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- Strength Of Materials - SOM
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S K Mondal’s Strength of Materials Contents Chapter – 1: Stress and Strain Chapter - 2 : Principal Stress and Strain Chapter - 3 : Moment of Inertia and Centroid Chapter - 4 : Bending Moment and Shear Force Diagram Chapter - 5 : Deflection of Beam Chapter - 6 : Bending Stress in Beam Chapter - 7 : Shear Stress in Beam Chapter - 8 : Fixed and Continuous Beam Chapter - 9 : Torsion Chapter-10 : Thin Cylinder Chapter-11 : Thick Cylinder Chapter-12 : Spring Chapter-13 : Theories of Column Chapter-14 : Strain Energy Method Chapter-15 : Theories of Failure Chapter-16 : Riveted and Welded Joint Er. S K Mondal IES Officer (Railway), GATE topper, NTPC ET-2003 batch, 12 years teaching experienced, Author of Hydro Power Familiarization (NTPC Ltd) Page 1 of 429

Note “Asked Objective Questions” is the total collection of questions from:20 yrs IES (2010-1992) [Engineering Service Examination] 21 yrs. GATE (2011-1992) and 14 yrs. IAS (Prelim.) [Civil Service Preliminary] Copyright © 2007 S K Mondal Every effort has been made to see that there are no errors (typographical or otherwise) in the material presented. However, it is still possible that there are a few errors (serious or otherwise). I would be thankful to the readers if they are brought to my attention at the following e-mail address: swapan_mondal_01@yahoo.co.in S K Mondal Page 2 of 429

1. Stress and Strain Theory at a Glance (for IES, GATE, PSU) 1.1 Stress (ı) When a material is subjected to an external force, a resisting force is set up within the component. The internal resistance force per unit area acting on a material or intensity of the forces distributed over a given section is called the stress at a point. x It uses original cross section area of the specimen and also known as engineering stress or conventional stress. Therefore, T x P A P is expressed in Newton (N) and A, original area, in square meters (m2), the stress ǔ will be expresses in N/ m2. This unit is called Pascal (Pa). x As Pascal is a small quantity, in practice, multiples of this unit is used. 1 kPa = 103 Pa = 103 N/ m2 1 MPa = 106 Pa = 106 N/ m2 (kPa = Kilo Pascal) =1 N/mm2 1 GPa = 109 Pa = 109 N/ m2 (MPa = Mega Pascal) (GPa = Giga Pascal) Let us take an example: A rod 10 mm q 10 mm cross-section is carrying an axial tensile load 10 kN. In this rod the tensile stress developed is given by Tt x P 10 kN 10q103 N 100N/mm2 100MPa 2 A 10 mm q10 mm 100 mm The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. x The force intensity on the shown section is defined as the normal stress. %F P T lim and Tavg %Al 0 %A A x Tensile stress (ıt) If ǔ > 0 the stress is tensile. i.e. The fibres of the component tend to elongate due to the external force. A member subjected to an external force tensile P and tensile stress distribution due to the force is shown in the given figure. Page 3 of 429

Chapter-1 x Stress and Strain S K Mondal’s Compressive stress (ıc) If ǔ < 0 the stress is compressive. i.e. The fibres of the component tend to shorten due to the external force. A member subjected to an external compressive force P and compressive stress distribution due to the force is shown in the given figure. x Shear stress ( U ) When forces are transmitted from one part of a body to other, the stresses developed in a plane parallel to the applied force are the shear stress. Shear stress acts parallel to plane of interest. Forces P is applied transversely to the member AB as shown. The corresponding internal forces act in the plane of section C and are called shearing forces. The corresponding average shear stress U P Area 1.2 Strain (İ) The displacement per unit length (dimensionless) is known as strain. x Tensile strain ( F t) The elongation per unit length as shown in the figure is known as tensile strain. ǆt = ƦL/ Lo It is engineering strain or conventional strain. Here we divide the elongation to original length not actual length (Lo + % L) Let us take an example: A rod 100 mm in original length. When we apply an axial tensile load 10 kN the final length of the rod after application of the load is 100.1 mm. So in this rod tensile strain is developed and is given by Ft x %L L Lo 100.1mm 100 mm 0.1mm 0.001 (Dimensionless)Tensile Lo Lo 100 mm 100 mm Compressive strain ( F c) If the applied force is compressive then the reduction of length per unit length is known as compressive strain. It is negative. Then İc = (-ǻL)/ Lo Let us take an example: A rod 100 mm in original length. When we apply an axial compressive load 10 kN the final length of the rod after application of the load is 99 mm. So in this rod a compressive strain is developed and is given by Page 4 of 429

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