Strain Energy Lecture No-5 J P Supale Mechanical Engineering Department SKN SITS LONAVALA Strength of Materials

What is Strain Energy? When a load is applied on a elastic body, it gets deformed i.e. the work is done by the load on the body. This work done is stored in a body is called as Strain energy or Resilience. When the load is removed, the body regains to its original position because of stored energy. It can be easily visualized in springs, when the spring is loaded, it gets compressed (work is done and getting stored) and regain to its original position immediately (because of stored energy), when the load is removed. Strength of Materials

Elastic and inelastic strain energy Recall loading – unloading of a prismatic bar The strain energy that recovers during unloading is called the elastic strain energy (triangle BCD) Area OABDO represents energy that is lost in the process of permanently deforming the bar. This energy is called inelastic strain energy Strength of Materials

Strain energy due to various loads The amount of strain energy stored in a body depends upon type of loads, they are as follow: Gradually Applied Load Suddenly Applied Load Impact Load Strength of Materials

Strain Energy Due to Axial Load Consider a bar of length L placed vertically and one end of it is attached at the ceiling. Let P=axially applied load L=length of bar A=Cross-sectional area of the bar δl=Deflection produced in the bar σ=Axial stress induced in the bar E=Modulus of elasticity of bar material Strength of Materials L δlδl P

Strain Energy Due to Axial Load Work done by the load is given as From Hooke’s law -On substituting δl, equation 1 becomes But, we know work done by load = strain energy, therefore Strength of Materials

Stress Due to Gradual Applied Load When a body subjected to gradual load, the stress increases from zero to maximum value. Let P=Gradually applied load δL=Deflection produced in the bar σ=maximum stress induced in the bar Strength of Materials Work done by load = Avg Load x Deformation Strain Energy stored in a body -  δLδL P

Stress Due to Gradual Applied Load But work done = Strain Energy stored Strength of Materials

Stress Due to Suddenly Applied Load When a body subjected to sudden load, the body gets deformed immediately after applying load. Let P=Sudden applied load δl=Deflection produced in the bar σ=maximum stress induced in the bar Strength of Materials Work done by load = Area under curve Strain Energy stored in a body -  δLδL P

Stress Due to Suddenly Applied Load But work done = Strain Energy stored Therefore, stress due to sudden load is twice to gradually applied load. Strength of Materials

Strain Energy due to Impact Load/Shock Load Consider the situation, when load P is allowed to fall freely on the collar attached to lower end of the bar. Leth=height through which the load falls on the collar Work done by the falling load is given as L δlδl P Collar h It is important to note here that the load does not change and the same load acts throughout the process.

Strain Energy due to Impact Load/Shock Load Strength of Materials Equating work done to strain energy stored in the bar, we have This is a quadratic equation in terms of σ. Its two values are given as

Strain Energy due to Impact Load/Shock Load Strength of Materials The effect of the load is to elongate the bar, hence it produces tensile stress in the bar. Therefore, possibility of negative stress does not arise. Equation 14 accordingly reduces to When h=0, then equation 15 becomes The equation is same as equation 10. Hence in case of zero height stress induced in the bar is equal to stress produced by same load but applied suddenly.

Numerical 1 A solid vertical prismatic steel bar of equilateral triangular section of side 30mm is firmly fixed at the top. A rigid collar at a distance 800mm from the top. Calculate the strain energy stored in each of following case:- a. when a pull of 15KN applied gradually b. when a force of 11KN is suddenly applied c. when a weight of 6KN falls through 180mm before it strikes collar take E= 200x10 3 MPa Strength of Materials

Area of cross Section A= ½ x base x height = ½ x 30 x 30 sin 60 = mm 2 Case A: Gradual Load of 15 KN Strength of Materials

CASE B- Sudden Load of 11 KN Strength of Materials

CASE C- Impact Load of 6KN Strain Energy Strength of Materials

Numerical 2 A load of 10N falls through a height of 2cm on to a collar rigidity attached to the lower end of vertical bar 1.5 m long and of 1.5cm 2 cross-sectional area. The upper end of the vertical bar is fixed. Determine 1) Maximum instantaneous stress induced in the vertical bar 2) Maximum instantaneous elongation 3) Strain energy stored in the vertical rod E=2x10 5 MPa. steps:- 1) calculate =60.23 MPa 2) Then =0.452mm 3) then

Workout Numerical 1 Strength of Materials A rod 12.5mm in diameter is stretched 3.2 under a steady load of 10kN. What stress would be produced in the bar by a weight of 700N, falling through 75 mm before commencing to stretch, the rod being initially unstressed ?. E=2.1 x 10 5 MPa steps:- 1) calculate length of rod (L) L=8246.7mm 2) then = MPa

Workout Numerical 2 A steel rope lowers a load of 20 kN at the rate 2m/sec. During the lowering of the load it is jammed. The unwound length of rope is 15m. What will be instantaneous stress developed and maximum instantaneous elongation? Assume diameter of rope 25mm. Take E=2x10 5 N/mm 2 and g =9.81 m/sec 2. Area of rope, A= mm 2. Kinetic energy of load = ½ mv 2 = Nm. Strain energy in rope, U=(σ 2 /2E)*V=18.41 σ 2 But, KE = Strain Energy from this calculate σ = N/mm 2 Deflection = σL/E = 1.116mm Strength of Materials