mechanical-properties-and-working-of-metals-and-alloys_compress

4.4 Linear Elastic Behaviour 183 symmetrical with respect to the centroidal axis of a beam can will be at a much higher level than the proportional limit be obtained from (4.11a) as moment given by the point ‘A’, while the offset yield strength at Y in tension remains at a little higher level than ðemaxÞpl¼ c ¼ chpl or; hpl ¼ L ðemax Þpl ð4:21Þ the tensile stress at the proportional limit, i.e. the point ‘A’. qpl L c Thus, the above offset construction is of no use for evalu- ation of the yield moment. Because of the small deviation On substituting for hpl from (4.21) into (4.20), we get from linearity in the bending moment–strain diagram, it is easier and more reliable to determine the elastic properties in Modulus of resilience, ðUBÞMR ¼ ðMBÞplðemaxÞpl ð4:22Þ bending by computation from the corresponding properties 2Ac obtained in uniaxial tension. Accordingly, the proportional limit moment ðMBÞpl can be calculated with quite accuracy Therefore, the modulus of resilience in pure bending is from the tensile proportional limit stress rpl by means of equal to the area under the linear elastic portion of the (4.10) or (4.14). Similarly with the same equation, the yield applied bending moment MB versus the maximum bending moment ðMBÞy can be calculated from the tensile yield strain emax diagram divided by the product A c : The modulus strength r0 ; measured in uniaxial tension, although the of resilience can be calculated from either of the above two value of ðMBÞy computed from this equations (valid for (4.20) and (4.22). For comparison of the values of the linear elastic region) will be approximate because of the modulus of resilience for beams of different materials, the involvement of a small amount of inelastic deformation in geometry of the beams must be maintained identical because yielding. the values of the modulus of resilience are dependent on the geometry of the beams. The other elastic properties of a 4.5.1 Discontinuous Yielding and Shape Factor material subjected to bending can be predicted from the equivalent properties in uniaxial tension. 4.5 Yielding The best example among the materials that yield discontin- uously is mild steel because it is widely used in application. Yielding in pure bending is identical to that in uniaxial The absolute maximum value of bending stress developed in tension, i.e. slip takes place in the direction of the maximum the extreme fibres of a beam made of mild structural steel shear stress on planes oriented at 45° with the axis of the subjected to pure bending reaches the uniaxial tensile upper beam. When the bending stresses developed on the extreme yield stress value ðr0ÞU just prior to the onset of yielding. At fibres of the beam reach the yield strength of the material, as this stage, a linear elastic stress distribution over the determined by uniaxial tensile test, these fibres farthest from cross-section of the beam is noted, as shown in Fig. 4.11a, the neutral axis of the beam yield locally producing a strain where the cross-section of the beam is assumed to be sym- equal to the yield strain in uniaxial tension. Afterwards with metrical with respect to its neutral axis. As soon as yielding the continuation of bending, yielding proceeds gradually begins at some point on the extreme surface, owing to an inward towards the neutral plane of the beam. imperfection, the stress in the extreme fibres of mild steel beam drops to the uniaxial tensile lower yield stress value Since initial local yielding near the outermost surfaces of ðr0ÞL. The onset of yielding causes a small slip band to form the beam changes the bending stress only in a small region at the outermost surface that starts from this surface and of the cross-section, its effect on the bending moment–strain propagate inward towards the neutral surface of the beam in diagram (Fig. 4.10) is minor that leads to a slight deviation the shape of a wedge. This propagating wedge acts like a of the diagram from its linear portion. This makes the notch and has stress concentration at its tip. This causes the position of the proportional limit moment very unreliable inner fibres of the mild steel beam to yield at the uniaxial and the detection of the initial yielding in bending difficult. tensile lower yield stress value ðr0ÞL: As a consequence, the At a strain slightly beyond the elastic range say e1; in elastic stress distribution curve of Fig. 4.11a changes Fig. 4.10, the bending moment–strain diagram will lie abruptly to the elastic and plastic stress distribution curve, as slightly below its corresponding elastic diagram, given by shown in Fig. 4.11b. Just like uniaxial tension, highly the extended linear elastic line ‘AB’. On the other hand, at localized yielding takes place at several points along the the same strain e1 ; the tensile stress–strain diagram for the beam until the entire beam has yielded. same material will lie substantially below its corresponding elastic diagram since there is an appreciable stress drop from After the inward progress of yielding up to some depth the elastic value at the onset of yielding in tension. The from the extreme surface of the beam, the actual elastic and value of yield moment in bending at Y0 in Fig. 4.10 obtained plastic stress distribution curve in mild steel appears as that by the construction for the same strain offset used in tension in Fig. 4.12a. The absolute maximum value of bending

184 4 Bending (a) (b) (–) Neutral surface (–) σx σx 0 0 (+) (+) (σ0)U (σ0)L y (σ0)U y Fig. 4.11 Discontinuous yielding in mild steel subjected to pure bending. a Linear elastic stress distribution over the cross-section of the beam just prior to the onset of yielding. b Elastic and plastic stress distribution just after initiation of yielding at the outermost surface of the beam (a) (b) (σ0)L (σ0)L Neutral surface (–) σx (–) 0 σx (+) 0 (σ0)L (+) y (σ0)L y Fig. 4.12 Stress distribution over the depth of a mild steel beam with surface of the beam. b Idealized stress distribution under fully plastic discontinuous yielding. a Actual elastic and plastic stress distribution condition of the beam after the inward progress of yielding up to some depth from the extreme stress over the depth of yielding is given by the uniaxial cross-section is 1.5, i.e. their ultimate moments are respec- tensile lower yield stress value ðr0ÞL that remains nearly tively 70 and 50% higher than their initial yield moments constant, as observed in uniaxial tension during the period of and therefore, the circular cross-section is preferred to the yield-point elongation. Ultimately, when the yielding pro- rectangular cross-section in the plastic design of steel gresses up to the neutral surface of the beam, the condition structures. becomes fully plastic. As long as the mild steel does not strain harden, the bending moment cannot rise above the 4.6 Nonlinear Stress–Strain Relations value that corresponds to this fully plastic condition, which is therefore called the ultimate moment, ðMBÞu: The ideal- Materials either deformed in the plastic range or exhibiting ized stress distribution curve under fully plastic condition nonlinear elastic deformation involve nonlinear stress–strain can be represented by two rectangles meeting at the neutral relationships. In either of the above cases, (4.9) or (4.10) or axis of the cross-section of beam and is shown in Fig. 4.12b. (4.14) cannot be applied because they are derived for the The magnitude of the ratio of the ultimate moment to initial conditions of linear elastic deformation. In such situations, yield moment, i.e. the ratio ðMBÞu=ðMBÞy; depends on the the stress distribution curve over the depth of the beam has cross-sectional shape of the beam and is called the shape to be established in order to determine the relation between factor (Richards 1961), the value of which will be obviously bending moment and stress for each stress distribution. above 1. The higher the value of the shape factor, the higher is the difference between the ultimate moment and the initial Let us first consider a beam having a vertical as well as a yield moment. The shape factor is applied in the plastic horizontal plane of symmetry, such as a circular, rectangular, design of structures made of mild steel because it determines I-shaped or any other shaped cross-section having two axes the allowance that remains beyond the initial yield moment. of symmetry. In this section, all diagrams are constructed Assuming that no strain hardening takes place, the shape assuming that the beam is bent concave upwards. Consider a factor for circular cross-section is 1.7 and that for rectangular ductile material stressed into the plastic range, whose stress–

4.6 Nonlinear Stress–Strain Relations 185 Fig. 4.13 Nonlinear stress– (a) (b)(i) (b)(ii) strain relation for a beam having a vertical as well as a horizontal (εx)min (σx)min plane of symmetry subjected to pure bending. a Stress–strain σx (σx)max (–) diagram identical in tension and Neutral compression. b Distributions of (+) surface (–) corresponding (i) strain and (ii) εx stress over the depth in the beam (εx)min 0 0 σx (–) y (+) (εx)max 0 εx (σx)min y (+) (σx)max (εx)max strain diagram in tension is identical to that in compression, Fig. 4.13b(i). If strains are determined for various values of depth from (4.23), the corresponding values of stresses rx as shown in Fig. 4.13a. The corresponding stress distribu- and rxC are then obtained from the stress–strain curve in Fig. 4.13a. Now, one can easily plot the stress distribution tion curve over the depth of the beam is shown in Fig. 4.13b curve over the depth of the beam, i.e. stress versus depth curve, as shown in Fig. 4.13b(ii). Since the stress–strain (ii), which has the same shape as that in Fig. 4.13a, but curve in tension is identical to that in compression for the material considered above, the neutral axis of the differs only in scale. In Fig. 4.13b(ii), the ordinate shows the cross-section coincides with the horizontal axis of symmetry and the stress distribution in tension will be identical to that depth distance y from the neutral axis of the cross-section in compression. Hence, the resultant of the tensile stresses while in Fig. 4.13a, the abscissa shows the strain ex along will be equal in magnitude to that of the compressive the neutral axis of the beam, i.e. in the x-direction; but in stresses and thus balance each other to satisfy the condition each figure, the other axis represents the stress. Since ex is of equilibrium. proportional to y in case of pure bending according to (4.2a) For materials with stress–strain curves that differ in ten- (where it is assumed that initially plane cross-sections sion and compression, as shown in Fig. 4.14a, the stress distributions are altered from those discussed above. For this remain plane beyond the elastic range), the corresponding unsymmetrical stress–strain diagram, if the neutral axis of strain ðexÞ distribution curve over the depth ðyÞ of the beam the cross-section would remain at the horizontal axis of is linear, as shown in Fig. 4.13b(i). symmetry, i.e. at the centre of the beam, as shown in Fig. 4.14b, then the resultant of the stress distribution in the If we know the maximum strain ðexÞmax (the tensile strain tensile part will differ in magnitude from that in the com- on the convex extreme face of the beam) and the minimum pressive region and hence imbalance will be created. To strain ðexÞmin; (the compressive strain on the concave meet the conditions required for equilibrium, the neutral axis extreme face of the beam), the stress and strain distribution of the cross-section must move towards the side of higher curves in Fig. 4.13b can be constructed from the stress– stress, as shown in Fig. 4.14c (in diagram, the compressive stress-side), so that the resultants of the stress distributions strain diagram in Fig. 4.13a in the following way. From on both sides of the neutral axis become equal to each other. With the continuation of bending, the neutral axis of the (4.2a) and (4.2b) cross-section must continuously shift its location to maintain equilibrium. A similar condition arises for the beam having ex ¼ y ; and exC ¼ Ày no horizontal plane of symmetry, such as a trapezoidal q q section or T-section. In both situations, the neutral axis of the cross-section moves away from the centroid with the Since from (4.15), continuation of bending. ðexÞmax ¼ ðexÞmin ¼ c Once the stress distribution curve over the depth of the q beam has been established as discussed above, the bending moment is determined as follows. The tensile and ) ex ¼ ðexÞmax y ð4:23aÞ c and exC ¼ ðexÞmin ðÀyÞ ð4:23bÞ c The above (4.23) is used to plot the strain distribution curve over the depth of the beam, i.e. strain versus depth curve, which is linear according to (4.23) and shown in

186 (b) 4 Bending (a) (c) σx (–) Neutral (–) surface 0 0 Centre σx of beam εx σx 0 (+) (+) yy Fig. 4.14 Nonlinear unsymmetrical stress–strain relation for a beam of symmetry, i.e. at the centre of the beam. c Actual stress distribution subjected to pure bending. a Stress–strain diagram differing in tension curve over the depth of beam, resulting from the shift of the neutral axis and compression. b Stress distribution curve over the depth of beam, if towards the side of higher stress in order to satisfy the conditions of the neutral axis of the cross-section would remain at the horizontal axis equilibrium Fig. 4.15 Typical stress C a 2 C 0 a σx distribution over the depth of a Neutral 0 beam of brittle material subjected Neutral surface to pure bending surface T 0a σx y (σmax)r 2 T y (σmax)r compressive segments of stress distribution curve will pro- elastic stress distribution represented by the dashed lines in duce, respectively, the resultant tensile force T and the Fig. 4.15 will have a rupture moment equal to ðMBÞr; from resultant compressive force C; both of which act through the which the modulus of rupture ðrmaxÞr is computed using the centroids of their respective stress distributions and will be flexure formula [(4.19)]. ðrmaxÞr is the maximum stress at of equal magnitude, as required for equilibrium. These equal rupture for the theoretical linear elastic stress distribution and opposite forces form a couple, the moment arm of which is the distance a between the centroids of their stress dis- and is always substantially higher than the real maximum tributions. The above can easily be visualized from stress at rupture ðrmaxÞrTrue : For example, ðrmaxÞr is greater Fig. 4.15, where T; C; and a have been shown and the stress than ðrmaxÞrTrue by 80% in case of cast iron and by 50–100% distributions are represented by shaded areas. The applied for concrete. bending moment must always be equal to the moment of this couple, i.e. T or absolute value of C multiplied by the Note that beams of brittle materials never undergo com- moment arm a and hence given by pressive fracture because brittle materials are stronger in MB ¼ Ta ¼ jCja ð4:24Þ compression than in tension. In beams of ductile materials, final failure generally includes either lateral buckling or excessive plastic deformation. The stress distributions usually developed by beams made 4.7 Shear Stresses in Elastically Bent Beam of brittle materials are like those of Fig. 4.15. In these fig- ures, the stress distributions under linear elastic condition, When the applied bending moment varies from one cross-section to another of a beam transversely loaded within which does not really exist, have also been superimposed by the elastic range, a shear force F is generally present at each the dashed lines so that the modulus of rupture ðrmaxÞr can cross-section along with the bending moment. This shear be compared with the true rupture stress ðrmaxÞrTrue : If the force F on any cross-section represents the resultant of a actual stress distribution at rupture is assumed to be repre- certain distribution of shear stresses s over that section. This sented by any of the diagrams in Fig. 4.15, the actual rupture moment ðMBÞrTrue will be given by T a : Whereas the linear

4.7 Shear Stresses in Elastically Bent Beam 187 τ τ b n' h m' τ τ n τ m z τ Enlarged view τ x τ F Fig. 4.16 Illustrating shear stresses in bending of a rectangular cross-section y distribution must satisfy the conditions of equilibrium for F various elements of the beam. mp For simplicity, consider a beam of rectangular cross-section of depth h and width b ; which may be imag- MB ined to be composed of infinitesimal strips parallel to the O neutral axis z of the cross-section as shown in Fig. 4.16. It may be reasonably assumed that on any one of such strips, MB + dMB the shear stress s is consistent across the width b of the beam nq and parallel to the depth direction of the beam, i.e. along y- axis. From the equilibrium conditions of the elemental strip, dx it can be established that such shear stress on one side of the F prismatic element mn must be accompanied by equal shear stresses on each of the other three sides of the element as Fig. 4.17 Figure to find a relationship between shearing force and exhibited in Fig. 4.16. Finally, there will be horizontal shear bending moment at any cross-section of a transversely loaded beam stresses between layers of beam parallel to the neutral plane between points of application of concentrated loads as well as transverse shear stresses between cross-sections, and these complementary shear stresses are always equal in For this, consider an element of the beam between two magnitude but opposite in sign at any point in the beam. The adjacent cross-sections mn and pq ; separated by a distance upper face of the topmost prismatic element, say m0n0 in dx; and subjected to varying bending moment as shown in Fig. 4.16, and the lower face of the bottommost prismatic Fig. 4.17. On the cross-section ‘mn’, shear force and bending element (not shown in figure) coincide respectively with the moment are represented, respectively, by F and MB (assumed top free surface and bottom free surface of the beam, where positive). If no external force is applied between the the shear stress s must vanish because any free surface of the cross-sections ‘mn’ and ‘pq’, the respective shear force and beam is free from stress. From this, we can conclude that the bending moment on the cross-section ‘pq’ will be F and shear stress s must vary with distance y along the depth from MB þ dMB as shown in Fig. 4.17, where dMB is the change in top to bottom of the beam and must be zero at y ¼ Æh=2 from the neutral axis z of the cross-section. Let us find a relationship between shearing force and bending moment at any cross-section of a transversely loaded beam between points of application of concentrated loads.

188 4 Bending bending moment between ‘mn’ and ‘pq’. For equilibrium of y from the neutral axis z of the cross-section, the normal the element, the algebraic sum of the moments of these forces force acting on dA will be rxdA; which can be related to about point ‘O’ must be equal to zero which is given below: bending moment MB from (4.9a) as follows: ÀMB þ ðMB þ dMBÞ À Fdx ¼ 0: rxdA ¼ MBy dA From the above, we get: Iz dMB ¼ F ð4:25Þ The sum of these normal forces over the left face m1n of dx the block becomes Thus, at any cross-section of the beam between points of Zc1 ð4:26aÞ application of concentrated loads, the shear force is equal to MBy dA the rate of change of bending moment with respect to the Iz separation distance x. y1 Again, between the above two adjacent cross-sections ‘mn’ and ‘pq’ separated by the distance dx; let us examine where y1 and c1 are, respectively, distances of the upper and the equilibrium of an elemental block m1np1q of the beam bottom faces of the elemental block m1np1q of the beam whose lower and upper faces are parallel to the neutral plane from the neutral plane, as shown in Fig. 4.18a. as shown in Fig. 4.18a. The shear stress on the bottom face of this block is s ¼ 0; because it coincides with the bottom Similarly, the sum of the normal forces over the right face free surface of the beam. A horizontal shear stress s acts p1q of the block will be upon the area bdx of the upper face of this block, which is located at an arbitrary distance y1 from the neutral plane of Zc1 ðMB þ dMBÞy dA ð4:26bÞ the beam. The left face of this block represented by ‘m1n’ is Iz shown by the shaded area in Fig. 4.18b. The free-body diagram of the complete elemental block is shown in y1 Fig. 4.18c. Since we are considering the general case of varying bending moment, Fig. 4.18c shows that the left face Since a horizontal shear stress s acts upon the area b dx of m1n and right face p1q of this block are respectively acted the upper face m1p1 of this block at an arbitrary distance y1 upon by the normal bending stresses rx and rx þ drx: from the neutral plane of the beam, the shear force on m1p1 Although shear stresses exist on these end faces m1n and becomes p1q, they are not shown on the free-body diagram in Fig. 4.18c, because they will not enter in the equation of sb dx ð4:26cÞ equilibrium of this block taken in the direction of the neutral axis x of the beam. The shear stress s present on the upper The forces, given by (4.26a), (4.26b) and (4.26c), must be face m1p1 of this block is shown in Fig. 4.18c, whereas the in equilibrium in the direction of the neutral axis x of the bottom face nq is free from stress. beam and hence the algebraic sum of these forces is equated to zero as follows: Referring to Fig. 4.18b, if we consider an elemental area dA on the left face m1n of the block at an arbitrary distance Zc1 ðMB þ dMBÞy dA À Zc1 MBy dA À s b dx ¼ 0 ð4:27Þ Iz Iz y1 y1 (a) (b) (c) dx p b m mm c2 h/2 x σx+dσx m1 p1 y1 x y1 m1 z m1 τ p1 h/2 m1 y c1 dA n σx q MB MB+ dMB nn nq y = bdy y Fig. 4.18 a Figure to examine the equilibrium of an elemental block elemental block, showing that the normal bending stresses rx and m1np1q of the beam between two adjacent cross-sections mn and pq; rx þ drx are acting, respectively, on the left face m1n and right face separated by a distance dx: b Representation of the left face of the block p1q, and the shear stress s is present on the upper face m1p1 whereas the m1np1q by the shaded area. c Free-body diagram of the complete bottom face nq is free from stress

4.7 Shear Stresses in Elastically Bent Beam 189 Prior to proceeding further, it is to be noted that if the variation of Q with y and let us do it with respect to a bending moment applied to the beam is of the same mag- rectangular cross-section as given in Fig. 4.18b, where dA ¼ nitude at the cross-sections ‘mn’ and ‘pq’, as observed in bdy and c1 ¼ h=2; so that case of pure bending, the distribution of the normal longi- tudinal fibre stresses on the end faces m1n and p1q of the Zc1 Zh b h2  elemental block will be identical and equal to rx : Then 2 24 y12 following the above (4.27), the equation for equilibrium of the block in the direction of the neutral axis x of the beam Q ¼ ydA ¼ b ydy ¼ À ð4:33Þ can be written as y1 y1 The area of the shaded portion m1m1nn of the ð4:34aÞ cross-section in Fig. 4.18b is given by Zc1 Zc1  sb dx ¼ MBy dA À MBy dA ¼ 0 h ð4:28Þ b 2 À y1 Iz Iz And the distance of its centroid from the neutral axis of y1 y1 the cross-section is given by From (4.28), it is concluded that s¼0 ð4:29Þ ðh=2Þ À y1 1  2 2 h Equation (4.29) clearly proves the fact that no shear y1 þ ¼ 2 þ y1 ð4:34bÞ stress is induced in the beam subjected to pure bending. It is to be noted that the product of (4.34a) and (4.34b) In case of non-uniform bending moment applied to a beam, the shearing stress s at any point of a cross-section of will result the same statical moment Q given by (4.33). the beam loaded within the elastic range can be obtained from (4.27) as follows: Substituting the value of Q from (4.33) into (4.32), we get Zc1 s ¼ F h2 À  ð4:35Þ s ¼ dMB 1 y dA 2Iz 4 y21 dx Izb ð4:30Þ It is clear from (4.35) that the variation of the shear stress y1 s with the distance y1 from the neutral plane is parabolic, as shown in Fig. 4.19. Further, from (4.35), at y1 ¼ 0; i.e. at the By using (4.25), (4.30) reduces to neutral axis of the cross-section, the maximum shear stress is Zc1 ð4:31Þ smax ¼ Fh2 ð4:36Þ s ¼ F y dA 8Iz Izb y1 The integral R c1 y dA represents the statical moment, And at y1 ¼ Æh=2; i.e. at the top and bottom surfaces of the beam, y1 about the neutral axis z of the cross-section, of the shaded area m1m1nn of the cross-section in Fig. 4.18b, i.e. of the part of the cross-section at an arbitrary distance y1 from the neutral plane of the beam asttawtichailchmtohme esnhteaRryc11 stress s is τmax required. If we denote the y dA ¼ Q; (4.31) takes the following form: s ¼ FQ ð4:32Þ Izb where Neutral axis F the rsetastuicltaalnmt sohmeaerntfo¼rceRyco11ny the cross-section; τ Q the dA; Fig. 4.19 Parabolic variation of shear stresses with distance from the Iz the moment of inertia of the cross-sectional area about neutral axis of cross-section of a rectangular beam, showing that the shear stress is the maximum at the neutral axis and zero at the top and its neutral axis z ; and bottom surfaces of the beam b the width of the cross-section To know the variation of this shear stress s with the distance y1 from the neutral plane, we have to examine the

190 4 Bending s¼0 ð4:37Þ application of concentrated external loads on the beam, where the shear force changes abruptly, it is theoretically investi- Since for rectangular cross-section Iz ¼ ðbh3Þ=12; so gated that the longitudinal strains in the fibres are not appre- h2=ð8IzÞ ¼ 3=ð2bhÞ; and as the total area of the cross-section ciably affected by the warping due to shear strain and thus, the A ¼ bh; so we obtain from (4.36) distribution of normal longitudinal fibre stresses in the beam is not substantially modified. Therefore, it is reasonable to apply 3  flexure formulas derived for pure bending in case of a beam 3F 2 F subjected to non-uniform bending moment. smax ¼ 2bh ¼ A ¼ 1:5 Â ðaverage shear stressÞ ð4:38Þ Therefore, the maximum horizontal or vertical shear 4.8 Solved Problems stress exists at the neutral axis of the cross-section, i.e. at 4.8.1. Compute the shape factor in pure bending for a y1 ¼ 0; and its value is 50% greater than the average shear transversely loaded flexure beam of mild steel having cir- stress F=A: cular cross-section. Assume that there is no strain hardening. The shear strain c is given by c ¼ s=G; where G is the Solution modulus of elasticity in shear, sometimes called the modulus The magnitude of the ratio of the ultimate moment to initial yield moment, i.e. ðMBÞu=ðMBÞy; in pure bending is to be of rigidity. Since there is a variation of the shear stress s from determined. Considering initial yielding at the uniaxial ten- the top to bottom of the beam and the shear strain c is pro- sile lower yield stress value ðr0ÞL of mild steel, the initial portional to the shear stress s; so c must also vary in a similar yield bending moment of a circular cross-section can be written from (4.14) as: manner. Thus, each cross-section of the beam that was initially ðMBÞy¼ ðr0ÞLIcz ¼ ðr0ÞLðpDD4=Þ2=64 ¼ ðr0ÞL pD3 ð4:39Þ plane before the deformation will become warped after it is 32 bent. The straight cross-sections before bending as repre- where D is the diameter of a circular cross-section; c is the sented by lines mn and pq in Fig. 4.20 will change respec- distance from the neutral surface of the symmetrical tively to curved lines m0n0 and p0q0 after bending. Since the shear stress s vanishes at the top and bottom surfaces of the cross-section to the extreme fibres, and c ¼ D=2 for a cir- beam, the shear strain c is zero at the points m0; n0; p0; q0; and hence, the curved cross-sections m0n0 and p0q0 remain normal cular cross-section; Iz is the moment of inertia of the circular to the upper and lower faces of the beam in its bent configu- ration. At the neutral plane, the maximum shear stress smax cross-sectional area about its neutral axis, i.e. centroidal axis will produce the maximum shearing strain given by cmax ¼ smax=G; which is equal to the angles between the tangents to and Iz ¼ ðpD4Þ=64: these curved cross-sections m0n0 and p0q0 and their respective As shown in Fig. 4.21, let us consider a thin strip of normal straight cross-sections mn and pq: As long as the shear thickness dy and length z at an arbitrary distance y from the force F remains uniform along the beam, the warping of each cross-section due to shear strain will be identical, i.e. mm0 ¼ z-axis, which is nothing but the neutral or centroidal axis of pp0; nn0 ¼ qq0: Thus, the shear stresses s do not contribute to the longitudinal fibre strains ex and the distribution of longi- the circular cross-section of the beam. Elementary area of the tudinal fibre stresses rx that is normal to the cross-section remains the same as in the case of pure bending. For a uni- strip is: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi formly distributed load applied to a beam, where the shear dA ¼ zdy ¼ ðD=2Þ2Ày2 dy; since for a circle; force F varies continuously along the beam, or even in case of z2 þ y2 ¼ ðD=2Þ2: m' m p' p Hence, using Fig. 4.21, we can obtain ultimate moment P as ZÂ Ã Z ðMBÞu¼ 4 Â ðr0ÞLdA y ¼ 4ðr0ÞL ydA n n' AA q q' ZD=2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 4ðr0ÞL y ðD=2Þ2Ày2 dy 0 Fig. 4.20 Demonstration of warping after bending of originally plane Let, ðD=2Þ2Ày2 ¼ w2; or; ydy ¼ Àwdw: cross-sections of the beam

4.8 Solved Problems 191 y through the centroids of their respective semicircular equal z dy cross-sectional areas, as required for equilibrium. The y z magnitude of T or C can be obtained from the multiplication of stress and the area over which the stress acts. Since the stress is the lower yield stress value ðr0ÞL of mild steel, and the area is the semicircle of diameter, say D, therefore T ¼ jCj ¼ ðr0ÞLðpD2=8Þ: Let a is the moment arm and yc is the distance between the centroid of the semicircle and its base, as shown in Fig. 4.22. Hence, a ¼ 2yc: Since there is no strain hardening, the ultimate moment ðMBÞu from (4.24) is: ðMBÞu¼ Ta ¼ jCja; or; ð4:41Þ Fig. 4.21 Circular cross-section of a flexure beam showing a thin strip ðMBÞu¼ ðr0ÞLpD8 2 2yc ¼ pðr0ÞLD2yc of thickness dy and length z at an arbitrary distance y from the z-axis 4 (neutral or centroidal axis of cross-section) We know that yc the area of the semicircle = summa- When y ¼ D=2; w ¼ 0; and when y ¼ 0; w ¼ D=2: tion of the moment oRf the elemental area about the base of the semicircle ¼ 2  A ydA (from Fig. 4.21). Z0 pffiffiffiffiffi ZD=2  pD2 ZD=2 ZD=2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 yzdy y ðD=2Þ2Ày2 dy ) ðMBÞu ¼ 4ðr0ÞL w2ðÀwdwÞ ¼ 4ðr0ÞL w2dw Or; yc ¼ 2 ¼ 2 D=2 0 00 ¼ 4ðr0ÞL ðD=2Þ3 ¼ ðr0ÞLD63 ð4:40Þ Let ðD=2Þ2À y2 ¼ w2; or; ydy ¼ Àwdw: 3 When y ¼ D=2; w ¼ 0; and when y ¼ 0; w ¼ D=2: ) Shape factor ¼ ðMBÞu ¼ Eq: ð4:40Þ 16 Z0 pffiffiffiffiffi 16 ZD=2 ðMBÞy Eq: ð4:39Þ pD2 w2ðÀwdwÞ pD2 ) yc ¼ ¼ w2dw ðr0ÞLðD3=6Þ ¼ ðr0ÞL½ðpD3Þ=32Š D=2 0 ¼ 32 ¼ 1:697 % 1:7: ¼ 16 ðD=2Þ3 ¼ 2D ð4:42Þ 6p pD2 3 3p Alternative way to determine ultimate moment: Substituting yc from (4.42) into (4.41), we get ultimate moment as Figure 4.22 shows that the resultant tensile force T and the resultant compressive force C of equal magnitude are acting ðMBÞu¼ pðr0ÞLD2  2D ¼ ðr0ÞLD63 ð4:43Þ 4 3p Equation (4.43) coincides with (4.40). ) Shape factor; ðMBÞu ¼ Eq: ð4:43Þ ¼ ðr0ÞLðD3=6Þ ðMBÞy Eq: ð4:39Þ ðr0ÞL½ðpD3Þ=32Š yc T a 32 C ¼ 6p ¼ 1:697 % 1:7: Fig. 4.22 Circular cross-section of the flexure beam in Fig. 4.21, 4.8.2. A prismatic beam of a brittle material having a showing that the resultant tensile force T and the resultant compressive trapezoidal cross-section, as shown in Fig. 4.23, is subjected force C of equal magnitude are acting through the centroids of their to pure bending in such a way that the top fibres experience respective semicircular equal cross-sectional areas and thus forming a compression. Determine the ratio of bases b1=b2 for maxi- couple, the moment arm of which is the distance a between the mum economy, if the allowable working stresses in tension centroids of the two semicircles and yc is the distance between the and compression are, respectively, 35 and 56 MPa. centroid of the semicircle and its base, where yc ¼ a=2:

192 4 Bending b1 Since Fig. 4.23 shows that c2 ¼ yc; so the value of c2 can h be equated to the distance yc to get the desired ratio of bases b1=b2: The distance yc from the base b2 to the centroid of the trapezoid can be determined as follows: c1 From the drawn Fig. 4.24, we can write z z ¼ ðb2 À b1Þ=2 ; or; z ¼ b2 À b1 y: yh 2h yc c2 Let us consider a thin strip of thickness dy at an arbitrary distance y from the base b2; as shown in Fig. 4.24. Hence, the length of the strip within the trapezoidal cross-section at b2 the distance y from the base b2 is: Fig. 4.23 Trapezoidal cross-section with height h; top edge length b1 b2 À 2z ¼ b2 À b2y À b1y ¼ b2h À b2y þ b1y : and bottom edge length b2 of a prismatic flexure beam, showing the h h distance from the z-axis (neutral or centroidal axis of cross-section) to the top edge as c1 and to the bottom edge as c2; which is equal to the Area of the elemental strip, distance yc from the base b2 to the centroid of the trapezoid  b2h À b2y þ b1y dy: Solution dA ¼ ðb2 À 2zÞdy ¼ h Given that the allowable working stresses in tension, Since area of trapezoid ¼ b1 þ b2 h; and area of trapezoid  ðrxÞmax¼ 35 MPa, and that in compression, ðrxÞmin¼ R2 56 MPa: Since the top fibres experience compression, so in given Fig. 4.23, c1 and c2 denote, respectively, the distances yc ¼ A ydA; therefore, from the neutral (centroidal) axis to the extreme fibres in compression and tension. Hence, from (4.10) we have 2 !Z ðb1 þ b2Þh yc ¼ ydA A 56 ¼ MBc1 ; and 35 ¼ MBc2 : ! Zh  À b2 y þ b1  Iz Iz 2 y b2h h y dy ¼ ðb1 þ b2Þh From the above, c1=c2 ¼ 56=35 ¼ 8=5: Again since c1 þ c2 ¼ h; therefore ¼ ðb1 !\"0 y2 h b2 y3h þ b1 y3h# 2 2 À h 30 h 30 þ b2Þh b2 0 2 ! 3b2h2 À 2b2h2 þ 2b1h2! c1 ¼ 8 h; and c2 ¼ 5 h: ¼ 13 13 ¼ ðb1 þ b2Þh! 6 2 !  h2ð2b1 þ b2Þ ¼ h 2b1 þ b2 ðb1 þ b2Þh 6 3 b1 þ b2 b2 – b1 b2 – b1 ð4:44Þ 2 2 b1 z Since yc ¼ c2; and c2 ¼ ð5=13Þh; so yc ¼ ð5=13Þh; and h from (4.44) we can write  z h 2b1 þ b2 ¼ 5 h; dy 3 b1 þ b2 13 or; 26b1 þ 13b2 ¼ 15b1 þ 15b2; Or; 11b1 ¼ 2b2; y ) b1 ¼ 2: b2 11 b2 4.8.3. A concentrated load P newton is applied transversely at the mid-section of a simply supported beam of length Fig. 4.24 Trapezoidal cross-section of the flexure beam in Fig. 4.23 1.2 m. The beam is of rectangular cross-section with width showing a thin strip of thickness dy and length ðb2 À 2zÞ within the of 150 mm and height of 250 mm. If the allowable working trapezoid at an arbitrary distance y from the base b2 stress in tension or compression is 15 MPa and that in shear is 2 MPa, what will be the safe value of the applied load P?

4.8 Solved Problems p 193 L/2 D/2 L P/2 z (Neutral axis) (a) –P/2 PL/4 (b) Fig. 4.26 Circular cross-section of a beam of diameter D; showing the z-axis (neutral or centroidal axis of cross-section) and the shaded Fig. 4.25 Centre loading of a simply supported beam of length L; with semicircular cross-section a transversely applied concentrated load P: a Shear force diagram. b Bending moment diagram Again, from (4.38), and given allowable value of smax; we get 3  3ÂP 2 F  37;500 Solution smax ¼ 2 N=mm2 ¼ Á A ¼ 4 For a rectangular beam, it is given that width, b ¼ 150 mm, ¼ P N=mm2 : 50;000 and height, h ¼ 250 mm: So, the cross-sectional area is A ¼ bh ¼ 150  250 ¼ 37;500 mm2; and the section modulus is ) Based on allowable shear stress: P ¼ 2  50;000 N ¼ 105 N ¼ 100 kN: (4.16b): Z ¼ bh2 ¼ 150  2502 ¼ 1;562;500 mm3: The length of 66 the beam is L ¼ 1200 mm: ðTrhxeÞmaalxlo¼waðbrlxeÞmwinork¼in1g5sMtrePsas in Since the load derived based on the allowable bending tension or compression is ¼ stress is lower than that based on the allowable shear stress, the safe value of applied load is P ¼ 78.125 kN. 15 N=mm2; and that in shear is smax ¼ 2 MPa ¼ 2 N=mm2: If a concentrated load P (in N) is applied transversely at the mid-section of a simply supported beam of length L ; 4.8.4. If a beam of circular cross-section, as shown in Fig. 4.26, has a total shear force equal to F; determine the then the resulting shear force and bending moment diagrams maximum shear stress induced at the neutral axis of the cross-section in terms of F and area, A; of the cross-section. are shown, respectively, in Fig. 4.25a, b. Solution From Fig. 4.25, it is clear that the maximum bending moment ðMBÞmax and the uniform shear force F are respectively: ðMBÞmax¼ PL ¼ 300 P N mm; and F ¼ P N: The moment of inertia of circular cross-section about the 4 2 neutral axis ðz-axisÞ is: From (4.14), and given allowable value of ðrxÞmax; we get Iz ¼ pD4=64: ðrxÞmax¼ 15 N=mm2 ¼ ðMBÞmax ¼ 300  P N=mm2 The statical moment about the neutral axis z of the shaded Z 1;562;500 semicircular cross-sectional area is: 3ÂP ¼ 15;625 N=mm2: Q ¼ area of the shaded semicircular cross-section  the distanceðycÞof the centroid of semicircle from ) Based on allowable bending stress: the neutral axis of the circular cross-section P ¼ 15  15;625 N ¼ 78;125 N ¼ 78:125 kN: ¼ ÀpD2=8Á  yc: 3