File Lecture5_Slides - Mechanics of Solids Lecture 5 by Dr...

Info icon This preview shows pages 1–13. Sign up to view the full content.

View Full Document Right Arrow Icon
Mechanics of Solids Lecture 5 by Dr Emre Erkmen Lecturer, School of Civil and Environmental Engineering, The University of Technology Sydney Office: 2.520 Phone:9514 9769 Email: [email protected] Lecture hours: Tuesdays 11:00-14:00, Wednesdays 15:00-18:00 Office hours: Tuesdays 15:00-17:00 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Mechanics of Solids - Lecture 5 Outline - Shear stress - Shear flow Built-up sections 2 -
Image of page 2
Mechanics of Solids - Lecture 5 Effect of longitudinal Shear stress 3
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Mechanics of Solids - Lecture 5 Effect of longitudinal Shear stress Slip between the unbonded boards 4 When boards are bonded together
Image of page 4
Mechanics of Solids - Lecture 5 Member under general loading The system of forces that may exist at a section of a beam include: SHEARING FORCE BENDING MOMENT 5 M V FBD x R A
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Mechanics of Solids - Lecture 5 Relation between moment and shear force Consider a segment Δ x isolated from a beam: M M + Δ M q 6 x A B Δ x Δ x V + Δ V V Since the segment is in equilibrium Σ M A = 0 A B
Image of page 6
Mechanics of Solids - Lecture 5 Relation between moment and shear force Δ x M M + Δ M V + Δ V V q By simplifying and neglecting the infinitesimals of higher order, the equation can be reduced to: ( ) ( ) 0 2 x M M M V V x q x Δ + Δ + Δ Δ − Δ = M V x Δ = Δ 7 Shear depends on the difference between the bending moments on the adjoining sections. If no shear acts at the adjoining sections of a beam, no change in the bending moment occurs. M V x Δ = Δ dM dx = 0 x Δ
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Mechanics of Solids - Lecture 5 Longitudinal Shear stress V V+ V M+ M t σ t t σ σ + Δ q t σ t t σ σ + Δ q F s My I σ = − y 8 x b b σ σ + Δ b σ
Image of page 8
Mechanics of Solids - Lecture 5 Shear force-Shear stress relation The normal stresses are due to the bending moment dA ( ) M M y I + Δ = − σ My I = − σ σ + Δ A’ Δ x 9 F s ( ) ' ' d d 0 s A A F A A σ σ σ + + Δ = ' ' d d s A A M M M F y A y A I I + Δ = ' d A M y A I Δ = s F b x τ = Δ ' d A M y A I Δ = ' d A M y A xbI τ Δ = Δ ' d A V y A bI = b
Image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon