Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To determine the deformation of axially loaded members. To determine the support reactions when these reactions cannot be determined solely from the equations of equilibrium. To analyze the effects of thermal stresses. Copyright © 2011 Pearson Education South Asia Pte Ltd

In-class Activities Reading Quiz Applications Elastic deformation in axially loaded member Principle of superposition Compatibility conditions ‘Force method’ of analysis Thermal Stress Stress Concentration Concept Quiz Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ The stress distributions at different cross sections are different. However, at locations far enough away from the support and the applied load, the stress distribution becomes uniform. This is due to Principle of superposition Inelastic property Poisson’s effect Saint Venant’s Principle Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) The principle of superposition is valid provided that The loading is linearly related to the stress or displacement The loading does not significantly change the original geometry of the member The Poisson’s ratio v ≤ 0.45 Young’s Modulus is small a, b and c a, b and d a and b only All Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 4) Stress concentrations become important in design if the material is brittle the material is ductile but subjected to fatigue loading the material is subjected to fatigue loadings to dynamic loading All of them Copyright © 2011 Pearson Education South Asia Pte Ltd

READING QUIZ (cont) 5) The principle of superposition is applicable to inelastic axial deformation residual stress evaluation large deformation None of the above Copyright © 2011 Pearson Education South Asia Pte Ltd

APPLICATIONS Most concrete columns are reinforced with steel rods; and these two materials work together in supporting the applied load. Are both subjected to axial stress? Copyright © 2011 Pearson Education South Asia Pte Ltd

APPLICATIONS (cont) Stress Concentration Thermal Stress
Inelastic Axial Deformation Copyright © 2011 Pearson Education South Asia Pte Ltd

ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
Provided these quantities do not exceed the proportional limit, we can relate them using Hooke’s Law, i.e. σ = E ε Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 1 The assembly shown in Fig. 4–7a consists of an aluminum tube AB having a cross-sectional area of 400 mm2. A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take Est = 200 GPa, Eal = 70 GPa. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 1 (cont) Solution Find the displacement of end C with respect to end B. Displacement of end B with respect to the fixed end A, Since both displacements are to the right, Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 A member is made from a material that has a specific weight and modulus of elasticity E. If it is in the form of a cone having the dimensions shown in Fig. 4–9a, determine how far its end is displaced due to gravity when it is suspended in the vertical position. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 (cont) Solution Radius x of the cone as a function of y is determined by proportion, The volume of a cone having a base of radius x and height y is Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 (cont) Solution Since , the internal force at the section becomes The area of the cross section is also a function of position y, Between the limits of y =0 and L yields Copyright © 2011 Pearson Education South Asia Pte Ltd

PRINCIPLE OF SUPERPOSITION
It can be used for simple problems having complicated loadings. This is done by dividing the loading into components, then algebraically adding the results. It is applicable provided the material obeys Hooke’s Law and the deformation is small. If P = P1 + P2 and d ≈ d1 ≈ d2, then the deflection at location x is sum of two cases, δx = δx1 + δx2 Copyright © 2011 Pearson Education South Asia Pte Ltd

COMPATIBILITY CONDITIONS
When the force equilibrium condition alone cannot determine the solution, the structural member is called statically indeterminate. In this case, compatibility conditions at the constraint locations shall be used to obtain the solution. For example, the stresses and elongations in the 3 steel wires are different, but their displacement at the common joint A must be the same. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 3 The bolt is made of 2014-T6 aluminum alloy and is tightened so it compresses a cylindrical tube made of Am 1004-T61 magnesium alloy. The tube has an outer radius of 10 mm, and both the inner radius of the tube and the radius of the bolt are 5 mm. The washers at the top and bottom of the tube are considered to be rigid and have a negligible thickness. Initially the nut is hand-tightened slightly; then, using a wrench, the nut is further tightened one-half turn. If the bolt has 20 threads per inch, determine the stress in the bolt. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 3 (cont) Solution Taking the 2 modulus of elasticity, Solving Eqs. 1 and 2 simultaneously, we get The stresses in the bolt and tube are therefore Copyright © 2011 Pearson Education South Asia Pte Ltd

FORCE METHOD OF ANALYSIS
It is also possible to solve statically indeterminate problem by writing the compatibility equation using the superposition of the forces acting on the free body diagram. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 4 The A-36 steel rod shown in Fig. 4–17a has a diameter of 10 mm. It is fixed to the wall at A, and before it is loaded there is a gap between the wall at B’ and the rod of 0.2 mm. Determine the reactions at A and Neglect the size of the collar at C. Take Est = 200 GPa. Copyright © 2011 Pearson Education South Asia Pte Ltd

THERMAL STRESS Ordinarily, the expansion or contraction δT is linearly related to the temperature increase or decrease ΔT that occurs. If the change in temperature varies throughout the length of the member, i.e. ΔT = ΔT (x), or if α varies along the length, then = linear coefficient of thermal expansion, property of the material = algebraic change in temperature of the member = original length of the member = algebraic change in length of the member Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 5 The rigid bar is fixed to the top of the three posts made of A-36 steel and 2014-T6 aluminum. The posts each have a length of 250 mm when no load is applied to the bar, and the temperature is T1 = 20°C. Determine the force supported by each post if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature is raised to T2 = 20°C. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 5 (cont) Solution From the free-body diagram we have The top of each post is displaced by an equal amount and hence, Final position of the top of each post is equal to its displacement caused by the temperature increase and internal axial compressive force. Copyright © 2011 Pearson Education South Asia Pte Ltd

STRESS CONCENTRATION The stress concentration factor K is a ratio of the maximum stress to the average stress acting at the smallest cross section; i.e. Copyright © 2011 Pearson Education South Asia Pte Ltd

STRESS CONCENTRATION (cont)
K is independent of the material properties K depends only on the specimen’s geometry and the type of discontinuity Copyright © 2011 Pearson Education South Asia Pte Ltd

INELASTIC AXIAL DEFORMATION
When a material is stressed beyond the elastic range, it starts to yield and thereby causes permanent deformation. Among various inelastic behavior, the common cases exhibit elastoplastic or elastic-perfectly-plastic behavior. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 6 The bar in Fig. 4–29a is made of steel that is assumed to be elastic perfectly plastic, with σY = 250 MPa. Determine (a) the maximum value of the applied load P that can be applied without causing the steel to yield and (b) the maximum value of P that the bar can support. Sketch the stress distribution at the critical section for each case. Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 6 (cont) Solution b) As P is increased to the plastic load it gradually changes the stress distribution from the elastic state to the plastic state. Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ 1) The assembly consists of two posts made from material 1 having modulus of elasticity of E1 and a cross-sectional area A1 and a material 2 having modulus of elasticity E2 and cross-sectional area A2. If a central load P is applied to the rigid cap, determine the force in each post. The support is also rigid. Copyright © 2011 Pearson Education South Asia Pte Ltd

CONCEPT QUIZ (cont) The value of stress concentration factor depends on the geometry. Which one of the following is true? Ka > Kb > Kc Ka > Kb > Kd A and B None of the above Copyright © 2011 Pearson Education South Asia Pte Ltd

Copyright © 2011 Pearson Education South Asia Pte Ltd

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