6. Non-uniform bending
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1 . Non-uniform bending Introduction Definition A non-uniform bending is te case were te cross-section is not only bent but also seared. It is known from te statics tat in suc a case, te bending moment in a member is not constant, ence te name nonuniform bending. Te case takes place as a result of te member loading, perpendicular to its longitudinal axis. Tis explains an alternative name of te case as a transverse loading. It is obvious tat te bending moment creates normal stress in te cross-section, wile te sear force creates searing stresses in tat section. Bernoulli s ypotesis We assume tat te cross-section, wic is a plane before loading, remains a plane after loading as result of te pure bending. However, it is only an approximate solution wen compared to real beams. It can be presented by means of simple consideration. Te plane cross-section means tat te straigt lines at te side surface of te beam remain straigt. Tus, te searing strains at te surface would be constant in te cross-section and due to linear pysical law (te Hooke s law). Te same refers to te sear stresses. Tis contradicts te static boundary condition, stating te sear stresses at te extreme fibers sould be zero, Fig..1. Fig..1. Warping due to sear force Fig..1.a) presents te plane cross-sections before loading and b) after te circular loading (te rigt angles remain rigt, tere is no searing strain). Te deformation of c) type is not allowed because of non-zero sear strains at extreme fibers. Te case d) represents te cross-section warping wit maximum value of sear strain at te neutral axis and zero at te extreme fibers. Te observation of te real beams deformations also sows te warping of te cross-sections. However, te solution of te teory of elasticity sows, tat if te sear force is constant, te distribution of normal stresses caused by te bending moment is te same as in te pure (circular) bending, i.e. it is te same as for te plane cross-sections. For te cases of non-constant sear force and common cross-section sapes, te error is quite acceptable 1. Normal stress In virtue of te Bernoulli s ypotesis te normal stress formula is similar to tat of te pure (circular) bending: M y x x z J, but te bending moment as well as te stress are functions of te axis coordinate. y 1 cf. for instance V.D. da Silva, Mecanics and Strengt of Materials, Springer 00, p.59
2 Sear stress As it as been sown, te computation of te searing stress always requires simplifying ypoteses, wic introduce errors, te importance of wic depends on te sape of te cross-section. From te equilibrium condition for an element cut from te beam under non-uniform bending, we get te formula for te searing stress in te cross-section plane: I Q x S y z xz J ybz, were: Q x is te cross-section sear force, S y I ( is te static moment of cut cross-section part, J y is te inertia moment of te wole cross-section, b ( is te widt of te cut line. Te sense and sign of te sear stress depends on te sense of te cross-section sear force and te assumed sense of te x-coordinate but te dependence is not as straigtforward as it seems. Let s consider an element cut from a beam wit positive sear force, Fig... a) N T b) N dx N-dN T dx N+dN Fig.. Searing stress for bending moment: a) positive, b) negative Positive sear force means te increasing moment wit bottom fibers tensioned or te decreasing moment wit bottom fibers compressed. In bot cases, te element s balance is ensured by te searing forces T of te same sense. Tese longitudinal sear forces cause in turn te searing stress, zx, wit te same direction and sense. Knowing tat te sear stresses act at te edge in pairs, zx xz, we can determine te sign of te sear stress. Te sign is not objective; it depends on te adopted coordinate axes. Usually, te maximum value of te sear stress is attained at te neutral axis for te normal stress. Design In most cases te dominant criterion in te design of a beam for strengt is te maximum value of te normal stress in te beam: M y max x R. W Next, te condition for te sear stress sould be verified: y
3 max xz R t were R t is te searing strengt, a material constant. Examples Example.1 Determine te sear stress distribution for te cross-sections: a) rectangular, b) triangular (isosceles triangle b) and c) circular. a) We calculate: Q b Q z z 1 4 xz. b b b Te maximum value is in te neutral axis: 1 Q xz ) xz ( z 0) A max(. Te stress distribution is parabolic wit te maximum 50 % greater tan te cross-section average. b) widt and eigt as z-functions: cross-section geometry term: b ) b( b z, ( z y ( 1 ( ( z z b( we searc a cut line wit te maximum value of te sear stress: zx 0 z z 0 9 z z Q Q Q Q max xz A J y 1 b 1 b Te maximum value is attained at te middle of te triangle eigt. c) static inertia moment: widt of te cut: so: S S I y r ( cos b ( r cos Q r cos 4 Q 4 Q xz cos 4 r r r r cos 4 and, finally, we find te maximum value at z = 0: 4 Q max xz. A z z 1 r Example. Determine te ratio of maximum values of te normal and sear stress in a cantilever wit a rectangular cross-section (b ), loaded by a point force at its free end (x = l).
4 te maximum value of te bending moment: M te sear force: Q P (constant) max Pl Pl te maximum value of te normal stress: max x b P te maximum value of te sear stress: max xz 1. 5 b max x Pl b l te ratio: 4. max xz b 1.5P As we can see, usually te normal stress is muc greater tan te sear stress. Example. Te welded profile IPES 00, made by te steelworks Pokój as te dimensions: te total eigt of 00 mm, te widt of te flanges: 0 mm, te eigt of te flanges: mm, te web tickness: 8 mm. Determine wat part of te cross-section sear force does te web carry. te static moment in te web: ( z z cm S y S y ( z 0) 177 cm S y ( z 77) 140 cm i.e. te diagram of te sear stress in te web is flat te static moment in te flange: ( y). 11 y 8.85.(11 y) cm S y S y ( y 0) 70 cm In Fig.. te diagram of te sear stress and te sear stress flow are sown. We calculate te sear force part carried by te web: Fig.. Sear stress in IPES 00 Q b i.e. te web carries 9.5% of te sear force. Example.4. QS y ( dz J b y Q 0.95Q, A simply supported beam wit a span lengt l = 4m and te rectangular cross-section aa is loaded by continuous loading q = 90 kn/m. Determine te value of parameter a of te cross-section if te acceptable values of normal and sear stress are R = 00 MPa and R t = 100 MPa, respectively. Draw te principal stresses lines.
5 te maximum value of te bending moment: te section modulus: max te design: max M y max x R Wy te verification of te sear stress te maximum value of te sear force: te sear stress: M y ql 8 W y a a 4 a a 0010 max Q 180 kn 180 knm a m max xz 1.5MPa Rt Te principal stresses: in te extreme fibers tere is te normal stress only, so te directions of te principal stress are vertical and orizontal, in te neutral axis tere is te sear stress only, so te directions of principal stress are rotated by 45 degrees in oter fibers at any z we ave: tg 1. 1 Te principal stress lines are drawn in Fig Two families of te principal stresses lines are perpendicular to eac oter. Trajectories of te tension principal stresses justify te location of reinforcement in RC beam. x zx Fig..4 Trajectories of te principal stress Review problems Problem.1 A timber beam wit te span of m and te widt 90 mm is to support tree concentrated forces sown in Fig..5. Knowing tat for te grade of timber used acc 1MPa and acc 0. 8MPa, determine te minimum required dept d of te beam. (Ans.: d = 1 cm) kn d
6 Fig..5 Beam wit load Problem. For te non-uniform bending of te cross-section of te beam in Fig.., determine te normal, sear and principal stresses at a point K in a section α α. Assume tat te cross-section dimensions are given in [cm]. (Ans.:σ x = 79.8 MPa, τ = 4.87 MPa, σ 1 = 0.1, σ = 79.9, α = 89 ) α 18 kn.5 m α.5 m Fig.. Beam and cross-section K 4 Problem. A timber beam of a rectangular cross-section carries a single concentrated load P in its midpoint, Fig..7. Determine te dept and te widt b of te beam, knowing tat R t = 1.5 MPa, R = 1 MPa. (Ans.: b = m) 40 kn 1 m 1 m b Fig..7 Beam wit loading Addendum Longitudinal sear flow Te longitudinal sear force (T in Fig..) per unit lengt is called te longitudinal sear flow. Te importance of tis internal force becomes visible wen compare a beam wit te same beam cut lengtways, Fig..8. Fig..8 Beam and beam cut lengtways In te cut beam te stresses and te curvature are and 4 times greater, respectively.
7 Sear centre ) Te equilibrium condition requires tat te action axis of te sear force as a position wic coincides wit te line of action of te resultant of te searing stress Terefore, te position of te action axis of te sear force is not arbitrary. Tere are two internal forces introducing searing stresses in te cross-section: te sear force and te torsional moment. Let s calculate te sear stresses in te cannel cross-section in Fig..9. Te resultants of te searing stress in te web (R a ) and in te flanges (R b ) reduce not only to te sear force but also to te torsion moment. To reduce te torsion moment, te sear force sould be applied at some distance d from te web. Fig..9 Cannel cross-section A sear centre is a point wic as te following property: if te line of action of te sear force passes troug tis point, it will not induce torsion of te bar. Te tin-walled cross-sections wit concurrent and straigt wall elements, Fig..10, are a particularly simple case of determination of te sear centre. Because te resultants of te searing stresses in te different wall elements pass troug te intersection of te centre lines, te moment of te searing stress in relation to tis point vanises, wic means tat it is te sear centre. Fig..10 Sear centre in particular cases. Glossary non-uniform bending, transverse loading zginanie poprzeczne longitudinal sear flow siła rozwarstwiająca strengt for sear stress wytrzymałość na ścinanie RC, reinforced concrete żelbet sear centre środek ścinania from: da Silva, op.cit.
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