Stress Analysis -MDP N161 Bending of Beams Stress and Deformation Chapter 6 Instructor: Dr. Chahinaz A.R.Saleh MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Lesson Overview Studying the effect of applied transverse force and moment on the beams. Calculation of stresses and deformations in beams. Normal stresses due to unsymmetric bending moment Equation of neutral axis Deflection of beam MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

Deformation of Straight Beam under Pure Bending Assumptions MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 For this case, top fibers are in compression and bottom in tension…..i.e. bending produces normal strains and hence normal stress MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Neutral surface – does not undergo a change in length MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 X = longitudinal axis Y = axis of symmetry MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Note that: Normal strain varies linearly with y Maximum and minimum strains at surfaces MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 But x = x/E  x = -Ey/ The stress varies linearly with y Remember E is the modulus of elasticity  is the radius of curvature MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Stress due to Bending Equilibrium Fx =0 x dA=0 -Ey/ dA =0  y dA =0 i.e. z is a centroidal axis MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Mz =0 M +x y dA=0 M =- x y dA=  Ey2/ dA =(E/)  y2 dA =E Iz / But x = -Ey/ M = -x Iz/y x =- M y/Iz MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Flexure Formula x =- M y/Iz MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Example 6-1 Determine the maximum stress for the shown loaded beam of rectangular cross section having width of 8.75 in and depth of 27 in. MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Solution 1- External reactions MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 2- Internal Reactions MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 3- cross section MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 4- Maximum stress MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Example 6-2 MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Solution 1- External reactions and maximum internal moment Mmax = 22.5 kN.m Acting at the middle of the beam MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Example 6-3 Determine the maximum stress for the shown loaded beam. MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Solution MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Deformation (bending angle ) For pure bending L= = L/ But 1/ = -x/Ey =-Lx/Ey =ML/EI   L MDP N161- Stress Analysis Fall 2009

Bending of Beams Unsymmetric Bending MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 In all previous cases which have been studied: Y was an axis of summetry..i.e. Y is a principal axis M acts about z Z is a central neutral axis MDP N161- Stress Analysis Fall 2009

Normal stress due to positive bending moment Mz According to right hand role; positive Moment (Mz )could be represented by an arrow pointed to the positive z dn. MDP N161- Stress Analysis Fall 2009

Normal stress due to positive bending moment My MDP N161- Stress Analysis Fall 2009

Arbitrarily Applied Moment If M is applied about a centroidal axis which is not a principal MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Then decompose M to Mz and MY, where Y and Z are the centroidal principal axis MDP N161- Stress Analysis Fall 2009

Normal Stress due to arbitrary applied moment MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 For this shape; axis 2 is not an axis of symmetry. If M is applied about axis 1; could we apply the flexure formula and say that  =- M1 y/I1 ? MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 What about this section? How can we get the stresses due to bending about axis 1? MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Conditions of application of flexure formula; z and y axes are centroidal principal axes M is about z which is a centroidal principal axis MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Then for these cases: MDP N161- Stress Analysis Fall 2009

Arbitrarily Applied Moment MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

Conditions of application the flexure formula z and y axes are centroidal principal axes Moments are about these centroidal principal axes MDP N161- Stress Analysis Fall 2009

Equation of neutral axis Equation of neutral axis is : MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009 Example MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009

MDP N161- Stress Analysis Fall 2009