Singularity Functions in beam loadings 1 Since the load is concentrated or discrete entities, discontinuous over the beam length, it is difficult to represent discrete functions with equation valid for entire beam length Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Singularity Functions in beam loadings 2 Different methods to solve for beams can be graphical, mathematical or computer based. Singularity functions to represent the loads is one of them (computing) Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Singularity Functions in beam loadings 3 Singularity functions are binomials <> where x is the point of interest in beam, and a is the place where in x the load begins to act. Depending on the type of load, provides value of x along the position as shown in the table Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 4 Load distribution Function Formula Conditions = 0, x a Quadratically distributed Unit parabolic function load = (x – a)2, x > a = 0, x a Linearly distributed load Unit ramp function = (x – a), x > a = 0, x < a Uniformly distributed load Unit step function = undefined, x = a = 1, x > a = 0, x < a Concentrated load Unit impulse function = , x = a = 0, x > a = 0, x < a Concentrated moment Unit doublet function = indeterminate, x = a = 0, x > a Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 5 Integrals of singularity functions Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 6 Integrating the loading function q to get shear V, and integrating shear (V) to get momentum (M), When x is either 0 or l, we get 0 for V and M thereby giving 4 boundary conditions • When x = 0, we get 0 for V and M thereby calculating the constants C1 and C2 as 0 • C1 and C2 will always be 0, if reaction forces and moments included in loading function, as diagrams should close at beam end Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. When x = l, we get 0 for V and M thereby calculating 7 the reaction forces R1 and R2 Formula Conditions = 0, x a = (x – a)2, x > a = 0, x a = (x – a), x > a = 0, x < a = undefined, x = a = 1, x > a Substituting C1 and C2 as 0 and R1 and R2 in equations for V and M for different values of x between 0 and l We can find the graph Vmax is at x=l Substituting V=0 to get Mmax Resulting in Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Mmax = 88.2 Vmax = -42 R1 = 18 and R2 = 42 Copyright ©2011 by Pearson Education, Inc. All rights reserved. 8 Integrating the loading function q to get shear V, and integrating shear (V) to get momentum (M), When x is either 0 or l, we get 0 for V and M thereby giving 4 boundary conditions When x = 0, we get 0 for V and M thereby calculating the constants C1 and C2 as 0, similar to previous situation Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Calculating reaction forces R1 and R2 substituting C1, C2, V and M as 0 @ x=l Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton 9 Copyright ©2011 by Pearson Education, Inc. All rights reserved. Substituting C1 and C2 as 0 and R1 and R2 in 10 equations for V and M for different values of x between 0 and l, We can find the graph where the Vmax is at x=l Substituting V=0 to get Mmax at X=7.4 Resulting in Mmax = 147.2 Vmax = -120 R1 = 56.7 and R2 = 123.3 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 11 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 12 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 13 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 14 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 15 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 16 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 17 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 18 This chapter is going to be a review of subjects you learned in mechanics of materials Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 23 Z X Y Along axis direction Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Opposite axis direction Copyright ©2011 by Pearson Education, Inc. All rights reserved. • Axis of the stresses is arbitrarily chosen for convenience. Normal and shear stresses at one point will vary with direction along the coordinate system. 24 • There will be planes where the shear stress is 0, and the normal stresses acting on these planes are principal stresses, and planes as principal planes • There will be planes where the shear stress components are 0, and the normal stresses acting on these planes are principal stresses, and planes as principal planes • Direction of surface normals to the planes is principal axes . • And the normal stresses acting in these directions are principal normal stresses • The principal shear stresses, act on planes at 45°to the planes of principal normal stresses Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. • The principal shear stresses from principal normal stresses can be calculated as 26 • Since usually 1>2>3, max is 13 and the direction of principal shear stresses are 45°to the principal normal planes and are mutually orthogonal Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 28 The Mohr’s Circle • Mohr circle is a graphical method find the principal stresses Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. The Mohr’s Circle Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton 29 Copyright ©2011 by Pearson Education, Inc. All rights reserved. The Mohr’s Circle Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton 30 Copyright ©2011 by Pearson Education, Inc. All rights reserved. The Mohr’s Circle Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton 31 Copyright ©2011 by Pearson Education, Inc. All rights reserved. 32 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 33 Modes of Failure • Tension - When object is in tension there is only one stress • More common cables, struts, bolts and any axially loaded members • Applied normal stress for axial tension is • Change in length Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 34 Modes of Failure • Shear – common loading in pins, bolted or riveted construction • Direct Shear is when there is no bending, there can be shear with bending too • For direct shear, there is no gap between the blade and the vice, which is very rare. when there is a small gap, most common, there is a moment creating bending stresses and shear • This is preferred because, the area of shear is doubled Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 35 Modes of Failure • Direct Bearing – the pin in hole can fail other than shear. Which is when the surfaces of the pin and hole or subjected to bearing stresses (compressive in nature) crushes the pin and hole than shear it. • Direct Shear is when there is no bending, there can be shear with bending too • Projected area of contact in case of tight fit • In case of loose/clearance fit, it approximated to • Length and dia needs to be carefully selected to avoid bearing failure Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 36 Modes of Failure • Tearout Failure – another failure can be the tearout of material surrounding the hole. Happens when hole is placed close to the edge. • Provide sufficient material around the holes to prevent this mode of failure • One pin diameter of material between the edge and the hole is good starting point for calculations Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Pure bending in Beams 37 • Pure Bending – it is rare for beam to be loaded in pure bending. It is useful though to understand the situation • Mostly shear loading and bending moments act together • Applying point loads P equidistant at the simply supported beam, absence of shear loading makes this pure bending • Assumptions used for analysis Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Pure bending in Beams Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton 38 Copyright ©2011 by Pearson Education, Inc. All rights reserved. Pure bending in Beams 39 • N-N along the neutral axis, no change in length • A-A shortents (in compression) and B-B lengthens (in tension • Bending stress is 0 at N-N and is linearly proportional to distance y away from N-N • Where M is the bending moment and I is the area moment of Inertia of the beam cross section at the neutral plane, y the distance from N-N • The max stress at outer plane is • Where c is distance from neutral plane (it should be same both sections if the beam is symmetrical about the neutral axis • C is taken as +ve initially and proper sign applied based on loading compression –ve and tension +ve Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Pure bending in Curved Beams 40 • Machines have curved beams like in C clamps, hooks etc, with a ROC) the first 6 assumptions still apply • If it has a significant curvature the neutral axis will not be coincident with the centroidal axis and the shift e is found from • Where A is area, rc is ROC of centroidal axis and r is the ROC of the differential area dA • For a rectangular beam this can be 𝒆 = 𝒓𝒄 -(𝒓𝒐 -𝒓𝒊 )/LN(𝒓𝒐 / 𝒓𝒊 ) Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Pure bending in Curved Beams 41 • Stress distribution is not linear but hyperbolic. Sign convention is +ve moment straightens the beam (tension inside and compression outside) • For pure bending loads stresses at inner and outer surface is • And if a force is applied on the curved beam then • The stresses will be Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Shear due to transverse loading 42 • Common case is both shear and bending moment on the beam. • Fig shows a point loaded beam shear and moment diagram • Cutting our segment P of width dx around A, cut from outer side at c upto depth y1 it is seen that the M(x1) to left < M(x2) to right and the difference being dM • Similarly for stresses (can be seen in fig b) since the stresses are proportion to moment Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Shear due to transverse loading 43 • Similarly for stresses (can be seen in fig b) since the stresses are proportion to moment. This stress imbalance is countered by shear stress component • Stress acting on left hand side of p at a distance y from neutral axis is stress times the differential area dA at that point • The total force acting on the left hand side will then be • Similarly for right hand side • Shear force on the top face is • Where bdx is the area of the top face of the element Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Shear due to transverse loading 44 • For equilibrium, the forces acting on p is 0 • Gives an expression for shear stress as a function of change in momentum wrt x • Since slope of dM/dx is the magnitude of the shear function V (dV/dx = d2M/dx2) • Assigning the integral as Q, then • Shear stresses vary across y • And becomes 0 when c=y1 • And maximum at neutral axis • A common rule of thumb is the shear stress due to transverse loading in a beam will be small enough to ignore if the length to depth ratio of the beam is 10 or more. Short beams below that ratio should be investigated for both transverse shear and bending. Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 45 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 46 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Torsion 47 • Isotropic, within elastic limits, pure torsion normal to its axis on straight bar • Fixed on one end and torsion applied at other, bar twists about its long axis • Element taken anywhere on the outer surface shears due to torsional loading • Shear stress varies from 0 at the center maximum on the outer element, as shown in ‘b’ • Max shear stress at outer radius r is • Where J is the polar moment of inertia of the CS and the angular deflection is • Where G is shear modulus, l is length and K is J which depends on the cross section Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. Torsion 48 • For non-circular sections, max and are • Where Q and K are functions of the Cross Section Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 49 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 50 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 51 Combined bending and torsional stresses • Most bars have combined stresses creating both normal and shear stresses. So these need to combined at certain locations to find the principal stresses and maximum shear stress • Limiting to bending of cantilever and in torsion • The shear and moment diagrams will be similar to a cantilever beam loaded at its end Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 52 Combined bending and torsional stresses • Torque applied at the end is Tmax = Fa = 8000 lb-in and is uniform over its length • Most heavily loaded CS is at the wall where M, T and are maximum • Looking at the CS diagram, bending normal stress is maximum at outer and 0 in the neutral axis, and shear it is reverse • Shear due to torsion is proportional to radius and is 0 at the center. While bending is along Y axis, torsion is uniformly distributed at outer fiber Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 53 Combined bending and torsional stresses • Choosing 2 points A and B at the outer end to get worst combination of stresses • A will have max bending and max torsion, and stresses in element A is shown in fig ‘b’ • The normal stress x acts on the X face in X direction while torsional shear xz acts on x face in +Z direction • At B, the torsional shear has same mag of point A but the direction is 90°hence xy • At B, the transverse shear stress is also maximum which is xy and both act in –y direction • Maximum normal bending and torsional shear stress on A Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 54 Combined bending and torsional stresses • Maximum shear stress and principal stresses that result from this combination • Shear due to transverse loading on B • Total shear stress on B is the sum of torsion and transverse • While point A has higher principal stress, Point B has higher shear which is principal. • Both points need to be checked Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 55 Spring Rates For a uniform bar in axial tension, 𝑘= 𝐹 𝑦 = 𝐹 𝐹𝐿 𝐴𝐸 = 𝐴𝐸 𝐿 For a uniform cross − section round bar in torsion, 𝑘= 𝑇 = 𝑇 𝑇𝐿 𝐽𝐺 𝐺𝐽 = 𝐿 For a cantilever beam with concentrated point load at end, 𝑘= 𝐹 𝑦 = 𝐹 𝐹𝑙3 3𝐸𝐼 3𝐸𝐼 = 3 𝑙 Note that k for a beam is unique to its manner of support and its loading distribution, since y depends on the beam’s deflection equation and point of load application. Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 56 Stress Concentration Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 57 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 58 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 59 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 60 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 61 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 62 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 63 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 64 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved. 65 Machine Design: An Integrated Approach, Fourth Edition Robert L. Norton Copyright ©2011 by Pearson Education, Inc. All rights reserved.