Lecture 10 bending stresses in beams
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This document discusses stresses in beams. It covers bending stresses, shear stresses, deflection in beams, and torsion in solid and hollow shafts. The key assumptions in beam bending theory are outlined. Bending stresses are explained, including the location of the neutral axis and how stresses vary through the beam cross-section based on the bending moment and geometry. Section modulus is defined as the ratio of the moment of inertia to the distance of the outermost fiber from the neutral axis. Composite beams made of different materials are also discussed.
![Composite beams
t t σ1 σ2
=
E1 E 2
y 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
€](https://image.slidesharecdn.com/lecture10bendingstressesinbeams-110823083017-phpapp01/85/Lecture-10-bending-stresses-in-beams-20-320.jpg)
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Lecture 10 bending stresses in beams
- 2. Unit 2- Stresses in Beams Topics 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.
- 3. 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.
- 4. Bending in beams Key Points: 1. Internal bending moment causes beam to deform. 2. For this case, top fibers in compression, bottom in tension.
- 5. 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.
- 6. 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
- 7. Bending Stress in beams Consider the simply supported beam below: Radius of Curvature, R P B Deflected A Shape Neutral Surface M M RA RB M M What stresses are generated within, due to bending?
- 8. Axial Stress Due to Bending: M=Bending Moment σx (Compression) M M Neutral 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
- 9. Bending Stress in beams
- 10. Bending Stress in beams
- 11. 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 €
- 12. 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 0 M M ∫ € y × dA = 0 Neutral axis coincides with the geometrical axis €
- 13. 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 R M M € ∫y 2 × dA = I E M E € M= I⇒ = R I R €
- 14. Flexure Formula M E σ = = I R y €
- 15. 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
- 16. 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 I Z= y max M σ = I y M σmax = I y max I M = σmax y max M = σmax Z €
- 17. Section Modulus of symmetrical sections Source:- http://en.wikipedia.org/wiki/Section_modulus
- 18. Section Modulus of unsymmetrical sections In case of symmetrical section neutral axis passes through geometrical center of the section. But in case of unsymmetrical section such as L and T neutral axis does not pass through geometrical center. The value of y for the outermost layer of the section from neutral axis will not be same.
- 19. Composite beams Composite beams consisting of layers with fibers, or rods strategically placed to increase stiffness and strength can be “designed” to resist bending.
- 20. Composite beams t t σ1 σ2 = E1 E 2 y 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 €