EA 3702 echanics & aterials Science (echanics of aterials) Chapter 4 Pure Bending
Pure Bending Ch 2 Aial Loading & Parallel Loading: uniform normal stress and shearing stress distribution Ch 3 Torsion: linear distributed shearing stress in the normal cross-section (and along longitude planes) Ch 4 Pure Bending: linearl distributed normal stress For the designing of beams and girders
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending ntroduction to Pure Bending Eample: bar for bar bell Free bod diagrams (FBD): For the entire bar bell: For the portion between the two hands: NO net forces in an direction Equal and opposite (bending) moments of force Pure Bending
Other Configurations Related to Bending Eccentric Loadings Transverse loadings n both cases, at cross-section CC, there is the need to consider (reaction) bending momentum of to balance the eternal bending moment due to P Both these two cases are NOT pure bending (discussed later) EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Smmetric ember in Pure Bending For a beam AB with a plane of smmetr subjected to bending moments & within that plane Sign conventions for pure bending moment : + Bending concave upward (ends upward) Bending concave down (ends downward)
Equations of Equilibrium for Pure Bending For a beam AB subject to pure bending w/ moments &, for a normal cross-section C with coordinates as defined, we have the following equations for that cross-section F = 0 = 0 = σ da = 0 σ da = 0 ( σ )da = n addition, as will be shown later, shearing stress Therefore, F da 0 = = 0 F da 0 da 0
Stress & Strain in a Prismatic ember in Pure Bending (1) Beam AB subjected to bending moments of - in the plane of smmetr (CAB) Two geometric considerations: Beam AB bends uniforml Line AB (or similarl lines parallel to the longitude ais) transforms into part of a circle centered at C. An cross-section perpendicular to the beam ais remains planar after bending and also passes through center C. EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending
Stress & Strain in a Prismatic ember in Pure Bending (2) Additional geometric considerations/observations: Side view Top view = 0 = 0 From Hooke s law, corresponding stress: = = 0 NO shearing stress between neighboring blocks as illustrated! EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending
Stress & Strain in a Prismatic ember in Pure Bending (3) From previous page = = 0 Additional considerations on other stress components: Net normal and shearing forces along and σ directions for an element τ at the surface should be σ ero for pure bending. Therefore = = = 0 For pure bending, of the si stress components, onl one - the aial normal stress is non-ero: For bending as illustrated, compression (σ < 0) in the upper section and tension (σ > 0) in the lower section EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 0
Neutral Surface & Neutral Ais for a Prismatic ember in Pure Bending Neutral Surface A plane or surface parallel to top (and bottom) plane for the beam where = 0 and = 0 Neutral Ais (N.A.) The intercept line between neutral surface and the normal crosssection (or the transverse section) Define origin O at the cross-point of aial ais and the neutral ais, with pointing upward ε > 0 ε < 0
Geometric Relationship in a Prismatic ember in Pure Bending (1) For a beam ABA B of original length L subjected to pure bending: Plane DOE is the neutral surface. Define the following terms: radius of curvature for the neutral surface the central angle After bending, arc length L L = ρθ Consider arc JK at distance awa from the neutral surface: its length after bending L will be: L = (ρ - )θ
Geometric Relationship in a Prismatic ember in Pure Bending (2) For arc JK, its original length is also L The absolute change in length δ for JK at from the neutral ais (N.A.) in bending is Therefore Aial strain in a given plane JK at a distance from the N.A. is δ = L - L δ = (ρ - )θ ρθ = - θ L
Geometric Relationship in a Prismatic ember in Pure Bending (3) Aial strain ε in pure bending is the ratio between distance from neutral plane and radius of curvature for the neutral surface For a given pure bending situation, ε increases linearl with distance from the neutral surface aimum aial strain occurs when reaches maimum value c (upper or lower surface) As a result, another wa to represent aial normal strain ε c ma ma c
Stresses and Deformation in the Elastic Range Within proportional limit, Hooke s law σ = Eε Aial normal Aial strain in stress in pure pure bending bending c Ec Since E Also, ma ma c ma For pure bending within proportional limit, aial normal stress increases linearl with distance from the neutral surface c ma ma E E Use when modulus E and ρ are known
Location of the Neutral Surface/Neutral Ais in Pure Bending Where is the neutral surface? Recall for pure bending F = 0 Aial stress in pure bending Therefore, We have da σ da = 0 c c da 0 ma ma da c ma da For pure bending within proportion limit, the neutral surface (& neutral ais) passes through the centroid of the crosssection. 0
Relationship between oment for Pure Bending and Normal Stress From previous, aial stress Recall the bending moment satisf Therefore, c c c ma ma 2 da da Define moment of inertia w.r.t. the centroid ais normal to the cross-section plane ma c Use when bending moment and geometr (c and ) are known ( σ )da = ma 2 da n pure bending, aial normal stress increases with ncreasing moment or distance from the neutral surface Decreasing moment of inertia (i.e., crosssection sie) w.r.t. the centroid ais.
Relationship between oment and Resulting Normal Stress Define elastic section modulus S aimum normal stress c due to bending is ma ma S For the same bending moment, beam cross-section could be optimied to obtain larger S or (i.e., /c) and smaller σ ma : Eample: Two prismatic beams with the same total rectangular cross-section area: The are subject to same bending moment in the vertical smmetr plane parallel to h S = c = 1 12 bh3 1 2 h = 1 6 bh2 = 1 6 Ah S The right one w/ larger h gives larger S and smaller σ ma c
Relationship between Bending Radius of Curvature & oment a normal strain We have ma c or 1 ma c Within proportional (elastic) limit, ε ma = σ ma Hooke s law E c Recall ma normal stress in bending ma 1 c Bending curvature ma ma 1 1 c E c Ec 1 E For pure bending, curvature increases with ncreasing bending moment Decreasing moment of inertia (i.e., crosssection sie) w.r.t. the centroid ais Decreasing elastic modulus E
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 2 in Class Eample (1) A metal bar with rectangular cross-section of 1 2 in 2 is subjected to bending moments in the vertical plane of smmetr. Calculate the bending moment that will cause the bar to ield (i.e., starting to deform plasticall) given ield strength Y = 36 ksi C aimum normal stress for pure bending: ma c Y 1 in Therefore, bending moment Y c 1 3 Y bh 12 Ybh 1 h 6 2 2 3610 3 lb in 6 2 1 2 2 in 3 24,000lb in
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 10 in Class Eample (2-1) Bending moment is applied in the plane of smmetr to a hollow rectangular metal allo beam (outer width b 2 = 5 inch and outer height h 2 = 10 inch, and uniform wall thickness of 1 inch). Please calculate (a) the maimum bending moment that can be applied if allowable normal stress is 12 10 3 psi; (b) the corresponding bending radius of curvature of the beam if the elastic modulus E = 1210 6 psi Knowing the cross-section moment of inertia 1 12 b 2 h 3 2 b h 1 3 1 C 5 in
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 10 in Class Eample (2-2) aimum normal stress ma c allowable 3 1210 psi C Therefore, bending moment ma_ allowable c ma_ allowable 1 12 1 h 2 b 2 h 3 2 b h 2 3 3 ma_ allowable b2h2 b1h 1 6h2 2 3 3 lb in 4 12000 510 38 in 692800lb in 610in 1 3 1 5 in
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 10 in Class Eample (2-3) Bending curvature 1 Bending radius of curvature E E C E 1 12 3 3 b h b h E 3 3 b h b h 2 2 1 1 2 2 12 1 1 5 in 1210 6 3 3 4 510 38 in 5000in 2 lb in 12692800lb in
Bending of embers ade of Several aterials (Composite Beam) (1) For uniform beam made of two different materials bonded together as illustrated Assume aial normal strain is still With proportional limit, aial normal stress satisf For material 1: σ 1 = E 1 ε = E 1 ρ For material 2: σ 2 = E 2 ε = E 2 ρ ε = ρ
Bending of embers ade of Several aterials (Composite Beam) (2) Treat the composite beam as one made of one material but with locall adjusted cross-section width based on ratio of modulus between different materials in the original beam E 1 E 2 For the transformed equivalenc, aial normal stress transf E 1 E 2 n E E 2 1 transf is the moment of inertia for the transformed cross-section assuming it is made of the same material (e.g., #1)
Bending of embers ade of Several aterials (Composite Beam) (3) transf Notes: Actual normal stress in material 1 1 = Actual normal stress in materials 2 2 = n, Curvature due be bending n E E 2 1 1 E1 transf
10 cm 10 cm Class Eample A composite bar is obtained b bonding a piece of metal 2 (E 2 = 200 GPa) between two pieces of metal 1 (E 1 = 100 GPa) plates with cross-section as shown. Please calculate the maimum stress in metal 1 and in metal 2 when the bar is subjected to pure bending moment = 5000 Nm in the plane of smmetr as illustrated 4 cm 2 1 2 cm Because n = E 2 /E 1 = 2, the transformed geometr for the composite bar will be 6 cm EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 1 4 cm
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 10 cm 10 cm Class Eample The maimum stress in metal 1 4 cm 6 cm ma (1) c transf 1 12 h 2 bh 3 6 2 bh 2 1 ma 65000N m ( 1) 2 3 0.060.1 m 50Pa 2 cm 4 cm The maimum stress in metal 2 ma ( 2) n ma (1) 250Pa 100Pa
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 10 cm 10 cm Class Eample Equivalentl, we ma also convert 1 to 2 with n = E 1 /E 2 = 0.5 4 cm 3 cm ma ma (2) c transf 1 12 h 2 bh 65000N m ( 2) 2 3 0.030.1 m 3 6 2 bh 100Pa 2 1 2 2 cm ma E1 ( 1) n ma (2) ma (2) 0.5100P 50Pa E 2
Deformation in a Transverse Cross-Section in Pure Bending (1) Geometric considerations in pure bending of a prismatic beam: The transverse cross section of a beam remains planar. However, the transverse cross-section would undergo in-plane deformation
Deformation in a Transverse Cross-Section in Pure Bending (2) ρ = radius of curvature 1/ = curvature Anticlastic curvature = 1 ρ = v ρ
Stress Concentrations in Bending Stress ma concentrate (i.e., show higher value) at certain location for bending Concentration factor K σ ma _actual = K c Local stress increases with sharper transition/local openings/cracks or larger ratio of dimension transition
Eccentric Aial Loading in a Plane of Smmetr (1) For eccentric aial loading as illustrated for member ADEB For section ADC, at a normal cross-section C, draw FBD: Balance of force along aial direction gives net internal force F = P Balance of moments gives internal moment = Pd EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending
Eccentric Aial Loading in a Plane of Smmetr (2) For straight section DCE, FBD is: At arbitrar cross-section C, FBD for section DC is: Overall load for DC viewed as superposition of two loads: Centric load of P-P Pure bending couples of - σ = (σ ) centric + (σ ) bending Recall epressions for normal aial stress due to centric loading and pure bending moment, we have σ = P A Non-uniform but linearl distributed normal stress for eccentric loading
Considerations Applicable within proportion limit & Saint-Venant s Principle Neutral ais usuall does not coincide with centroidal ais Eccentric Aial Loading in a Plane of Smmetr (3) σ = (σ ) centric + (σ ) bending - still use value for pure bending case σ = P A Uniform stress distribution for centric load Linear stress distribution for pure bending & overall
Class Eample An open link chain is made with uniform wire of 0.4 inch diameter as illustrated. Knowing the chain carries a load of 100 lb. Please calculate (a) the largest tensile and compressive stresses in the straight portion of the link, and (b) the distance between centroidal ais and neutral ais of the cross-section. Knowing moment of inertia Draw FBD of half link n the straight portion: nternal aial tensile force P = 100 lb Bending moment of = 100 lb 0.5 in = 50 lbin EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 1 r 4 4 0.5 in P = 100 lb 100 lb 100 lb 100 lb 0.4 in = 50 lbin 0.4 in
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Class Eample Normal stress in the straight section comes from two contributions P A P A Normal stress due to centric tensile force P 100lb 3.14160.2 in 2 2 796 psi P = 100 lb aimum normal stress due to pure bending moment = 50 lbin 100 lb 0.4 in c r 1 r 4 4 4 450lb in 7958 psi 3 3 3 r 3.14160.2 in
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Class Eample aimum tensile stress ( Tension) 796 psi 7958psi 8754 psi ma aimum compressive stress ( Compression) 796 psi 7958psi 7162 psi ma P = 100 lb = 50 lbin 0.4 in For the distance δ between the centroidal ais and the neutral ais P A P P 4 A 1 0 4 r 4 A r 4 <0 C O 100 lb δ P A r 4 4 P r 2 r 4 4 4 P r 2 2 100lb0.2 in 450lb in 2 0.02in >0
0.8 in 0.4 in ore Comple Beam (1) A bending moment is applied at the centroid of a T-shaped beam in the plane of smmetr, as illustrated. Neglecting the fillets effects, please calculate the maimum tensile and compressive stress EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 3482 lbin 1.8 in C 0.6 in First, determine the location of centroid: Divide the cross-section into two sections: Section 1 (top): 1.8 in A 1 = 1.8 in 0.4 in = 0.72 in 2, 1_ave = 1.0 in (from bottom) Section 2 (bottom): C A 2 = 0.6 in 0.8 in = 0.48 in 2, 2 2_ave = 0.4 in (from bottom) Y ave = (A 1 1_ave +A 2 2_ave )/(A 1 +A 2 ) = (0.721.0+0.480.4)/(0.72+0.48) in = 0.76 in (from bottom) 1_ave 2_ave 0.6 in Y ave 1 0.8 in 0.4 in
ore Comple Beam (1) The total moment of inertia: 2 A d i oment of inertia for section i A i Area for section i C 2 d i Distance between the centroid of section i to the centroid ais for the entire cross-section Therefore, 1 3 1 3 4 1.80.4 0.720.24 0.60.8 0.480.36 0.139in 12 12 aimum tensile stress occurs at the bottom surface ct 3482lb in0.76in 3 ma ( t) 19.010 psi 4 0.139in aimum compressive stress occurs at the top surface cc 3482lb in0.44in 3 ma ( c) 11.010 psi 4 0.139in EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending i i i i 1_ave 2_ave 1.8 in 0.6 in Y ave 1 0.8 in 0.4 in
General Case of Eccentric Aial loading For general eccentric aial loading P-P not in plane of smmetr, as illustrated, equivalent loading contain: 1. Centric aial force P P 2. Bending moment around = Pa 3. Bending moment around = Pb From Saint-Venant s principle, overall aial normal stress: P A
General Case of Eccentric Aial loading For the location of the neutral ais (N.A.), 0 A P A P or A straight line in the - plane
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Class Eample A load P is applied to the end of a S1025.4 rolled steel member. Knowing the compressive stress in the member should not eceed 12 ksi, please calculate the largest permissible load P For standard beam of S1025.4, from 4.75 in P Table (Appendi C), the following D E information is obtained: 1.5 in Cross-section area A = 7.45 in 2 Section moduli: S = /C = 24.6 in 3 and S = /C = 2.89 in 3 The equivalent moment for bending around and ais: = P (4.75 in) = P (1.5 in) A C B
Class Eample Total aial normal stress has three contributions Aial normal stress due to centric loading P: P P A P 7.46in 0. 1340 2 P in aimum aial normal stress due to bending moment : S EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 2 4.75in P 2 0. 1931P in 3 24.6in aimum aial normal stress due to bending moment : S P 1.5in P 2 0. 519P in 3 2.89in A D C B 4.75 in P E 1.5 in
Class Eample A B The total aial stress at each of the corner points are obtained b eamining the sign for each factor 2 2 A P ( 0.1340P 0.1931P 0.5190P) in 0.578P in 2 2 B P ( 0.1340P 0.1931P 0.5190P) in 0.460P in 2 2 D P ( 0.1340P 0.1931P 0.5190P) in 0.192P in 2 2 E P ( 0.1340P 0.1931P 0.5190P) in 0.846P in Largest compressive stress <12ksi E P 0.846P in 2 1210 EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 3 lb in 2 P D C 3 1210 lb 0.846 P E 1.5 in 4.75 in 14.210 3 lb
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Unsmmetric Bending (1) Smmetric bending: ember has plane(s) of smmetr and moment is in that smmetr plane. Eample: bend around ais & ais Unsmmetric bending: The bending moment is NOT in a smmetr plane, or The member has does not have plane(s) of smmetr
Unsmmetric Bending (2) f couple vector for bending moment is directed along one of the principle centroidal aes of the crosssection: The neutral ais (N.A.) will coincide with the ais of the couple and the equations for smmetric members could be applied. The selection of principle centroidal ais will satisf C N.A. da 0 Principle of superposition could be used to determine stress in general case of unsmmetric bending. C N.A.
Unsmmetric Bending (3) Resolve moment along and = Cosθ σ = = Sinθ σ = + Left view Within proportional limit & Saint-Venant s principle is applicable
Unsmmetric Bending (4) Linear distribution of Neutral ais (N.A.) generall NOT coincide with bending couple. To determine eact location of N.A cos sin 0 = tanθ Define as the angle between N.A. and : tan tan
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Homework 4.0 Read tetbook for chapter 4.1 to 4.5, and 4.7 to 4.9 and give a statement confirm reading
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending 100 mm Homework 4.1 For pure bending moment of 5 knm on hollow beam with uniform wall thickness of 10 mm and cross-section dimension as shown, please calculate the stress at point E and point F, respectivel. E C F 50 mm
40 mm 20 mm Homework 4.2 The beam shown is made of polmer which has allowable tensile stress of 25 Pa and compression stress of 30 Pa. Please calculate the largest value for the bending couple that can be applied to the plane of smmetr to the beam. 50 mm 25 mm EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Homework 4.3 A flat 10 mm wide long strip of steel is bent into part of a circle with radius of curvature of 100 mm b two bending couples as shown. Calculate (a) the maimum thickness of the steel strip if allowable stress is 400 Pa, (b) with the dimension designed, the corresponding moment applied to reach maimum stress of 400 Pa knowing E = 200 GPa. A A A A 10 mm
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Homework 4.4 A composite bar having aluminum plate (E Al = 70 GPa, allowable stress = 100 Pa) sandwiched between two copper plates (E Brass = 105 GPa, allowable stress = 150 Pa). Please calculate the largest permissible bending moment when the composite bar is subject to bending moment about a horiontal ais, as illustrated. 1 cm 2 cm 1 cm 4 cm
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Homework 4.5 A plastic clinder support with radius of 4 in is subjected to 5000 lb eccentric aial force as shown. Determine the aial normal stress at point B when (a) a = 0, (b) a = 2 in. Knowing moment of inertia for the cross-section is 1 r 4 4 B 5000 lbs a
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Homework 4.6 A member is subject to loading force in the vertical plane of smmetr as illustrated. The allowable stress in horiontal crosssection FGH is 100 Pa. Please calculate the largest force P that can be applied. 5 cm P P 2 cm F G G H 2 cm 4 cm H 5 cm P F G
EA 3702 echanics & aterials Science Zhe Cheng (2018) 4 Pure Bending Homework 4.7 The bending moment couple is applied to a beam cross-section in a plane forming an angle =30 o from the horiontal plane. Please calculate stress at points of E, F, and G. = 500 lbin E G F 0.5 in 0.5 in 0.5 in