IV Energy Methods

Strain Energy: definition Strain energy is a form of potential energy. Strain energy is defined as the increase in energy associated with the deformation of the member. Strain energy is equal to the work done by slowly increasing load applied to the member. Work done to distort an elastic member is stored as strain energy. Some energy may be lost in plastic deformation of the member and some may be converted into heat instead of stored as strain energy, but the rest is recoverable. A spring is an example of a storage device for strain energy.

Strain Energy A uniform rod is subjected to a slowly increasing load The elementary work done by the load P as the rod elongates by a small dx is which is equal to the area of width dx under the load-deformation diagram. The total work done by the load for a deformation x1, which results in an increase of strain energy in the rod. In the case of a linear elastic deformation,

Strain Energy Density The strain-energy density of a material is defined as the strain energy per unit volume. To eliminate the effects of size, evaluate the strain- energy per unit volume, The total strain energy density resulting from the deformation is equal to the area under the curve to e1. As the material is unloaded, the stress returns to zero but there is a permanent deformation. Only the strain energy represented by the triangular area is recovered. Remainder of the energy spent in deforming the material is dissipated as heat.

Strain-Energy Density The strain energy density resulting from setting e1 = eR is the modulus of toughness. The energy per unit volume required to cause the material to rupture is related to its ductility as well as its ultimate strength. If the stress remains within the proportional limit, The strain energy density resulting from setting s1 = sY is the modulus of resilience.

Strain Energy under Axial Loading In an element with a nonuniform stress distribution, Under axial loading, For values of u < uY , i.e., below the proportional limit, For a rod of uniform cross-section,

Comparison of Energy Stored in Straight and Stepped bars L/2 L/2 L Da nA A A P P (a) (b) Note for n=2; case (b) has U= which is 3/4 of case (a)

Problem 1 One of the two bolts need to support a sudden tensile loading. To choose it is necessary to determine the greatest amount of strain energy each bolt can absorb. Bolt A has a diameter of 20 mm for 50 mm length and a root diameter of 18 mm for 6 mm length. Bolt B has 18 mm diameter throughout the length. Take E = 210 GPa and y= 310 Mpa.

Solution

Strain Energy in Bending For a beam subjected to a bending load, Setting dV = dA dx, For an end-loaded cantilever beam,

Sample Problem 1 SOLUTION: Determine the reactions at A and B from a free-body diagram of the complete beam. Develop a diagram of the bending moment distribution. Taking into account only the normal stresses due to bending, determine the strain energy of the beam for the loading shown. Evaluate the strain energy knowing that the beam is a W250x67, P = 160kN, L = 3.6m, a = 0.9m, b = 2.7m, and E = 200GPa. Integrate over the volume of the beam to find the strain energy. Apply the particular given conditions to evaluate the strain energy. 4- 11

Sample Problem 1 SOLUTION: Determine the reactions at A and B from a free-body diagram of the complete beam. Develop a diagram of the bending moment distribution.

Sample Problem 1 Integrate over the volume of the beam to find the strain energy.

Problem 2

Solution:

Problem 4

Solution: Finding the support reactions Finding the internal moment

Solution (Contd.)

Problem 5

Solution:

Problem 6

Solution:

Strain Energy For Shearing Stresses For a material subjected to plane shearing stresses, For values of txy within the proportional limit, The total strain energy is found from

Strain Energy in Torsion For a shaft subjected to a torsional load, Setting dV = dA dx, In the case of a uniform shaft,

Strain Energy for Transverse Shear For transverse shear , and hence, The Integral inside the parenthesis can be replaced with f, the form factor, which is dimension less and unique for a specific cross section

Strain Energy for Transverse Shear (cont) Substituting the form factor, fs Form factor for rectangular Cross Section: t=b A=bh I=bh3 /12 Substituting in the form factor equation and simplify , f=6/5 Strain Energy for Transverse Shear (cont) NA

Problem 1

Solution:

Problem 2

Problem 3 Rod AC is made of aluminum (G = 73 GPa) and is subjected to torque T applied at end C. Knowing that portion BC of the rod is hollow and has an inside diameter of 16 mm, determine the strain energy of the rod for a maximum shearing stress of 120 MPa.

Solution

Annoucement Assignment 2 – Author – due 2 weeks Test 1 Marks – ready (tampal kat luar bilik c16-101-11) Test 2 – 13 Dec 2011, 1.5hours, buckling, energy 4 - 32

Impact Loading Impact loading is a dynamic loading i.e. vary with time It occurs when one object strikes the other (collision) Consider a block released from rest falling h distance and strikes the spring to compress it max before it comes to rest Energy of falling is transformed in to strain energy of spring Work done on falling of weight over a distance (h+ max ) is work done on spring to displace it max . Therefore Ue=Ui h max k

Impact Loading (Contd..) Solving the above quadratic equation, the maximum root is , if ,then

IMPACT LOADINGS (cont) If the weight W is applied statically (or gradually to the spring, the end displacement of the spring is ∆st = W/k. Using this simplification, the above equation becomes Once ∆max is computed, the maximum force applied to the spring can be determined from Note: if h = 0; then ∆max = 2 ∆st Copyright © 2011 Pearson Education South Asia Pte Ltd

Impact Loading To determine the maximum stress sm Assume that the kinetic energy is transferred entirely to the structure, Assume that the stress-strain diagram obtained from a static test is also valid under impact loading. Consider a rod which is hit at its end with a body of mass m moving with a velocity v0. Maximum value of the strain energy, Rod deforms under impact. Stresses reach a maximum value sm and then disappear. For the case of a uniform rod,

Example 1 SOLUTION: Due to the change in diameter, the normal stress distribution is nonuniform. Find the static load Pm which produces the same strain energy as the impact. Evaluate the maximum stress resulting from the static load Pm Body of mass m with velocity v0 hits the end of the nonuniform rod BCD. Knowing that the diameter of the portion BC is twice the diameter of portion CD, determine the maximum value of the normal stress in the rod.

Example 1 Find the static load Pm which produces the same strain energy as the impact. Evaluate the maximum stress resulting from the static load Pm SOLUTION: Due to the change in diameter, the normal stress distribution is nonuniform.

Example 2 SOLUTION: The normal stress varies linearly along the length of the beam as across a transverse section. Find the static load Pm which produces the same strain energy as the impact. Evaluate the maximum stress resulting from the static load Pm A block of weight W is dropped from a height h onto the free end of the cantilever beam. Determine the maximum value of the stresses in the beam.

Example 2 Find the static load Pm which produces the same strain energy as the impact. For an end-loaded cantilever beam, SOLUTION: The normal stress varies linearly along the length of the beam as across a transverse section. Evaluate the maximum stress resulting from the static load Pm

Problem 4 The aluminum pipe( OD= 60 mm, t=10 mm) shown is used to support a load of 600 kN. Determine the maximum displacement at the top of the pipe when the load is applied (a) gradually and (b) suddenly from h =0. Take E =70x103N/mm2 when h=0, then =1.1906 mm 240 mm h 600 kN

Problem 5 The cylindrical block E has a speed o= 5 m/s when it strikes squarely the yoke BD that is attached to the 22 mm diameter rods AB and CD. Knowing that the rods are made of a steel for which Y = 345 MPa and E = 200 GPa, determine the weight of the block E for which the factor of safety is five with respect to permanent deformation of the rods.

Solution

Solution (Contd.) Substituting the data

Problem 6

Solution l

Problem 7 The composite aluminium bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum axial stress developed in the bar if the 5kg collar is dropped from a height of h= 100 mm. Eal= 70 GPa and Y=410 MPa

Problem 8 The composite aluminium bar is made from two segments having diameters of 5 mm and 10 mm. Determine the maximum h from which the 5kg collar is dropped so that it produces a maximum axial stress in the bar of max=300 MPa. Eal= 70 GPa and Y=410 MPa 4 - 48

Problem 9 The composite aluminium 2014-T6 bar is made from two segments having diameters of 7.5 mm and 15 mm. Determine the maximum axial stress developed in the bar if the 10 kg collar is dropped from a height of h=100 mm. Eal= 73.1 GPa and Y=414 MPa

Problem 10 The composite aluminium 2014-T6 bar is made from two segments having diameters of 7.5 mm and 15 mm. Determine the maximum height h from which the 10 kg collar should be dropped so that it produces a maximum axial stress in the bar of max=300 MPa. Eal= 73.1 GPa and Y=414 MPa

Castigliano’s Theorem Strain energy for any elastic structure subjected to two concentrated loads, Differentiating with respect to the loads, * Castigliano’s Theorem is named after the Italian engineer, Alberto Castigliano (1847-1884) Castigliano’s theorem: For an elastic structure subjected to n loads, the deflection xj of the point of application of Pj, measured along the line of action of Pj, can be expressed as

Deflections by Castigliano’s Theorem Application of Castigliano’s theorem is simplified if the differentiation with respect to the load Pj is performed before the integration or summation to obtain the strain energy U. In the case of a beam, For a truss,

Problem 7 Determine the displacement of point B on the beam shown Apply an external force at B Using method of section the internal moment and partial derivative are to be determined w L A B P wx V M x

Solution (Contd.)

Problem 8 ( )

Solution (Contd.)

Problem 9

Solution (Contd.)

Solution (Contd.)

Sample Problem SOLUTION: For application of Castigliano’s theorem, introduce a dummy vertical load Q at C. Find the reactions at A and B due to the dummy load from a free-body diagram of the entire truss. Apply the method of joints to determine the axial force in each member due to Q. Members of the truss shown consist of sections of aluminum pipe with the cross-sectional areas indicated. Using E = 73 GPa, determine the vertical deflection of the joint C caused by the load P. evaluate the derivative with respect to Q of the strain energy of the truss due to the loads P and Q. Setting Q = 0, evaluate the derivative which is equivalent to the desired displacement at C.

Sample Problem SOLUTION: Find the reactions at A and B due to a dummy load Q at C from a free-body diagram of the entire truss. Apply the method of joints to determine the axial force in each member due to Q.

Sample Problem Combine with the results of Sample Problem 11.4 to evaluate the derivative with respect to Q of the strain energy of the truss due to the loads P and Q. Setting Q = 0, evaluate the derivative which is equivalent to the desired displacement at C.