# Consider an elastic spring as shown in the Fig.2.4. When the spring is slowly

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1 .3 Strain Energy Consider an elastic spring as shown in the Fig..4. When the spring is slowly pulled, it deflects by a small amount u 1. When the load is removed from the spring, it goes back to the original position. When the spring is pulled by a force, it does some work and this can be calculated once the load-displacement relationship is known. It may be noted that, the spring is a mathematical idealization of the rod being pulled by a force P axially. It is assumed here that the force is applied gradually so that it slowly increases from zero to a maximum value P. Such a load is called static loading, as there are no inertial effects due to motion. Let the load-displacement relationship be as shown in Fig..5. Now, work done by the external force may be calculated as, 1 1 W ext = Pu 1 1 = ( force displacement) (.7)

2 The area enclosed by force-displacement curve gives the total work done by the externally applied load. Here it is assumed that the energy is conserved i.e. the work done by gradually applied loads is equal to energy stored in the structure. This internal energy is known as strain energy. Now strain energy stored in a spring is 1 U = P1u 1 (.8)

3 Work and energy are expressed in the same units. In SI system, the unit of work and energy is the joule (J), which is equal to one Newton metre (N.m). The strain energy may also be defined as the internal work done by the stress resultants in moving through the corresponding deformations. Consider an infinitesimal element within a three dimensional homogeneous and isotropic material. In the most general case, the state of stress acting on such an element may be as σ, σ and σ and shear stresses shown in Fig..6. There are normal stresses ( x y z) ( xy, yz and zx) τ τ τ acting on the element. Corresponding to normal and shear stresses we have normal and shear strains. Now strain energy may be written as, U 1 T = σ ε dv (.9) v T in which σ is the transpose of the stress column vector i.e., T T { σ} = ( σx, σ y, σz, τxy, τ yz, τzx ) and { ε} ( εx, εy, εz, εxy, εyz, εzx ) = (.1)

4 The strain energy may be further classified as elastic strain energy and inelastic strain energy as shown in Fig..7. If the force P is removed then the spring shortens. When the elastic limit of the spring is not exceeded, then on removal of load, the spring regains its original shape. If the elastic limit of the material is exceeded, a permanent set will remain on removal of load. In the present case, load the spring beyond its elastic limit. Then we obtain the load-displacement curve OABCDO as shown in Fig..7. Now if at B, the load is removed, the spring gradually shortens. However, a permanent set of OD is till retained. The shaded area BCD is known as the elastic strain energy. This can be recovered upon removing the load. The area OABDOrepresents the inelastic portion of strain energy. The area OABCDOcorresponds to strain energy stored in the structure. The area OABEO is defined as the complementary strain energy. For the linearly elastic structure it may be seen that Area OBC = Area OBE i.e. Strain energy = Complementary strain energy This is not the case always as observed from Fig..7. The complementary energy has no physical meaning. The definition is being used for its convenience in structural analysis as will be clear from the subsequent chapters.

5 Usually structural member is subjected to any one or the combination of bending moment; shear force, axial force and twisting moment. The member resists these external actions by internal stresses. In this section, the internal stresses induced in the structure due to external forces and the associated displacements are calculated for different actions. Knowing internal stresses due to individual forces, one could calculate the resulting stress distribution due to combination of external forces by the method of superposition. After knowing internal stresses and deformations, one could easily evaluate strain energy stored in a simple beam due to axial, bending, shear and torsional deformations..3.1 Strain energy under axial load Consider a member of constant cross sectional area A, subjected to axial force P as shown in Fig..8. Let E be the Young s modulus of the material. Let the member be under equilibrium under the action of this force, which is applied through the centroid of the cross section. Now, the applied force P is resisted by P uniformly distributed internal stresses given by average stress σ = as shown A by the free body diagram (vide Fig..8). Under the action of axial load P applied at one end gradually, the beam gets elongated by (say) u. This may be calculated as follows. The incremental elongation du of small element of length dx of beam is given by, σ P du = ε dx = dx = dx (.11) E AE Now the total elongation of the member of length L may be obtained by integration L P u = dx (.1) AE

6 1 Now the work done by external loads W = Pu (.13) In a conservative system, the external work is stored as the internal strain energy. Hence, the strain energy stored in the bar in axial deformation is, U 1 = Pu (.14) Substituting equation (.1) in (.14) we get,

7 U L P = dx (.15) AE.3. Strain energy due to bending Consider a prismatic beam subjected to loads as shown in the Fig..9. The loads are assumed to act on the beam in a plane containing the axis of symmetry of the cross section and the beam axis. It is assumed that the transverse cross sections (such as AB and CD), which are perpendicular to centroidal axis, remain plane and perpendicular to the centroidal axis of beam (as shown in Fig.9).

8

9 Consider a small segment of beam of length ds subjected to bending moment as shown in the Fig..9. Now one cross section rotates about another cross section by a small amount d θ. From the figure, M d θ = 1 ds = ds (.16) R EI where R is the radius of curvature of the bent beam and EI is the flexural rigidity of the beam. Now the work done by the moment M while rotating through angle dθ will be stored in the segment of beam as strain energy du. Hence, 1 du = M dθ (.17) Substituting for d θ in equation (.17), we get, du 1 M = ds (.18) EI Now, the energy stored in the complete beam of span L may be obtained by integrating equation (.18). Thus, U L M = ds (.19) EI

10 .3.3 Strain energy due to transverse shear

11 The shearing stress on a cross section of beam of rectangular cross section may be found out by the relation VQ τ = (.) bi ZZ where Q is the first moment of the portion of the cross-sectional area above the point where shear stress is required about neutral axis, V is the transverse shear force, b is the width of the rectangular cross-section and I zz is the moment of inertia of the cross-sectional area about the neutral axis. Due to shear stress, the angle between the lines which are originally at right angle will change. The shear stress varies across the height in a parabolic manner in the case of a rectangular cross-section. Also, the shear stress distribution is different for different shape of the cross section. However, to simplify the computation shear stress is assumed to be uniform (which is strictly not correct) across the cross section. Consider a segment of length ds subjected to shear stressτ. The shear stress across the cross section may be taken as V τ = k A in which A is area of the cross-section and k is the form factor which is dependent on the shape of the cross section. One could write, the deformation du as where Δγ is the shear strain and is given by du = Δγ ds (.1) τ V Δ γ = = k (.) G AG Hence, the total deformation of the beam due to the action of shear force is L V u = k ds (.3) AG Now the strain energy stored in the beam due to the action of transverse shear force is given by, 1 L kv U = Vu = ds (.4) AG

12 The strain energy due to transverse shear stress is very low compared to strain energy due to bending and hence is usually neglected. Thus the error induced in assuming a uniform shear stress across the cross section is very small..3.4 Strain energy due to torsion Consider a circular shaft of length L radius R, subjected to a torque T at one end (see Fig..11). Under the action of torque one end of the shaft rotates with respect to the fixed end by an angle d φ. Hence the strain energy stored in the shaft is, 1 U = Tφ (.5)

13 Consider an elemental length ds of the shaft. Let the one end rotates by a small amount d φ with respect to another end. Now the strain energy stored in the elemental length is, 1 du = Tdφ (.6) We know that Tds d φ = (.7) GJ where, G is the shear modulus of the shaft material and J is the polar moment of area. Substituting for d φ from (.7) in equation (.6), we obtain T du = ds (.8) GJ Now, the total strain energy stored in the beam may be obtained by integrating the above equation. = L T U ds (.9) GJ Hence the elastic strain energy stored in a member of length s (it may be curved or straight) due to axial force, bending moment, shear force and torsion is summarized below. P 1. Due to axial force U1 = ds AE M. Due to bending U = ds EI V 3. Due to shear U 3 = ds AG T 4. Due to torsion U 4 = ds GJ s s s s

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