Theory of simple bending (assumptions) Material of beam is homogenous and isotropic => constant E in all direction Young s modulus is constant in compression and tension => to simplify analysis Transverse section which are plane before bending before bending remain plain after bending. => Eliminate effects of strains in other direction (next slide) Beam is initially straight and all longitudinal filaments bend in circular arcs => simplify calculations Radius of curvature is large compared with dimension of cross sections => simplify calculations Each layer of the beam is free to expand or contract => Otherwise they will generate additional internal stresses.
Bending in beams Key Points: 1. Internal bending moment causes beam to deform.. For this case, top fibers in compression, bottom in tension.
Bending in beams Key Points: 1. Neutral surface no change in length.. Neutral Axis Line of intersection of neutral surface with the transverse section. 3. All cross-sections remain plane and perpendicular to longitudinal axis.
Bending in beams Key Points: 1. Bending moment causes beam to deform.. X = longitudinal axis 3. Y = axis of symmetry 4. Neutral surface does not undergo a change in length
Bending Stress in beams Consider the simply supported beam below: Radius of Curvature, R P Neutral Surface A B Deflected Shape R A M M R B M M What stresses are generated within, due to bending?
Axial Stress Due to Bending: M=Bending Moment Neutral Surface M σ x (Compression) σ x =0 σ x (Tension) M Beam stress generated due to bending: σ x is NOT UNIFORM through the section depth σ x DEPENDS ON: (i) Bending Moment, M (ii) Geometry of Cross-section
Bending Stress in beams
Bending Stress in beams
Stresses due to bending R Strain in layer EF = y R A C N N E F B D E = Stress_in _ the _ layer _ EF Strain _in _ the _ layer _ EF E = σ y R σ y = E R σ = E R y
Neutral axis da dy y force on the layer=stress on layer*area of layer = σ da = E R y da N A σ x σ x Stress diagram Total force on the beam section E = R y da = E R y da M x M For equilibrium forces should be 0 y da = 0 Neutral axis coincides with the geometrical axis
Moment of resistance da dy y force on the layer=stress on layer*area of layer = σ da = E R y da N A Moment of this force about NA σ x σ x Stress diagram = E R y da y = E R y da M x M Total moment M= E R y da = E y R y da = I da M = E R I M I = E R
Flexure Formula M I = E R = σ y
Beam subjected to BM In this case beam is subjected to moments in two directions y and z. The total moment will be a resultant of these moments. You can apply principle of superposition to calculate stresses. (topic covered in unit 1). Resultant moments and stresses
Section Modulus Section modulus is defined as ratio of moment of inertia about the neutral axis to the distance of the outermost layer from the neutral axis Z = M I M I I y max = σ y = σ max y max M = σ max M = σ max Z I y max
Section Modulus of symmetrical sections Source:- http://en.wikipedia.org/wiki/section_modulus
BEAMS: COMPOSITE BEAMS; STRESS CONCENTRATIONS Slide No. 1 Bending of In the previous discussion, we have considered only those beams that are fabricated from a single material such as steel. However, in engineering design there is an increasing trend to use beams fabricated from two or more materials.
Slide No. Steel Bending of These are called composite beams. They offer the opportunity of using each of the materials employed in their construction advantage. Concrete Aluminum Steel Slide No. 3 Foam Core with Metal Cover Plates Consider a composite beam made of metal cover plates on the top and bottom with a plastic foam core as shown by the cross sectional area of Figure 6. The design concept of this composite beam is to use light-low strength foam to support the load-bearing metal plates located at the top and bottom.
Slide No. 4 Foam Core with Metal Cover Plates Foam Core t m Figure 6 Metal Face Plates b h f t m Slide No. 5 Foam Core with Metal Cover Plates The strain is continuous across the interface between the foam and the cover plates. The stress in the foam is given by σ f = E f ε 0 (53) The stress in the foam is considered zero because its modulus of elasticity E f is small compared to the modulus of elasticity of the metal.
Slide No. 6 Foam Core with Metal Cover Plates Assumptions: Plane sections remain plane before and after loading. The strain is linearly distributed as shown in Figure 7. Slide No. 7 Foam Core with Metal Cover Plates y M Compressive Strain Neutral Axis x Tensile Strain Figure 7
Slide No. 8 Foam Core with Metal Cover Plates Using Hooke s law, the stress in the metal of the cover plates can be expressed as y σ m = εem = Em (53) ρ but E σ m m = / ρ = M / I My I x x, therefore (54) Slide No. 9 Foam Core with Metal Cover Plates The relation for the stress is the same as that established earlier; however, the foam does not contribute to the load carrying capacity of the beam because its modulus of elasticity is negligible. For this reason, the foam is not considered when determining the moment of inertia I x.
Slide No. 10 Foam Core with Metal Cover Plates Under these assumptions, the moment of inertia about the neutral axis is given by hf tm btm I ( ) NA Ad = btm f m = h + t (55) Combining Eqs 54 and 55, the maximum stress in the metal is computed as ( hf + tm ) ( h + t ) M = max σ (56) bt m f m Slide No. 11 Foam Core with Metal Cover Plates The maximum stress in the metal plates of the beam is given by Foam Core Metal Face Plates h f t m σ max M = bt m ( h f + tm ) ( h + t ) f m (56) b t m
Slide No. 1 Example 1 A simply-supported, foam core, metal cover plate composite beam is subjected to a uniformly distributed load of magnitude q. Aluminum cover plates 0.063 in. thick, 10 in. wide and 10 ft long are adhesively bonded to a polystyrene foam core. The foam is 10 in. wide, 6 in. high, and 10 ft long. If the yield strength of the aluminum cover plates is 3 ksi, determine q. Slide No. 13 Example 1 (cont d) The maximum moment for a simply supported beam is given by ql q( 10 1) M max = = = 1800q 8 8 When the composite beam yields, the stresses in the cover plates are σ max = F y = 3,000 psi
Slide No. 14 Example 1 (cont d) Substituting above values for M max and σ max into Eq. 56, we get M ( hf + tm ) σ max = btm( h f + tm ) 1800q( 6 + 0.063) 3,000 = 10( 0.063)[ 6 + 0.063] Or lb lb q = 67. = 806 in ft Bending of Members Made of Several Materials The derivation given for foam core with metal plating was based on the assumption that the modulus of elasticity E f of the foam is so negligible,that is, it does not contribute to the load-carrying capacity of the composite beam.
Slide No. 16 Bending of Members Made of Several Materials When the moduli of elasticity of various materials that make up the beam structure are not negligible and they should be accounted for, then procedure for calculating the normal stresses and shearing stresses on the section will follow different approach, the transformed section of the member. Slide No. 17 Transformed Section Consider a bar consisting of two portions of different materials bonded together as shown in Fig. 8. This composite bar will deform as described earlier. Thus the normal strain ε x still varies linearly with the distance y from the neutral axis of the section (see Fig 8b), and the following formula holds: y ε x = (57) ρ
Slide No. 18 Transformed Section y y M 1 y ε x = ρ σ = 1 E 1 y ρ N.A ε x σ x σ = E y ρ Figure 8 (a) (b) (c) Slide No. 19 Transformed Section Because we have different materials, we cannot simply assume that the neutral axis passes through the centroid of the composite section. In fact one of the goal of this discussion will be to determine the location of this axis.
Slide No. 0 Transformed Section We can write: σ 1 E1 y E1ε x = ρ = (58a) E y σ = Eε x = ρ From Eq. 58, it follows that df 1 E1 y σ1da = da ρ E y df = σ da = da ρ (58b) = (59a) (59b) Slide No. 1 Transformed Section But, denoting by n the ratio E /E 1 of the two moduli of elasticity, df can expressed as ( ne1 ) y E1 y df = da = ( nda) (60) ρ ρ Comparing Eqs. 59a and 60, it is noted that the same force df would be exerted on an element of area n da of the first material.
Slide No. Transformed Section This mean that the resistance to bending of the bar would remain the same if both portions were made of the first material, providing that the width of each element of the lower portion were multiplied by the factor n. The widening (if n>1) and narrowing (n<1) must be accomplished in a direction parallel to the neutral axis of the section. Slide No. 3 Transformed Section b 1 N.A E n = E 1 = b Figure 9 b n b
Slide No. 4 Transformed Section Since the transformed section represents the cross section of a member made of a homogeneous material with a modulus of elasticity E 1,the previous method may be used to find the neutral axis of the section and the stresses at various points of the section. Figure 30 shows the fictitious distribution of normal stresses on the section. Slide No. 5 Transformed Section y y C σ = x N.A. My I σ x Figure 30. Distribution of Fictitious Normal Stress on Cross Section
Slide No. 6 Stresses on Transformed Section 1. To obtain the stress σ 1 at a point located in the upper portion of the cross section of the original composite beam, the stress is simply computed from My/I.. To obtain the stress σ at a point located in the upper portion of the cross section of the original composite beam, stress σ x computed from My/I is multiplied by n. Slide No. 7 Example A steel bar and aluminum bar are bonded together to form the composite beam shown. The modulus of elasticity for aluminum is 70 GPa and for streel is 00 GPa. Knowing that the beam is bent about a horizontal axis by a moment M = 1500 N- m, determine the maximum stress in (a) the aluminum and (b) the steel.
Slide No. 8 Example (cont d) M Steel 0 mm Aluminum 40 mm 30 mm Slide No. 9 Example (cont d) First, because we have different materials, we need to transform the section into a section that represents a section that is made of homogeneous material, either steel or aluminum. We have n E = s Ea 00 = = 70.857
Slide No. 30 Example (cont d) Steel 0 mm 30 mm n = 85.71 mm Aluminum Aluminum 40 mm Aluminum 30 mm Figure 31a 30 mm Figure 31b Slide No. 31 Example (cont d) 10 y = I C NA Consider the transformed section of Fig. 31b, therefore ( 85.71 0) + 40( 30 40) = ( 85.71 0) + ( 30 40) 3 ( ) ( 85.71 30)(.353 0) 85.71.353 = 3 30 + 3 ( 40 + 0.353) 3 4 9 4 3.353 mm from top 3 = 85.4 10 3 mm = 85.4 10 m
Slide No. 3 Example (cont d) 85.71 mm y C =.353 mm N.A. C 0 mm 40 mm 30 mm Slide No. 33 Example (cont d) a) Maximum normal stress in aluminum occurs at extreme lower fiber of section, that is at y = -(0+40-.353) = -37.65 mm. 3 My 1500( 37.65 10 ) 6 σ al = = = 66.53 10 Pa 9 I 85.4 10 = + 66.53 MPa (T)
Slide No. 34 Example (cont d) b) Maximum normal stress in stelel occurs at extreme upper fiber of the cross section, that is. at y =+.353 mm. 3 (.353 10 ) My 1500 n = (.867) I 85.4 10 = 11.8 MPa (C) σ St = 9 = 11.8 10 6 Pa Slide No. 35 Reinforced Concrete Beam An important example of structural members made of different materials is demonstrated by reinforced concrete beams. These beams, when subjected to positive bending moments, are reinforced by steel rods placed a short distance above their lower face as shown in Figure 33a.
Slide No. 36 Reinforced Concrete Beam Figure 3 Dead and Live Loads M Slide No. 37 Reinforced Concrete Beam Figure 33 d b d - x x b C 1 x N.A. σ F x n A s (a) (b) (c)
Slide No. 38 Reinforced Concrete Beam Concrete is very weak in tension, so it will crack below the neutral surface and the steel rods will carry the entire tensile load. The upper part of the concrete beam will carry the compressive load. To obtain the transformed section, the total cross-sectional area A s of steel bar is replaced by an equivalent area na s. Slide No. 39 Reinforced Concrete Beam The ratio n is given by Modulus of Elasticity for Steel E n = = Modulus of Elasticity for Concrete E The position of the neutral axis is obtained by determining the distance x from the upper face of the beam (upper fiber) to the centroid C of the transformed section. s c
Slide No. 40 Reinforced Concrete Beam Note that the first moment of transformed section with respect to neutral axis must be zero. Since the the first moment of each of the two portions of the transformed section is obtained by multiplying its area by the distance of its own centroid from the neutral axis, we get Slide No. 41 Reinforced Concrete Beam x ( bx) ( d x)( na ) = or 1 bx + nas x nasd = 0 (61) Solving the quadratic equation for x, both the position of the neutral axis in the beam and the portion of the cross section of the concrete beam can be obtained. s 0
Slide No. 4 Reinforced Concrete Beam The neutral axis for a concrete beam is found by solving the quadratic equation: b 1 bx + na x na d b = 0 s s (6) d d - x x C 1 x n A s Slide No. 43 Example 3 5 -in A concrete floor slab is reinforced by 8 diameter steel rods placed 1 in. above the lower face of the slab and spaced 6 in. on centers. The modulus of elasticity is 3 10 6 psi for concrete used and 30 10 6 psi for steel. Knowing that a bending moment of 35 kip in is applied to each 1-ft width of the slab, determine (a) the maximum stress in concrete and (b) the stress in the steel.
Slide No. 44 Example 3 (cont d) M = 35 kip in 4 in. 1 in. 5 in. 6 in. 6 in. 6 in. 6 in. 4 in. 5 in. Slide No. 45 Example 3 (cont d) Transformed Section Consider a portion of the slab 1 in. wide, in 5 which there are two -in diameter rods having a 8 total cross-sectional area 4 in. x 4 - x 1 in. C N.A. A s 5 π 8 = = 0.614 in 4 E n = E s s c na = 10 30 10 = 6 3 10 6 ( 0.614) = 6.14 in = 10
Example 3 (cont d) Neutral Axis The neutral axis of the slab passes through the centroid of the transformed section. Using Eq. 6: Slide No. 46 Quadratic Formula b ± x = b 4ac a x1 = 1.575 take x =.599 1 bx + nas x nas d = 0 1 6x + 6.14x 4.56 = 0 ( 1) x + 6.14x 6.14( 4) = 0 x =1.575 in Slide No. 47 Example 3 (cont d) Moment of Inertia The centroidal moment of inertia of the transformed section is 1 in. 4 in. 1.575.45 C N.A. 1 1.575 I = 3 ( ) 3 + 6.14 4 (.45) = 51.7 in 6.14 in
Example 3 (cont d) Maximum stress in concrete: 35( 1.575) σ c = My I = Stress in steel: σ s = n My I 51.7 (.45) = 1.066 ksi (C) 35 = ( 10) = + 16.4 ksi (T) 51.7 Stress Concentrations a b Slide No. 49 Stress concentrations may occur: in the vicinity of points where the loads are applied in the vicinity of abrupt changes in cross section Mc σ m = K I Figure 33
Stress Concentrations Slide No. 50 Example 4 Grooves 10 mm deep are to be cut in a steel bar which is 60 mm wide and 9 mm thick as shown. Determine the smallest allowable width of the grooves if the stress in the bar is not to exceed 150 MPa when the bending moment is equal to 180 N m. Stress Concentrations Slide No. 51 Example 4 (cont d) Figure 34
Stress Concentrations Slide No. 5 Example 4 (cont d) From Fig. 34a: 1 1 d = ( ) d = 60 10 = 40 mm c = ( 40) = 0 mm The moment of inertia of the critical cross section about its neutral axis is given by 3 3 3 9 4 ( 9 10 )( 40 10 ) = 48 10 m 1 3 1 I = bd 1 = 1 Stress Concentrations Slide No. 53 Example 4 (cont d) Therefore, the stress is σ Mc 180 = I 48 10 = 9 3 ( 0 10 ) = 75 MPa Using Mc σ m = K I 150 = K = ( 75) K Also D 60 = = 1.5 d 40
Stress Concentrations Slide No. 54 Example 4 (cont d) From Fig. 33b, and for values of D/d = 1.5 and K =, therefore r = 0.13 d r = 0.13 ( d ) = 0.13( 40) = 5. mm Thus, the smallest allowable width of the grooves is ( 5.) 10.4 mm r = = Stress Concentrations a b Slide No. 55 Stress concentrations may occur: in the vicinity of points where the loads are applied in the vicinity of abrupt changes in cross section Mc σ m = K I Figure 33