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Cofounder at EbyT Technologies + Looking for opportunities in Robotics Autonomous Navigation and SLAM

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- 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. 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. Bending in beamsKey Points:1. Internal bending moment causes beam to deform.2. For this case, top fibers in compression, bottom in tension.
- 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. 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. 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. 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. Bending Stress in beams
- 9. Bending Stress in beams
- 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. 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. 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. Flexure Formula M E σ = = I R y€
- 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. 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. Section Modulus of symmetrical sectionsSource:-http://en.wikipedia.org/wiki/Section_modulus
- 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. Composite beamsComposite beams consisting of layers with fibers, or rods strategically placed toincrease stiffness and strength can be “designed” to resist bending.
- 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 €

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