Asymmetrical Bending Normal stress

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Asymmetrical Bending Normal stress

• Bending moment
• (i) about x

(ii) about y

• Strain due to rotation
• (i) about x (shown)

(ii) about y (not shown)
Total strain
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Asymmetrical Bending Normal stress

• Elasticity

• Bending moment curvature relations

• Solving the above relations for the curvatures

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Asymmetrical Bending Normal stress

• Substituting into the stress-curvatures relation

• Neutral axis szz 0

• Using Eq. (2) to eliminate My from Eq. (1)

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Asymmetrical Bending Normal stress

• Plane of loading making an angle f with the x axis

• 2nd moments of area reative to new axes n-s
• In Ixcos2q Iysin2q 2Ixycosq sinq
• Is Ixsin2q Iycos2q 2Ixycosq sinq

• Product moment of area
• Ins (Ix Iy )cosq sinq Ixy(cos2q sin2q )

• Axis n makes angle q with the x axis (positive
  counterclockwise)

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Asymmetrical Bending Normal stress

• Example (Problem 7.28, Boresi Schmidt, p.291)
• Cantilever beam subjected to a point load at the
  free end
• Bending moments from the FBDs on the right
• Critical section at the clamped end

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Asymmetrical Bending Normal stress

• Alternatively, considering plane of loading at
  angle f 100º to x axis
• Mmax 1?1.25 1.25 kNm
• Mx Mmax cos 10º 1.231 kNm,
• My Mmax sin 10º 0.217 kNm
• Neutral axis

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Asymmetrical Bending Normal stress

• Spreadsheet calculations of cross sectional
  properties
• Ix 184,333.33 mm4, Iy 189,333.33 mm4, Ixy
  66,000 mm4,
• Substituting into the formula for a
• tana 0.1885
• ? a 10.68º
• Extreme values of szz at points
• A(12, 23) and B(18, 17)
• szz (A) 181 MPa
• szz (B) 146 MPa

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Asymmetrical Bending Deflection

• Total deflection d is perpendicular to the
  neutral axis
• Differential equation for the deflection

• Mn is the bending moment about the neutral axis
• In is the second moment of area about the neutral
  axis obtained using the given transformation
  formula with q a.

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Asymmetrical Bending Deflection

• Vertical deflection v can be determined using the
  differential equation

• Again, using Eq. (2) to eliminate My

• Then

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Bending Shear stress in thin-wall open sections

• Shear develops along the middle line. Shear flow
• q t t
• Consider shear stress at distance s from top end
  of section (area A? st)
• Force equilibrium of element (ds)?(dz) in z
  direction
• (szz dszz)dA? - szzdA?
• (q dq)dz - qdz 0

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Bending Shear stress in thin-wall open sections

• Since

the bending stress formula gives

• Substituting Eq. (4) into (Eq. (3) generates the
  integrals 1st moments of area A? with respect
  to x and y axes

• Final formula for shear flow/stress

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Bending Shear stress in thin-wall open
sectionsShear Centre

• Shear flow in channel section
• Due to symmetry w.r.t. x-axis
• Ixy 0
• Consider plane of loading parallel to y-axis
• Vx 0

For any A? in the top flange,
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Bending Shear stress in thin-wall open sections

• Total shear force in each flange

• Torque due to the pair of F forces

• This torque will cause twist unless balanced by
  eccentricity of applied load. If plane of loading
  is moved by distance e in the x direction

• Point C Shear centre

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Bending Shear stress in thin-wall open sections

• Z-section
• Only vertical load Vx 0
• Asymmetric section assuming t ltlt d, t ltlt b

• Shear flow in the top flange

where
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Bending Shear stress in thin-wall open sections

• Substituting into the shear flow equation

• Shear force on the flange

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Bending Shear stress in thin-wall open sections

• Example 8.1 (Boresi Schmidt)
• Cross section symmetrical w.r.t. x-axis
• Ixy 0
• Only vertical loading
• Vx 0

• 2nd moment of area of oblique sub-section w.r.t.
  axis parallel to x through its centre (use
  transformation formula and assume t ltlt L)

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Bending Shear stress in thin-wall open sections

• Applying parallel axes theorem

• The shear flow formula is applied at a distance y
  from the x axis for which

• Shear force on the web

• Moments about D (F1 not required)
• (DB)F2 (DB) eVy ? e 30.1 mm