notes_12_bending_wo_t

# notes_12_bending_wo_t - Lecture 3 2003 Kollbruner Section...

This preview shows pages 1–5. Sign up to view the full content.

Lecture 3 - 2003 Kollbruner Section 5.2 Characteristics of Thin Walled Sections and . . Kollbruner Section 5.3 Bending without Twist thin walled => (cross section shape arbitrary and thickness can vary) axial stresses and shear stress along center of wall govern normal (to curved cross section) stress neglected position determined by curvelinear coordinate s along center line of cross section St. Venant torsion not a player as K ~ t^3 use shear flow: z => ζ x => ξ a) equilibrium of wall element: y , η z, ζ x x s ds dx t q σ q + dq / ds*ds τ = τ xs = τ sx σ + d σ dx*dx στ σ + d σ dx t ds σ⋅ t ds + q + d d s qds dx qdx = 0 using q = τ⋅ t dx as it includes τ (s) and t(s) d q + d σ t = 0 ds dx 1 notes_12_bending_wo_twist.mcd

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

View Full Document
s ds b) compatibility (shear strain) d u + d v = γ ds dx v is displacement in s direction u is displacement in x direction du/ds*ds=du du/ds dv/dx*dx=dv dv/dx dx view looking to surface c) tangential displacement ( δ v) in terms of η , ζ and φ dv α d ζ z component α dv d η y component v is displacement in s direction η is displacement in y direction ζ is displacement in z direction and φ is rotation all at point s on cross section. δ v is differential over distance dx δη is component of δ v in y δζ is component of δ v in z direction direction and hp* δφ is component due to differential rotation between x and x+dx dv d φ p d φ hp . hp*d φ rotation component superposition => dv d η cos α () d ζ sin α + h p d φ + = η , ζ and φ depend on s and x while α and hp are independent of x (prismatic section) rewrite as: 2 notes_12_bending_wo_twist.mcd x
δ v δη () + δζ sin α = cos α + h p δφ δ x δ x δ x δ x further assumptions: 1) preservation of cross section shape => ζ = ζ (x); η = η (x) φ = φ (x) 2) shear though finite is small ~ 0 => 3) Hooke's law holds => σ = E δ u δ x equilibrium for the cross section: σ dA = N x σ⋅ y dA = M z z dA = M y τ⋅ h p dA = qh p ds = T p dA = q cos α cos α ds = V y dA = q sin α sin α ds = V z 5.3 Bending without twist d u = d v ds dx axial stress Nx = axial force Mz and My are bending moments wrt y and z respectively note integral expressions Tp is torsional moment wrt cross sectional point P Qx and Qy shear forces σ dA = 0 p ds = 0 possible only if lateral loads pass through P 3 notes_12_bending_wo_twist.mcd

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

View Full Document
δ v δη () + δζ sin α = cos α + h p δφ from above, δ x δ x δ x δ x using d u = d v with no twist => δφ = 0 ds dx δ x cos α which can be integrated to become: becomes: δ u = d η d ζ sin α δ s dx dx ds − ζ ' ⋅ sin α where = ζ ' (prime is control F7) u = −η ' cos α ds + u 0 x d ζ dx δη and is η
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 10

notes_12_bending_wo_t - Lecture 3 2003 Kollbruner Section...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online