1.8 Strain energy due to axial load, bending and shear
(i) Axial Loading
Let us consider a straight bar of length L., having uniform cross-sectional area A. If an axial load P is applied gradually, and if the bar undergoes a deformation A, the work done, stored as strain energy (U) in the body, will be equal to average force (1/2 P ) multiplied by the deformation ∆.
Thus
If, however, the bar has variable area of cross section, consider a small element of length dx and area of cross-section A, . The strain energy dU stored in this small element of length dr will be, from above Eq.
The total strain energy,U can be obtained by integrating the above expression over the length of the bar.
Thus
(ii) Flexural Loading (Moment or Couple) :
Let us now consider a member of length L subjected to uniform bending moment M. Consider an element of length dr, and let di be the change in the slope of the element due to applied moment M. If M is applied gradually, the strain energy stored in the small clement will be,
But
Or
Integrating this over the entire length, we get the total strain energy stored in the member. Thus,
Expression of strain energy due to axial load bending moment and shear forces :
Strain energy under axial load
Consider a member of constant cross sectional area A, subjected to axial force Pas shown in Figure. 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 uniformly distributed internal stresses given by average stress σ = P / A as shown by the free body diagram. 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,
(1)Now the total elongation of the member of length L may be obtained by integration
(2)
Now the work done by external loads (3)
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,
(4)
Substituting equation (2) in (4) we get,
(5)
Strain energy due to bending
Consider a prismatic beam subjected to loads as shown in the Figure. 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 Figure).
Consider a small segment of beam of length ds subjected to bending moment as shown in the Figure. Now one cross section rotates about another cross section by a small amount dθ. From the figure,
(1)
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,
(2)
Substituting for dθ in equation (2), we get,
(3)
Now, the energy stored in the complete beam of span L may be obtained by integrating equation (3). Thus,
Strain energy due to transverse shear
The shearing stress on a cross section of beam of rectangular cross section may be found out by the relation
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, is the width of the rectangular cross-section and IZZ 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 stress τ. The shear stress across the cross section may be taken as
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
Hence, the total deformation of the beam due to the action of shear force is
Now the strain energy stored in the beam due to the action of transverse shear force is given by,
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.
1.9 Theorem of minimum potential energy
The theorem of minimum potential energy can be stated as:
Of all the kinematically admissible displacement functions the actual displacement function is the one that minimizes the potential energy function at stable equilibrium.
Corollary
• The better approximation of displacement function is the one that yields a lower potential energy.
• The greater the degrees of freedom, lower will be the potential energy for a given set of kinematically admissible functions.
Procedure
Let the displacement u(x) be represented by a series of kinematically admissible functions fi (x),
Substitute in Potential Energy: Ω Ω C1 C2 C3……Cn = ( ) , ,
Minimize Potential energy:
Set of n-algebraic equations in the unknown Ci .
1.10 Law of conservation of energy
Law of Conservation of Energy For structural analysis this can be stated as: Consider a structural system that is isolated such it neither gives nor receives energy; the total energy of this system remains constant. The isolation of the structure is key: we can consider a structure isolated once we have identified and accounted for all sources of restraint and loading. For example to neglect the self-weight of a beam is problematic as the beam receives gravitational energy not accounted for possibly leading to collapse. In the spring and force example, we have accounted for all restraints and loading (for example we have ignored gravity by having no mass). Thus the total potential energy of the system Π is constant both before and after the deformation. In structural analysis the relevant forms of energy are the potential energy of the load and the strain energy of the material. We usually ignore heat and other energies.
This is essentially a basic law of physics energy is neither created nor destroyed. For the purpose of structural analysis, the law can be stated in the following form: If a structure and external loads acting on it arc isolated, such that these neither receive nor give out energy then the total energy of the system remains constant. A typical application of the law of conservation of energy can be illustrated by referring to a bar subjected to an axial pull P gradually applied as shown in Figure. When equilibrium is reached it will be found that the bar has ex-tended by an amount S. Considering that the process is adiabatic (heat is neither supplied nor taken out), according to the law of conservation of energy, loss of potential energy P&Q must appear is elsewhere. In this case it is found in the form of strain energy stored figure. in the bar. This is given by
The minus sign for external work is to take into account the loss of potential energy
1.11 Principle of virtual work
Virtual work is the work done during virtual (small) displacements by real forces or the work done by hypothetical forces (imaginary forces) during real small displacements. Virtual displacement is a small displacement given to the body consistent with the boundary conditions of the body.
1.11.1 Problem:
Let us discuss the problem based on the vertical deflection at the free end of the cantilever truss shown in figure.Take cross sectional area of compression members are 850 mm2 and tension members as 1000 mm2. Modulus of elasticity E = 210 GPa for all the members.
Joint c:
– PDc sin 18.43 – 40 = 0
– PDC = – 3.16 W
– PBC – PDC cos 18.43 = 0
– PBC = – 3 W
PBC = 3 W
Joint E:
ED = 0
BE = EC = 3W.
Joint B:
PBD cos 18.43 + Pbe = 0
Joint A:
Pad cos = 0
V0 = 0
Member P I a
BE 3 W 3 3 .1000 27 kN
EC 3 W 3 3 .1000 27 kN.
BD -3.16 W -3.16 3.16 .850 37.12 kN
DC -3.16 W -3.16 3.16 .850 37.12 M
AD 0 – – –
ED 0 – – – = 5.63 × 10-4 mm.