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# Lecture 10 bending stresses in beams

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### Lecture 10 bending stresses in beams

1. 1. Unit 2- Stresses in BeamsTopics Covered  Lecture -1 – Review of shear force and bending moment diagram  Lecture -2 – Bending stresses in beams  Lecture -3 – Shear stresses in beams  Lecture -4- Deflection in beams  Lecture -5 – Torsion in solid and hollow shafts.
2. 2. 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.
3. 3. Bending in beamsKey Points:1.  Internal bending moment causes beam to deform.2.  For this case, top fibers in compression, bottom in tension.
4. 4. Bending in beams Key Points: 1.  Neutral surface – no change in length. 2.  Neutral Axis – Line of intersection of neutral surface with the transverse section. 3.  All cross-sections remain plane and perpendicular to longitudinal axis.
5. 5. Bending in beams Key Points: 1.  Bending moment causes beam to deform. 2.  X = longitudinal axis 3.  Y = axis of symmetry 4.  Neutral surface – does not undergo a change in length
6. 6. Bending Stress in beams Consider the simply supported beam below: Radius of Curvature, R P B Deflected A ShapeNeutral Surface M M RA RB M M What stresses are generated within, due to bending?
7. 7. Axial Stress Due to Bending: M=Bending Moment σx (Compression) M MNeutral Surface σx=0 Beam σx (Tension) stress generated due to bending: σx is NOT UNIFORM through the section depth σx DEPENDS ON: (i) Bending Moment, M (ii) Geometry of Cross-section
8. 8. Bending Stress in beams
9. 9. Bending Stress in beams
10. 10. Stresses due to bending y R Strain in layer EF = R A’ C’ N’ N’ Stress _ in _ the _ layer _ EF E F E= B’ D’ Strain _ in _ the _ layer _ EF σ E= € ⎛ y ⎞ ⎜ ⎟ ⎝ R ⎠ σ E E = σ= y y R R €
11. 11. Neutral axis dA force on the layer=stress on layer*area of layer dy = σ × dA y E = × y × dA R N A Total force on the beam section € E σx σx = ∫ R × y × dA Stress diagram E = R ∫ y × dA x For equilibrium forces should be 0M M ∫ € y × dA = 0 Neutral axis coincides with the geometrical axis €
12. 12. Moment of resistance dA force on the layer=stress on layer*area of layer dy = σ × dA y E = × y × dA R N A Moment of this force about NA € E = × y × dA × y σx σx R Stress diagram E = × y 2 × dA R E E x Total moment M= ∫ R × y 2 × dA = ∫ y 2 × dA RM M € ∫y 2 × dA = I E M E € M= I⇒ = R I R €
13. 13. Flexure Formula M E σ = = I R y€
14. 14. Beam subjected to 2 BM In this case beam is subjected to moments in two directions y and z. The total moment will be a resultant of these 2 moments. You can apply principle of superposition to calculate stresses. (topic covered in unit 1). Resultant moments and stresses
15. 15. Section ModulusSection modulus is defined as ratio of moment of inertia about the neutral axis tothe distance of the outermost layer from the neutral axis I Z= y max M σ = I y M σmax = I y max I M = σmax y max M = σmax Z €
16. 16. Section Modulus of symmetrical sectionsSource:-http://en.wikipedia.org/wiki/Section_modulus
17. 17. Section Modulus of unsymmetrical sectionsIn case of symmetrical section neutral axis passes through geometrical center ofthe section. But in case of unsymmetrical section such as L and T neutral axisdoes not pass through geometrical center.The value of y for the outermost layer of the section from neutral axis will not besame.
18. 18. Composite beamsComposite beams consisting of layers with fibers, or rods strategically placed toincrease stiffness and strength can be “designed” to resist bending.
19. 19. Composite beams t t σ1 σ2 = E1 E 2y E1 σ1 = σ E2 2 d = mσ 2 m=modular ratio σ M= I y € M = M1 + M 2 b σ σ = 1 I1 + 2 I2 y y σ = 2 [ mI1 + I2 ] y Equivalent I (moment of inertia)= mI1 + I2 €