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STRESS
Stress is the internal resistance offered by the body to the external load applied to it per unit cross sectional area. Stresses are normal to the plane to which they act and are tensile or compressive in nature. As we know that in mechanics of deformable solids, externally applied forces acts on a body and body suffers a deformation. From equilibrium point of view, this action should be opposed or reacted by internal forces which are set up within the particles of material due to cohesion. These internal forces give rise to a concept of stress. Consider a rectangular rod subjected to axial pull P. Let us imagine that the same rectangular bar is assumed to be cut into two halves at section XX. Each portion of this rectangular bar is in equilibrium under the action of load P and the internal forces acting at the section XX has been shown. Now stress is defined as the force intensity or force per unit area. Here we use a symbol ๐ to represent the stress.
Where A is the area of the X โ X section
Here we are using an assumption that the total force or total load carried by the rectangular bar is uniformly distributed over its cross โ section. But the stress distributions may be for from uniform, with local regions of high stress known as stress concentrations. If the force carried by a component is not uniformly distributed over its cross โ sectional area, A, we must consider a small area, โฮดAโ which carries a small load โฮดPโ, of the total force โP', Then definition of stress is As a particular stress generally holds true only at a point, therefore it is defined mathematically as UNITS The basic units of stress in S.I units i.e. (International system) are N / m^2 (or Pa) MPa = 10^6 Pa GPa = 10^9 Pa KPa = 10^3 Pa Sometimes N/mm^2 units are also used, because this is an equivalent to MPa. While US customary unit is pound per square inch psi. TYPES OF STRESSES There two basic stresses;
Tensile or compressive Stresses The normal stresses can be either tensile or compressive whether the stresses acts out of the area or into the area Bearing Stress When one object presses against another, it is referred to a bearing stress ( They are in fact the compressive stresses ). Sign convections for Normal stress Direct stresses or normal stresses
- Compressive โ ve (negative). Shear Stresses Let us consider now the situation, where the cross โ sectional area of a block of material is subject to a distribution of forces which are parallel, rather than normal, to the area concerned. Such forces are associated with a shearing of the material, and are referred to as shear forces. The resulting stress is known as shear stress. The resulting force intensities are known as shear stresses, the mean shear stress being equal to ๐ = ๐ ๐ด Where P is the total force and A the area over which it acts. As we know that the particular stress generally holds good only at a point therefore, we can define shear stress at a point as The Greek symbol ๐ (tau, suggesting tangential) is used to denote shear stress.
MEMBERS IN UNI โ AXIAL STATE OF STRESS
Introduction [For members subjected to uniaxial state of stress]. For a prismatic bar loaded in tension by an axial force P, the elongation of the bar can be determined as Suppose the bar is loaded at one or more intermediate positions, then equation (1) can be readily adapted to handle this situation, i.e. we can determine the axial force in each part of the bar i.e. parts AB, BC, CD, and calculate the elongation or shortening of each part separately, finally, these changes in lengths can be added algebraically to obtain the total charge in length of the entire bar.
When either the axial force or the cross โ sectional area varies continuously along the axis of the bar, then equation (1) is no longer suitable. Instead, the elongation can be found by considering a deferential element of a bar and then the equation (1) becomes i.e. the axial force Px and area of the cross โ section Ax must be expressed as functions of x. If the expressions for Px and Ax are not too complicated, the integral can be evaluated analytically, otherwise Numerical methods or techniques can be used to evaluate these integrals. 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. EXAMPLES Problem 1: A steel bar of rectangular cross-section, 3 cm by 2 cm, carries an axial load of 30 KN. Estimate the average tensile stress over a normal cross-section of the bar.
Or, ๐ = ๐ฟ๐ ๐ Elasticity The property of material by virtue of which it returns to its original shape and size upon removal of load is known as elasticity. Hooks Law It states that within elastic limit stress is proportional to strain. Mathematically ๐ธ =
Where, E = Youngโs Modulus Hooks law holds good equally for tension and compression. Poissonโs Ratio The ratio lateral strain to longitudinal strain produced by a single stress is known as Poissonโs ratio. Symbol used for Poissonโs ratio is ฮผ or 1/ m. Modulus of Elasticity (or Youngโs Modulus) Youngโs modulus is defined as the ratio of stress to strain within elastic limit. Deformation of a body due to load acting on it We know that youngโs modulus ๐ธ =
Or, strain ๐ =
=
Now, strain ๐ =
๐ฟ๐ ๐
So, deformation ๐ฟ๐ =
๐๐ฟ ๐ด๐ธ FURTHER EXAMPLES Problem 3 : A cylindrical block is 30 cm long and has a circular cross-section 10 cm in diameter. It carries a total compressive load of 70 kN, and under this load it contracts by 0. cm. Estimate the average compressive stress over a normal cross-section and the compressive strain
Problem 5 : A circular bar of diameter 2.50 cm is subjected to an axial tension of 20 kN. If the material is elastic with a Young's modulus E = 70 GN/m^2 , estimate the percentage elongation. Problem 6: The piston of a hydraulic ram is 40 cm diameter, and the piston rod 6 cm diameter. The water pressure is 1 MN/mm^2. Estimate the stress in the piston rod and the
elongation of a length of 1 m of the rod when the piston is under pressure from the piston-rod side. Take Young's modulus as E = 200 GN/mm^2.
Problem 8: A circular, metal rod of diameter 1 cm is loaded in tension. When the tensile load is 5kN, the extension of a 25 cm length is measured accurately and found to be 0. cm. Estimate the value of Youngโs modulus, E, of the metal.
Problem 9: A straight, uniform rod of length L rotates at uniform angular speed ๐ about an axis through one end and perpendicular to its length. Estimate the maximum tensile stress generated in the rod and the elongation of the rod at this speed. The density of the material is ๐ and Youngโs modulus is E.
TUTORIAL QUESTIONS
- The piston rod of a double-acting hydraulic cylinder is 20 cm diameter and 4 m long. The piston has a diameter of 40 cm, and is subjected to 10 MN/m2 water pressure on one side and 3 MN/m2 on the other. On the return stroke these pressures are interchanged. Estimate the maximum stress occurring in the piston-rod, and the change of length of the rod between two strokes, allowing for the area of piston-rod on one side of the piston. Take E = 200 GN/m^2. [31. 0 MN/m^2 ( compressive); 0.098 cm].
- A uniform steel rope 250 m long hangs down a shaft. Find the elongation of the first 125 m at the top if the density of steel is 7840 kg/m3 and Young's modulus is 200 GN/m^2. [0.902 cm]
- A steel wire, 150 m long, weighs 20 N per metre length. It is placed on a horizontal floor and pulled slowly along by a horizontal force applied to one end. If this force measures 600 N, estimate the increase in length of the wire due to its being towed, assuming a uniform coefficient of friction. Take the density of steel as 7840 kg/m3a nd Young's modulus as 200 GN/m^2. [0.865 cm]
- The hoisting rope for a mine shaft is to lift a cage of weight W. The rope is of variable section so that the stress on every section is equal to (T when the rope is fully extended. If p is the density of the material of the rope, show that the cross-sectional area A at a height z above the cage is
