑳 = 𝟏 𝟐 . 𝑾𝑳 𝑨𝑬 Deformation of uniformly tappering circular bar We know ∆𝐿 = 𝑃𝐿 ( 𝜋 4 )𝑑2𝐸 Change in diameter with respect to length 𝑘 = 𝑑1 − 𝑑2 𝐿 Diameter in the small section = 𝑑1 − 𝑑1 − 𝑑2 𝐿 𝑥 = 𝑑1 − 𝑘𝑥 Deformation in small section = 𝑃𝑑𝑥 𝜋 4 𝑑1 − 𝑘𝑥 2𝐸 L x dx d1 d2 dx L x d
  • 10.
    DEFORMATION Deformation of thebar ∆𝐿 = 4𝑃 𝜋𝐸 0 𝐿 𝑑𝑥 𝑑1 − 𝑘𝑥 2 = − 4𝑃 𝜋𝑘𝐸 𝑑1 𝑑2 𝑑𝑧 𝑧2 = 4𝑃 𝜋𝐸𝑘 1 𝑧 𝑑1 𝑑2 = 4𝑃 𝜋𝐸 𝑑1 − 𝑑2 𝐿 × 1 𝑑2 − 1 𝑑1 = 4𝑃 𝜋𝐸 𝑑1 − 𝑑2 𝐿 × 𝑑1 − 𝑑2 𝑑1𝑑2 So the deformation of tapering circular bar ∆𝑷 = 𝟒𝑷𝑳 𝝅𝑬𝒅𝟏𝒅𝟐 Deformation of uniformly tappering flat bar Change in width with respect to length 𝑎 − 𝑏 𝐿 = 𝑘 Width at the small section 𝑎 − 𝑘𝑥 From the definition of deformation at small section = 𝑃𝑑𝑥 𝐸 𝑎 − 𝑘𝑥 𝑡 Deformation for bar ∆𝐿 = 𝑃 𝐸𝑡 0 𝐿 𝑑𝑥 𝑎 − 𝑘𝑥 ∆𝐿 = − 𝑃 𝐸𝑡 𝑎 − 𝑏 𝐿 𝑎 𝑏 𝑑𝑧 𝑧 ∆𝐿 = − 𝑃 𝐸𝑡 𝑎 − 𝑏 𝐿 ln 𝑏 𝑎 = 𝑷𝑳 𝒕𝑬 𝒂 − 𝒃 𝐥𝐧 𝒂 𝒃 a b t
  • 11.
    STRESS Stress on compositebar A composite bar may be defined as a bar made up of two or more different materials, joined together, in such a manner that the system extends or contracts as one unit, equally, when subjected to tension or compression. In case of composite bars, the following points should be kept in view: • The extension or contraction of the bar being equal, the strain i.e. deformation per unit length is also equal. ∆𝐿1 = ∆𝐿2 • The total external load on the bar is equal to the sum of the loads carried by different materials. 𝑃 = 𝑃1 + 𝑃2 𝐴1 𝐸1 𝑃1 ∆𝐿1 𝐴2 𝐸2 𝑃2 ∆𝐿2 ∆𝐿1 = ∆𝐿2 𝑃1𝐿 𝐸1𝐴1 = 𝑃2𝐿 𝐸2𝐴2 → 𝑃1 = 𝑃2𝐸1𝐴1 𝐸2𝐴2 From the assumption 𝑃 = 𝑃1 + 𝑃2 𝑃 = 𝑃2𝐸1𝐴1 𝐸2𝐴2 + 𝑃2 = 𝑃2 1 + 𝐸1𝐴1 𝐸2𝐴2 𝑷 = 𝑷𝟐 𝟏 + 𝑬𝟏𝑨𝟏 𝑬𝟐𝑨𝟐 𝑷 = 𝑷𝟏 𝟏 + 𝑬𝟐𝑨𝟐 𝑬𝟏𝑨𝟏 Again 𝑃1𝐿 𝐸1𝐴1 = 𝑃2𝐿 𝐸2𝐴2 We know 𝜎 = 𝑃 𝐴 Putting these value 𝜎1 𝐸1 = 𝜎2 𝐸2 → 𝜎1 = 𝐸1 𝐸2 𝜎2 The ratio E1 / E2 is known as modular ratio of the two materials.
  • 12.
    STRESS 2m 2m 1.5m 25 N Diameter2 cm for all bar 𝐸𝑠 = 2.05 𝐺𝑁 𝑚2 𝐸𝑐 = 1.1 𝐺𝑁 𝑚2 We know, ∆𝐿𝑠 = ∆𝐿𝑐 = ∆𝐿𝑠 𝑃 = 𝑃𝑠 + 𝑃𝑐 + 𝑃𝑠 = 2𝑃𝑠 + 𝑃𝑐 So, ∆𝐿𝑠 = ∆𝐿𝑐 = ∆𝐿𝑠 𝑃𝑠𝐿𝑠 𝐸𝑠𝐴𝑠 = 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 = 𝑃𝑠𝐿𝑠 𝐸𝑠𝐴𝑠 𝑃𝑠𝐿𝑠 𝐸𝑠𝐴𝑠 = 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 𝑃𝑠 = 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 × 𝐸𝑠𝐴𝑠 𝐿𝑠 𝐴𝑐 = 𝜋 4 0.02 2 = 3.1415 × 10−4 𝑚2 𝐴𝑠 = 𝜋 4 0.02 2 = 3.1415 × 10−4 𝑚2 𝐴𝑐 = 𝐴𝑠 S C S We get 𝑃 = 2𝑃𝑠 + 𝑃𝑐 𝑃 = 2 𝑃𝑐𝐿𝑐 𝐸𝑐𝐴𝑐 × 𝐸𝑠𝐴𝑠 𝐿𝑠 + 𝑃𝑐 25 = 2 𝑃𝑐1.5 1.1 × 109 × 2.05 × 109 2 + 𝑃𝑐 25 = 2.79𝑃𝑐 + 𝑃𝑐 𝑃𝑐 = 6.59 𝑁 Now 𝑃 = 2𝑃𝑠 + 𝑃𝑐 𝑃𝑠 = 25 − 6.59 2 = 9.21 𝑁 Stress 𝑆𝑠 = 𝑃𝑠 𝐴𝑠 = 9.21 3.1415 × 10−4 = 29.317 𝑘𝑁/𝑚2
  • 13.
    STRESS Thermal stresses insimple bar The thermal stresses or strains, in a simple bar, may be found out as discussed below: • Calculate the amount of deformation due to change of temperature with the assumption that bar is free to expand or contract. • Calculate the load (or force) required to bring the deformed bar to the original length. • Calculate the stress and strain in the bar caused by this load. ∆𝐿 𝜎 𝐿 Let L = Original length of the body T = Increase in temperature ⍺ = Coefficient of linear expansion We know that increase in length due increase in temperature ∆𝐿 = 𝐿. 𝛼. 𝑡 If the ends of the bar are fixed to rigid supports, so that its expansion is prevented, then compressive strain induced in the bar. 𝜺 = ∆𝑳 𝑳 = 𝑳. 𝜶. 𝒕 𝑳 = 𝜶. 𝒕 Stress 𝝈 = 𝑬. 𝜺 = 𝜶. 𝒕. 𝑬 If the supports yeild by Δ then 𝜺 = ∆𝑳 − ∆ 𝑳 = 𝑳. 𝜶. 𝒕 − ∆ 𝑳 = 𝜶. 𝒕 − ∆ 𝑳 Stress 𝝈 = 𝑬 𝜶. 𝒕 − ∆ 𝑳
  • 14.
    STRESS Thermal stresses inbars of circular Tapering bar Let L = Original length of the body T = Increase in temperature ⍺ = Coefficient of linear expansion We know that increase in length due increase in temperature ∆𝐿 = 𝐿. 𝛼. 𝑡 P = Load required to bring the bar to the orginal length We know that decrase in length due to load P ∆𝐿 = 4𝑃𝐿 𝜋. 𝐸. 𝑑1. 𝑑2 𝐿. 𝛼. 𝑡 = 4𝑃𝐿 𝜋. 𝐸. 𝑑1. 𝑑2 𝑃 = 𝛼. 𝑡. 𝜋. 𝐸. 𝑑1. 𝑑2 4 Max stress 𝜎𝑚𝑎𝑥 = 𝑃 𝐴 = 𝛼. 𝑡. 𝜋. 𝐸. 𝑑1. 𝑑2 4 𝜋 4 𝑑2 2 = 𝜶. 𝒕. 𝑬. 𝒅𝟏 𝒅𝟐 If we substitute 𝑑1 = 𝑑2 we get 𝝈 = 𝜶. 𝒕. 𝑬 𝑑1 𝑑2 𝐿 𝐿1 𝐿2 Thermal stresses in bars of varying section We know that as a result of the increase in temperature, the bar will tend to expand. But since it is fixed at its ends, therefore it will cause some compressive stress in the body. Moreover, as the thermal stress is shared equally by both the portions, therefore 𝜎1𝐴1 = 𝜎2𝐴2 Total deformation of the body ∆𝐿 = ∆𝐿1 + ∆𝐿2 = 𝜎1𝐿1 𝐸1 + 𝜎2𝐿2 𝐸2
  • 15.
    NUMERICAL 1. A rod2m long at a temperature of 10℃. Find the expansion of the rod when the temparature rise to 80℃ . If the expansion is prevented. Find the stress induceed in the material. 𝛼 = 0.000012℃; 𝐸 = 1.0 × 105 𝑀𝑁/𝑚2 ∆𝐿 = 𝛼. 𝑡. 𝐿 = 0.000012 80 − 10 2 = 1.68 × 10−3 𝑚 Stress induced 𝜎 = 𝐸. 𝑡. 𝛼 = 1.0 × 1011 × 70 × 0.000012 = 8.4 × 107 𝑁 𝑚2 2. Steel rod of 3cm diameter and 5m long is connected to two bricks. And the rod is maintained at temperature of 95℃. Determine the stress and pull excerted when temperature falls to 30℃. i. The ends do not yield ii. The ends yeild 0.12 cm Stress 𝜎 = 𝛼. 𝐸. 𝑡 = 3 × 10−2 × 2 × 1011 × 95 − 30 × 11 × 10−6 = 4.29 × 106 𝑁 𝑚2 Stress when ends yield 𝜎 = 𝐸. 𝛼. 𝑡 − ∆ 𝐿 = 2 × 1011 × 11 × 10−6 × 65 − 1.2 × 10−3 3 × 10−2 𝜎 = 7.85 × 109 𝑁/𝑚2
  • 16.
    PRINCIPLE OF SUPERPOSITION The principle of superposition states that when there are numbers of loads are acting together on an elastic material, the resultant strain will be the sum of individual strains caused by each load acting separately.
  • 17.
    50 kN P kN120 kN 40 kN 60 kN 0.5 m 0.5 m 0.5 m 0.5 m 50 kN 50 kN 60 kN 60 kN 20 kN 20 kN 140 kN 140 kN From the above figure 𝑃 = 90 𝑘𝑁 𝐸 = 2 × 105 𝑁 𝑚𝑚2 = 2 × 1011 𝑁 𝑚2 We know ∆𝐿1 = 𝑃1𝐿1 𝐴1𝐸 = 50000 × 0.5 𝜋 4 0.1 2 × 2 × 1011 = 1.6 × 10−5 𝑚 ∆𝐿2 = 𝑃2𝐿2 𝐴2𝐸 = 140000 × 0.5 𝜋 4 0.2 2 × 2 × 1011 = 1.1 × 10−5 𝑚 ∆𝐿3 = 4𝑃𝐿 𝜋𝐸𝑑1𝑑2 = 4 × 0.5 × 20000 𝜋 × 2 × 1011 × 0.2 × 0.4 = 7.96 × 10−7 𝑚 ∆𝐿4 = 𝑃4𝐿4 𝐴4𝐸 = 0.5 × 60000 𝜋 4 0.4 2 × 2 × 1011 = 1.19 × 10−6 𝑚 Now ∆𝐿 = ∆𝐿1 + ∆𝐿2 + ∆𝐿3 + ∆𝐿4 = 1.6 × 10−5 + 1.1 × 10−5 + 7.96 × 10−7 + 1.19 × 10−6 = 2.8989 × 10−5 𝑚 10 cm 20 cm 40 cm 20 cm 40 cm PRINCIPLE OF SUPER POSITION
  • 18.
    ELASTICITY Elasticity, ability ofa deformed material body to return to its original shape and size when the forces causing the deformation are removed. Hooke's law The modern theory of elasticity generalizes Hooke's law to say that the strain (deformation) of an elastic object or material is proportional to the stress applied to it. Mathematically 𝐸 = 𝐹 𝐴 ∆𝐿 𝐿 = 𝐹𝐿 𝐴∆𝐿 E is called elastic constant or young modulus 𝐿 ∆𝐿 + 𝐿 Relation between K, G and E The relation among different elastic modulas/constant • 𝐾 = 𝑚𝐸 3(𝑚−2) • 𝐺 = 𝑚𝐸 2(1+𝑚) Here • m = Poisson’s ratio • E = Elasticity • G = Rigidity modulus • K = Bulk modulus
  • 19.
    THIN CYLINDRICAL ANDSPHERICAL SHELLS In engineering field, we daily come across vessels of cylindrical and spherical shapes containing fluids such as tanks, boilers, compressed air receivers etc. Generally, the walls of such vessels are very thin as compared to their diameters. These vessels, when empty, are subjected to atmospheric pressure internally as well as externally. In such a case, the resultant pressure on the walls of the shell is zero. But whenever a vessel is subjected to internal pressure (due to steam, compressed air etc.) its walls are subjected to tensile stresses. In general, if the thickness of the wall of a shell is less than 1/10th to 1/15th of its diameter, it is known as a thin shell. Failure of a cylindrical shell due to an internal pressure if these stresses exceed the permissible limit, the cylinder is likely to fail in any one of the following two ways as shown in Fig. 1. It may split up into two troughs 2. It may split up into two cylinder Stresses in thin cylindrical shells A little consideration will show that the walls of the cylindrical shell will be subjected to the following two types of tensile stresses: 1. Circumferential stress or Hoop stress 2. Longitudinal stress.
  • 20.
    THIN CYLINDRICAL ANDSPHERICAL SHELLS Circumferential stress Consider a thin cylindrical shell subjected to an internal pressure as shown in Fig. We know that as a result of the internal pressure, the cylinder has a tendency to split up into two troughs as shown in the figure. Let • L = Length of the shell, • d = Diameter of the shell, • t = Thickness of the shell and • p = Intensity of internal pressure. Total pressure along the diameter (say X-X axis) of the shell, 𝑃 = Intensity of the pressure × area = 𝑝. 𝑑𝑥. 𝐿 and circumferential stress in the shell, 𝜎𝑐 = Total pressure Resisting section = 𝑝. 𝑑. 𝐿 2𝑡𝐿 = 𝑝𝑑 2𝑡 Longitudinal stress Consider the same cylindrical shell, subjected to the same internal pressure as shown in Fig. We know that as a result of the internal pressure, the cylinder also has a tendency to split into two pieces as shown in the figure. Total pressure along its length (say Y-Y axis) of the shell 𝑃 = Intensity of the pressure × area = 𝑝. 𝜋 4 . 𝑑2 Longitudinal stress in the shell 𝜎𝑙 = Total pressure Resisting section = 𝑝. 𝜋 4 . 𝑑2 𝜋. 𝑡. 𝑑 = 𝑝𝑑 4𝑡 = 1 2 𝜎𝑐
  • 21.
    THIN CYLINDRICAL ANDSPHERICAL SHELLS Change in dimension due to internal pressure We know, 𝜎𝑐 = 𝑝𝑑 2𝑡 ; 𝜎𝑙 = 𝑝𝑑 4𝑡 ; poison′ s ratio = 1 m ; ΔL = Change in length; Δd = Change in diameter Change in length ΔL = 𝜀𝐿. 𝐿 → εL = ΔL L = 𝜎𝐿 𝐸 − 1 𝑚 . 𝜎𝑐 𝐸 = 1 𝐸 𝑝𝑑 4𝑡 − 1 𝑚 . 𝑝𝑑 2𝑡 ΔL L = 1 𝐸 . 𝑝𝑑 2𝑡 1 2 − 1 𝑚 𝚫𝐋 = 𝒑𝒅𝑳 𝟐𝒕𝑬 𝟏 𝟐 − 𝟏 𝒎 𝜺𝑳 = 𝒑𝒅 𝟐𝒕𝑬 𝟏 𝟐 − 𝟏 𝒎 Change in diameter Δd = 𝜀𝑑. 𝑑 → εd = Δd d = σc E − 1 𝑚 . 𝜎𝐿 𝐸 = 𝑝𝑑 2𝑡𝐸 − 1 𝑚 . 𝑝𝑑 4𝑡𝐸 Δd d = 𝑝𝑑 2𝑡𝐸 1 − 1 2𝑚 𝚫𝐝 = 𝒑𝒅𝟐 𝟐𝒕𝑬 𝟏 − 𝟏 𝟐𝒎 𝜺𝒅 = 𝒑𝒅 𝟐𝒕𝑬 𝟏 − 𝟏 𝟐𝒎 Change in volume ΔV = 𝑉 + ΔV − V ∆𝑉 = 𝜋 4 𝑑 + Δ𝑑 2 𝐿 + ∆𝐿 − 𝜋 4 𝑑2 . 𝐿 ∆𝑉 = 𝜋 4 𝑑2 ∆𝐿 + 2∆𝑑. 𝑑. 𝐿 ∆𝑉 𝑉 = 𝜋 4 𝑑2 ∆𝐿 + 2∆𝑑. 𝑑. 𝐿 𝜋/4 𝑑2. 𝐿 = ∆𝐿 𝐿 + 2 ∆𝑑 𝑑 = 𝜀𝐿 + 2𝜀𝑐 . 𝑉
  • 22.
    NUMERICAL 1. Given that 𝑑= 800 𝑚𝑚 = 0.8 𝑚; 𝑡 = 10 𝑚𝑚 = 0.01; 𝑃 = 2.5 𝑀𝑃𝑎 = 2.5 × 106 𝑁/𝑚2 𝜎𝑐 = 𝑃𝑑 2𝑡 = 2.5 × 109 × 0.8 2 × 0.01 = 1 × 108 𝑁/𝑚2 𝜎𝑙 = 0.5 × 𝜎𝑙 = 1 × 1011 × 0.5 = 5 × 107 𝑁/𝑚2 2. Given that 𝑑 = 40 𝑚𝑚 = 0.04 𝑚; 𝑡 = 5 𝑚𝑚 = 0.005; 𝜎𝑙 = 30 𝑀𝑃𝑎 = 30 × 109 𝑁/𝑚2 𝜎𝑙 = 𝑃𝑑 2𝑡 → 30 × 106 = 𝑃 × 0.04 2 × 0.005 → 𝑃 = 30 × 106 4 = 7.5 × 106 𝑁/𝑚2 3. Given that 𝑑 = 800 𝑚𝑚 = 0.8 𝑚; 𝑡 = 10 𝑚𝑚 = 0.01; 𝑃 = 2.5 𝑀𝑃𝑎 = 2.5 × 106 𝑁/𝑚2 𝐿 = 4𝑚; 1 𝑚 = 0.25; 𝐸 = 200 × 109 𝑁/𝑚2 Δd = 𝑃𝑑2 2𝐸𝑡 1 − 1 2𝑚 = 2.5 × 106 × 0.8 2 2 × 200 × 109 × 0.01 1 − 0.25 2 Δd = 3.5 × 10−4𝑚 = 0.35 𝑚𝑚 ∆𝑙 = 𝑃𝑑𝑙 2𝑡𝐸 1 2 − 1 𝑚 = 2.5 × 106 × 4 × 0.8 2 × 200 × 109 × 0.01 0.5 − 0.25 ∆𝑙 = 5 × 10−4 = 0.5 𝑚𝑚
  • 23.
    THIN CYLINDRICAL ANDSPHERICAL SHELLS Thin spherical shell Consider a thin spherical shell subjected to an internal pressure as shown in Fig. We know that as a result of the internal pressure, the shell is likely to be torn away along the centre of the sphere. Let, • d = Diameter of the shell, • t = Thickness of the shell and • p = Intensity of internal pressure. 𝑃 = Intensity of the pressure × area = 𝑝. 𝜋 4 . 𝑑2 Stress in the shell, 𝜎 = Total pressure Resisting section = 𝑝. 𝜋 4 . 𝑑2 𝜋. 𝑡. 𝑑 = 𝑝𝑑 4𝑡
  • 24.
    TORTION OF CIRCULARSHAFT Shear stress in a circular shaft subjected to torsion Consider a circular shaft fixed at one end subjected to a torque at the other end as shown in fig. We know, Shear strain = Deformation per unit length Φ = 𝐴𝐴′ 𝐶𝐴 = 𝐴𝐴′ 𝑙 = 𝑅. 𝜃 𝑙 Also Shear stress = 𝜏; Modulas of rigidity = C 𝐶 = 𝜏 Φ 𝜏 = 𝐶. Φ = C. 𝑅. 𝜃 𝑙 𝜏 𝑅 = 𝐶. 𝜃 𝑙 If 𝜏𝑥 is the intensity of shear stress at a distance 𝑥 from the centre of the shaft 𝜏𝑥 𝑥 = 𝜏 𝑅 = 𝐶. 𝜃 𝑙 Maximum torque transmitted by a circular solid shaft We know, Shear stress at x 𝜏𝑥 𝑥 = 𝜏 𝑅 Shear stress at that section, 𝜏𝑥 = 𝜏 𝑅 . 𝑥 Turning force at that section 𝐹 = Stress × Area 𝐹 = 𝜏 𝑅 . 𝑥 2𝜋𝑥. 𝑑𝑥 𝐹 = 𝜏 𝑅 . 2𝜋𝑥2 . 𝑑𝑥
  • 25.
    TORTION OF CIRCULARSHAFT 𝐹 = 𝜏 𝑅 . 2𝜋𝑥2 . 𝑑𝑥 Turning moment at this section 𝑑𝑇 = 𝜏 𝑅 . 2𝜋𝑥2. 𝑑𝑥 . 𝑥 𝑑𝑇 = 𝜏 𝑅 . 2𝜋𝑥3 . 𝑑𝑥 0 𝑇 𝑑𝑇 = 0 𝑅 𝜏 𝑅 . 2𝜋𝑥3 . 𝑑𝑥 𝑇 = 2𝜋𝜏 𝑅 𝑥4 4 0 𝑅 = 2𝜋𝜏 𝑅 𝑅4 4 𝑇 = 𝜋𝜏𝑅3 2 = 𝜋 16 × 𝐷3 × 𝜏 Polar moment of inertia Reletion between angle of twist and torque 𝑇 = 𝜋 16 × 𝜏 × 𝐷3 → 𝜏 = 𝑇 × 16 𝜋 × 1 𝐷3 Also 𝜏 = 𝐶. 𝑅. 𝜃 𝑙 Equiting the both equation 𝑇 × 16 𝜋 × 1 𝐷3 = 𝐶. 𝑅. 𝜃 𝑙 𝑇 × 16 𝜋 × 1 𝐷3 × 1 𝑅 = 𝐶. 𝜃 𝑙 𝑇32 𝜋𝐷4 = 𝐶𝜃 𝑙 → 𝑇 𝜋 32 𝐷4 = 𝐶𝜃 𝑙 We consider 𝐽 = 𝜋 32 𝐷4 This is called polar moment of inertia. Then we get 𝑻 𝑱 = 𝑪𝜽 𝒍 = 𝝉 𝑹
  • 26.
    TORTION OF CIRCULARSHAFT Maximum torque transmitted by a circular hollow shaft 𝑇 = 𝜋 16 × 𝜏 × 𝐷4 − 𝑑4 𝐷 Polar moment of inertia 𝐽 = 𝜋 16 𝐷4 − 𝑑4 Power transmitted by a shaft We have already discussed that the main purpose of a shaft is to transmit power from one shaft to another. 𝑃 = 2𝜋𝑁𝑇 60 The unit will be watt T = Torque N = r. p. m
  • 27.
    BENDING MOMENT ANDSHEAR FORCE Shear force Shear force is the force that acts perpendicular to the longitudinal axis of a structural member, such as a beam or a column, and tends to cause the member to slide or shear along its cross section. In other words, it is the force that is exerted on a section of a structure perpendicular to the plane of the section. Shear force is usually denoted by the symbol V and its unit is Newtons (N) or pounds (lb). • When a force acts downwards on the left-hand side of the section, the shear force is positive (+) and acts upwards on the right-hand side of the section. Conversely, when a force acts upwards on the left-hand side of the section, the shear force is negative (-) and acts downwards on the right-hand side of the section. • The shear at any given point of a beam is positive when the external forces (loads and reactions) acting on the beam tend to shear off the beam at that point as indicated in Fig. • In other words, the sign convention of shear force is such that it is positive when it tends to create a clockwise moment on the section, and negative when it tends to create an anticlockwise moment on the section.
  • 28.
    BENDING MOMENT ANDSHEAR FORCE Bending moment Bending moment, on the other hand, is the moment that tends to cause a structural member to bend or deform around a neutral axis. It is the algebraic sum of the moments about the neutral axis of all the forces acting on one side of the section. Bending moment is usually denoted by the symbol M and its unit is Newton-meters (Nm) or pound-feet (lb-ft). • The sign convention of bending moment is based on the direction of the moment acting on a section of a structure. If the moment is causing clockwise rotation on a section, it is considered negative, and if it is causing counter clockwise rotation, it is considered positive. • The bending moment at any given point of a beam is positive when the external forces acting on the beam tend to bend the beam at that point as indicated in Fig. • For example, if a beam is supported at both ends and a load is applied at the midpoint of the beam, the bending moment will be maximum at the midpoint and will be negative on the right side of the section and positive on the left side of the section, as the moment is causing clockwise rotation on the right side and counter clockwise rotation on the left side.
  • 29.
    BENDING MOMENT ANDSHEAR FORCE Shear force and bending moment diagram A shear force diagram (SFD) shows the variation of shear force along the length of a structural member. The diagram is plotted with shear force on the y-axis and the length of the member on the x-axis. The diagram shows positive values of shear force as upward or rightward forces and negative values of shear force as downward or leftward forces. A bending moment diagram (BMD) shows the variation of bending moment along the length of a structural member. The diagram is plotted with bending moment on the y-axis and the length of the member on the x-axis. Simply supported beam with a point load at its mid point: 𝐴 𝐵 𝑊 𝐿 𝑅𝐴 = 𝑊 2 𝑅𝐵 = 𝑊 2 Shear force at A and B Since the load is in the mid point, the reaction force will be, 𝑅𝐴 = 𝑅𝐵 = 𝑊 2 Bending at poin C with respect to A and B is same 𝑀𝑐 = 𝑅𝐴 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑊 2 × 𝐿 2 = 𝑊𝐿 4 𝐶 𝐶 𝑊𝐿 4 Simply supported beam with a point load at any point: 𝐴 𝐵 𝑊 𝐿 𝑅𝐴 = 𝑊𝑏 𝐿 𝑅𝐵 = 𝑊𝑎 𝐿 𝐶 𝑏 𝑎 Reaction force at A 𝑅𝐴 × 𝐿 − 𝑊 × 𝑏 = 0; 𝑅𝐴 = 𝑊𝑏 𝐿 Reaction force at B 𝑅𝐵 × 𝐿 − 𝑊 × 𝑎 = 0; 𝑅𝐵 = 𝑊𝑎 𝐿 Bending moment at C 𝑀𝐶 = 𝑅𝐴 × 𝑎 = 𝑊𝑎𝑏 𝐿 ; 𝑀𝐶 = 𝑅𝐵 × 𝑏 = 𝑊𝑎𝑏 𝐿 𝐶 𝑊𝑎𝑏 𝐿
  • 30.
    BENDING MOMENT ANDSHEAR FORCE Reaction force at A 𝑅𝐴 × 2.5 − 2 × 1.5 − 4 × 1 = 0; 𝑅𝐴 = 7 2.5 = 2.8 Reaction force at B 𝑅𝐵 × 2.5 − 4 × 1.5 − 2 × 1 = 0; 𝑅𝐵 = 3.2 Bending moment at C 𝑀𝐶 = 𝑅𝐴 × 1 = 2.8 × 1 = 2.8 Bending moment at D 𝑀𝐷 = 𝑅𝐴 × 1.5 = 2.8 × 1.5 − 2 × 0.5 = 3.2 Shear force diagram 𝐹𝐴 = +𝑅𝐴 = +2.8; 𝐹𝐶 = 2.8 − 2 = 0.8; 𝐹𝐷 = 0.8 − 4 = −3.2; 𝐹𝐵 = −3.2; Bending moment diagram 𝑀𝐴 = 𝑀𝐵 = 0; 𝑀𝐶 = 2.8; 𝑀𝐷 = 3.2
  • 31.
    BENDING MOMENT ANDSHEAR FORCE Simply suppoted beam with uniformly distributed load 𝐴 𝐵 𝑤 = 𝑢𝑛𝑖𝑡 𝑙𝑜𝑎𝑑 𝐿 𝑅𝐴 = 𝑤𝐿 2 𝑅𝐵 = 𝑤𝐿 2 𝐶 𝐶 𝑤𝐿2 8 𝑥 Reaction force at A and B 𝑅𝐴 = 𝑅𝐵 = 𝑤𝐿 2 Shear force diagram 𝐹𝐴 = +𝑅𝐴 = + 𝑤𝐿 2 ; 𝐹𝐶 = 𝑤𝐿 2 − 𝑤𝐿 2 = 0; 𝐹𝐵 = − 𝑤𝐿 2 ; 𝐹𝑥 = −𝑅𝐵 + 𝑤𝑥 = 𝑤𝑥 − 𝑤𝐿 2 Bending moment diagram 𝑀𝐴 = 𝑀𝐵 = 0; 𝑀𝐶 = 𝑅𝐴. 𝐿 2 = 𝑤𝐿2 4 − 𝑤𝐿 2 . 𝐿 4 = 𝑤𝐿2 8 ; 𝑀𝑥 = 𝑅𝐵. 𝑥 = 𝑤𝐿𝑥 2 − 𝑤𝑥. 𝑥 2 = 𝑤𝐿𝑥 2 − 𝑤𝑥2 2 𝐴 𝐵 𝑤 = 1000 𝑘𝑔/𝑚 4𝑚 𝐶 𝐸 𝑅𝐴 = 𝑅𝐵 = 1000 × 4 2 = 2000 SFD 𝐹𝐴 = 2000; 𝐹𝑐 = 2000 − 1000 × 2 = 0: 𝐹𝐵 = −2000; BMD 𝑀𝐴 = 𝑀𝐵 = 0 𝑀𝐷 = 2000 × 1 − 1000 × 1 × 1 2 = 1500; 𝑀𝐶 = 2000 × 2 − 1000 × 2 × 2 2 = 2000; 𝐷 1 1 𝐶 𝑤𝐿2 8
  • 32.
    BENDING MOMENT ANDSHEAR FORCE 𝐴 𝐵 𝑤 = 500 𝑘𝑔/𝑚 4 𝐶 𝐸 𝐷 2 2 1000 𝑁 1000 𝑁 Reaction force at A and B 𝑅𝐴 × 8 = 500 × 4 × 4; 𝑅𝐴 = 4000 8 = 1000 𝑅𝐵 = 1000 𝑘𝑔 SFD 𝐹𝐴 = +1000; 𝐹𝐷 = +1000; 𝐹𝐶 = 1000 − 500 × 2 = 0; 𝐹𝐸 = −500; 𝐹𝐵 = −2000; BMD 𝑀𝐴 = 𝑀𝐵 = 0; 𝑀𝐷 = 1000 × 2 = 2000; 𝑀𝐶 = 1000 × 4 − 500 × 2 × 2 2 = 3000; 𝑀𝐸 = 1000 × 6 − 500 × 4 × 4 2 = 2000; 𝐶 𝐸 𝐷 𝐴 𝐵 𝑤 = 500 𝑘𝑔/𝑚 4 𝐶 𝐸 𝐷 2 2 1000 𝑁 1000 𝑁 Reaction force at B (taking moment at A) 𝑅𝐵 × 4 + 1000 × 2 = 500 × 4 × 2 + 1000 × 6 𝑅𝐵 = 8000 4 = 2000 𝑘𝑔 𝑅𝐴 = 𝑅𝐵 = 2000 𝑘𝑔 SFD 𝐹𝑐 = −1000; 𝐹𝐴 = −1000 + 2000 = 1000 𝐹𝐸 = 1000 − 500 × 2 = 0 𝐹𝐵 = 0 − 500 × 2 + 2000 = 1000 𝐹𝐷 = 1000 − 1000 = 0 BMD 𝑀𝑐 = 0; 𝑀𝐴 = −1000 × 2 = −2000 𝑀𝐸 = −1000 × 4 + 2000 × 2 − 500 × 2 × 1 = −1000 𝑀𝐵 = −1000 × 2 = −2000
  • 33.
    Reaction force atB taking moment at A 𝑅𝐵 × 5 = 0.5 × 2 × 4 + 1.5 × 2 × 2 + 0.5 × 1 × 0.5 = 10.25 𝑅𝐵 = 2.05 𝑘𝑁 𝑅𝐴 = 0.5 × 1 + 1.5 × 2 + 0.5 × 2 − 2.05 = 2.45 𝑘𝑁 SFD 𝐹𝐴 = 2.45 𝑘𝑁; 𝐹𝐶 = +2.45 − 0.5 × 1 = 1.95 𝑘𝑁; 𝐹𝐷 = 1.95 − 1.5 × 2 = −1.05 𝑘𝑁; 𝐹𝐵 = −2.05 BMD 𝑀𝐴 = 0; 𝑀𝐶 = 2.45 × 1 − 0.5 × 1 × 0.5 = 2.2 𝑀𝐷 = 2.05 × 2 − 0.5 × 2 × 1 = 3.1 𝑀𝐵 = 0 The maximum bending moment is at M, x distance from c where shear force changes sign. 1.95 𝑥 = 1.05 2 − 𝑥 → 𝑥 = 1.3 𝑚 𝑀𝑀 = 2.45 × 2.3 − 0.5 × 1 × (0.5 + 1.3) − 1.5 × 1.3 × 0.65 = 3.47
  • 34.
    Reaction force atA by taking moment at B 𝑅𝐴 × 5 = 10 × 2 × 3 𝑅𝐴 = 12 𝑘𝑁 Reaction force at A 𝑅𝐵 = 10 × 2 − 12 = 8 𝑘𝑁 Reaction force at A by taking moment at B 𝑅𝐴 × 4 = 4 × 2.5 + 2 × 1 × 2 𝑅𝐴 = 3.5 𝑘𝑁 Reaction force at A 𝑅𝐵 = (2 × 1 + 4) − 3.5 = 2.5 𝑘𝑁 Reaction force at A by taking moment at B 𝑅𝐴 × 6 = 4 × 1.5 × 5.25 + 2 × 3 × 1.5 + (5 × 1.5) 𝑅𝐴 = 8 𝑘𝑁 Reaction force at A 𝑅𝐵 = (4 × 1.5 + 5 × 3 + 5) − 8 = 18 𝑘𝑁
  • 35.
    Reaction force atA by taking moment at B 𝑅𝐴 × 3 = 4.5 × 3 × 1.5 − 4.5 × 1 × 0.5 𝑅𝐴 = 6 𝑘𝑁 Reaction force at A 𝑅𝐵 = 4.5 × 4 − 6 = 12 𝑘𝑁
  • 36.
    Cantilever with pointload at free end BENDING MOMENT AND SHEAR FORCE We know that shear force at any section X from a distance x from free end, is equal to the total unbalance vertical force 𝐹𝑥 = 𝑤 𝑀𝑥 = −𝑤. 𝑥 0.5 0.5 0.5 1000 1000 1000 SFD 𝐹𝐵 = 1000; 𝐹𝑐 = 2000; 𝐹𝐷 = 3000 BMD 𝑀𝐵 = 0; 𝑀𝐶 = −1000 × 0.5 = −500 𝑀𝐷 = −1000 × 1 − 1000 × 0.5 = −1500 𝑀𝐴 = −1000 × 1.5 − 1000 × 1 − 1000 × 0.5 = −3000 B C D
  • 37.
    COLUMNS AND STRUTS Introduction Astructural member, subjected to an axial compressive force, is called a strut. As per definition, a strut may be horizontal, inclined or even vertical. But a vertical strut, used in buildings or frames, is called a column. Failure of a column or struts t has been observed, that when a column or a strut is subjected to some compressive force, then the compressive stress induced, 𝜎 = 𝑃 𝐴 • A little consideration will show, that if the force or load is gradually increased the column will reach a stage, when it will be subjected to the ultimate crushing stress. Beyond this stage, the column will fail by crushing. The load corresponding to the crushing stress, is called crushing load. It has also been experienced that sometimes, a compression member does not fail entirely by crushing, but also by bending i.e., buckling. This happens in the case of long columns. It has also been observed that all the short columns fail due to their crushing. But, • If a long column is subjected to a compressive load, it is subjected to a compressive stress. If the load is gradually increased, column will reach a stage, when it will start buckling. The load, at which the column just buckles is called buckling load, critical load or crippling load and the column is said to have developed an elastic instability. • A little consideration will show that for a long column, the value of buckling load will be less than the crushing load. Moreover, the value of buckling load is low for long columns and relatively high for short columns.
  • 38.
    EULER’S COLUMN THEORY Assumption 1.Initially the column is perfectly straight and the load applied is truly axial. 2. The cross-section of the column is uniform throughout its length. 3. The column material is perfectly elastic, homogeneous and isotropic and thus obeys Hooke's law. 4. The length of column is very large as compared to its cross- sectional dimensions. 5. The shortening of column, due to direct compression (being very small) is neglected. 6. The failure of column occurs due to buckling alone. Sign convention Types of end conditions of column 1. Both ends hinged, 2. Both ends fixed, 3. One end is fixed and the other hinged, and 4. One end is fixed and the other free.
  • 39.
    COLUMN WITH BOTHENDS HINGED Consider a long column AB with a length of l is hinged at its both ends. Its carrying a critical load P at B. As a result of loading the column deflect into a curve AX1B as shown in Fig. Now consider a secttion X at a distance x from A. 𝑃 = Load on the column 𝑦 = Deflection of the column at X Moment due to critical load 𝑀 = −𝑃. 𝑦 𝐸𝐼 𝑑2 𝑦 𝑑𝑥2 = −𝑃. 𝑦 𝐸𝐼 𝑑2𝑦 𝑑𝑥2 + 𝑃. 𝑦 = 0 𝑑2𝑦 𝑑𝑥2 + 𝑃. 𝑦 𝐸𝐼 = 0 The general solution for above differential equation, 𝑦 = 𝐶1. cos 𝑥. 𝑃 𝐸𝐼 + 𝐶2 sin 𝑥. 𝑃 𝐸𝐼 We know, when 𝑥 = 0, 𝑦 = 0, then C1 = 0 When 𝑥 = 𝑙, 𝑦 = 0 then sin 𝑙. 𝑃 𝐸𝐼 = 0 sin 𝑙. 𝑃 𝐸𝐼 = sin 0 = sin 𝜋 = sin 2𝜋 Taking the least significant value 𝑙. 𝑃 𝐸𝐼 = 𝜋 → 𝑃 = 𝝅𝟐𝑬𝑰 𝒍𝟐
  • 40.
    EULER’S FORMULA ANDEQUIVALENT LENGTH In the previous articles, we have derived the relations for the crippling load under various end conditions. Sometimes, all these cases are represented by a general equation called Euler's formula, 𝑃𝐸 = 𝜋2 𝐸𝐼 𝐿𝑒 2 Where 𝐿𝑒 equivalent length. The equivalent length for all the condition are given below. Conditions Relation between 𝒍 and 𝑳𝒆 Formula Both end hinged 𝐿𝑒 = 𝑙 𝑃 = 𝜋2𝐸𝐼 𝑙2 Both end fixed 𝐿𝑒 = 𝑙 2 𝑃 = 4𝜋2𝐸𝐼 𝑙2 One end hinged, other end fixed 𝐿𝑒 = 𝑙 2 𝑃 = 2𝜋2𝐸𝐼 𝑙2 One end fixed, other end free 𝐿𝑒 = 2𝑙 𝑃 = 𝜋2𝐸𝐼 4𝑙2 Slenderness ratio We know that, 𝑃𝐸 = 𝜋2𝐸𝐼 𝐿𝑒 2 Now substituting 𝐼 = 𝐴𝑘2 where 𝐴 is cross sectional area and 𝑘 is the least redius of gyration. 𝑃𝐸 = 𝜋2 𝐸(𝐴𝑘2 ) 𝐿𝑒 2 = 𝑃𝐸 = 𝜋2 𝐴 𝐿𝑒 𝑘 2 Where 𝐿𝑒 𝑘 is called the slenderness ratio.
  • 41.
    PRINCIPAL STRESSES ANDSTRAIN Principal planes Principal planes are the planes within the material such that the resultant stresses across them are completely normal stresses or planes across which the shear stresses is zero. Principal stress Principal stresses are those stresses which are acting on the principal planes. • The plane carrying the maximum normal stress is called the major principal plane and the stress acting on it is called major principal stress. • The plane carrying minimum normal stress is known as minor principal plane and the stress acting on it is called as minor principal stress. Normal stress and shear stress on obli
  • 42.
    • আখলাক অর্থকক? • ভাললা আখলাক এর উধারণ দাও? ৫টি • সৃটির সসবা কক? • সৃটির সসবায় উধাহরন দাও? ৫টি • "রাহমাতুকিল আলামীন" অর্ থকক? • তুকম ককভালব সদশ সক ভাললাবাসলব কলখ? • ক্ষমা কার গুণ? • মহানবী (স) সকান শহলর দাওয়াত কদলত সেলয় লাকিত হলয়কিললন? • ক্ষমার গুরুত্ব কলখ? • সততা কক? • সতামালদর মলধে সলব থ াত্তম সসই বোক্তি,
  • 43.