Energy_02_Elastic_Strain_Energy

Energy_02_Elastic_Strain_Energy - Section 5.2 5.2 Elastic...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 5.2 Solid Mechanics Part I Kelly 180 5.2 Elastic Strain Energy The strain energy stored in an elastic material upon deformation is calculated below for a number of different geometries and loading conditions. These expressions for stored energy will then be used to solve some elasticity problems using the energy methods mentioned in the previous section. 5.2.1 Strain energy in deformed Components Bar under axial load Consider a bar of elastic material fixed at one end and subjected to a steadily increasing force P , Fig. 5.2.1. The force is applied slowly so that kinetic energies are negligible. The initial length of the bar is L . The work dW done in extending the bar a small amount Δ d is 1 Δ = Pd dW (5.2.1) Figure 5.2.1: a bar loaded by a constant stress It was shown in §4.3.2 that the force and extension Δ are linearly related through EA PL / = Δ , Eqn. 4.3.5, where E is the Young’s modulus and A is the cross sectional area. This linear relationship is plotted in Fig. 5.2.2. The work expressed by Eqn. 5.2.1 is the white region under the force-extension curve (line). The total work done during the complete extension up to a final force P and final extension Δ is the total area beneath the curve. The work done is stored as elastic strain energy U and so EA L P P U 2 2 1 2 = Δ = (5.2.2) If the axial force (and/or the cross-sectional area and Young’s modulus) varies along the bar, then the above calculation can be done for a small element of length dx . The energy stored in this element would be EA dx P 2 / 2 and the total strain energy stored in the bar would be 1 the small change in force during this small extension may be neglected P L Δ d Δ
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Section 5.2 Solid Mechanics Part I Kelly 181 = L dx EA P U 0 2 2 (5.2.3) Figure 5.2.2: force-displacement curve for uniaxial load The strain energy is always positive, due to the square on the force P , regardless of whether the bar is being compressed or elongated. Note the factor of one half in Eqn. 5.2.2. The energy stored is not simply force times displacement because the force is changing during the deformation. Circular Bar in Torsion Consider a circular bar subjected to a torque T . The torque is equivalent to a couple: two forces of magnitude F acting in opposite directions and separated by a distance r 2 as in Fig. 5.2.3; Fr T 2 = . As the bar twists through a small angle φ Δ , the forces each move through a distance Δ = r s . The work done is therefore ( ) Δ = = Δ T Fs W 2 . Figure 5.2.3: torque acting on a circular bar It was shown in §4.4 that the torque and angle of twist are linearly related through Eqn. 4.4.10, GJ TL / = , where L is the length of the bar, G is the shear modulus and J is the polar moment of inertia. The angle of twist can be plotted against the torque as in Fig. 5.2.4. The total strain energy stored in the cylinder during the straining up to a final angle of twist is the work done, equal to the shaded area in Fig. 5.2.4, leading to GJ L T T U 2 2 1 2 = = (5.2.4) Δ d P force-extension curve dW Δ r F Δ
Background image of page 2
Section 5.2 Solid Mechanics Part I Kelly 182 Figure 5.2.4: torque – angle of twist plot for torsion
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

Energy_02_Elastic_Strain_Energy - Section 5.2 5.2 Elastic...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online