AERSP 301 Shear of beams (Open Cross-section) Jose Palacios

Shear of Open and Closed Section Beams Megson – Ch. 17 Open Section Beams Consider only shear loads applied through shear center (no twisting) Torsion loads must be considered separately Assumptions Axial constraints are negligible Shear stresses normal to beam surface are negligible Near surface shear stress = 0 Walls are thin Direct and shear stresses on planes normal to the beam surface are const through the thickness Beam is of uniform section Thickness may vary around c/s but not along the beam Thin-Walled Neglect higher order terms of t (t2, t3, …) Closed Section Beams Consider both shear and torsion loading

Force equilibrium: General stress, Strain, and Displacement Relationships S – the distance measured around the c/s from some convenient origin σz – Direct stress (due to bending moments or bending action of shear loads)  – Shear stresses due to shear loads or torsion loads (for closed section) σs – Hoop stress, usually zero (non-zero due to internal pressure in closed section beams) zs = sz =  shear flow; shear force per unit length q =  t (positive in the direction of s)

Force equilibrium (cont’d) From force equilibrium considerations in z-direction: Force equilibrium in s-direction gives

Stress Strain Relationships Direct stress: z and s  strains z and s Shear stress:   strains  (= zs = zs) Express strains in terms of displacements of a point on the c/s wall vt and vn: tangential and normal displacements in xy plane  Not used (1/r: curvature of wall in x-y plane)

Stress Strain Relationships To obtain the shear strain, consider the element below: Shear strain:

Origin O of axes chosen arbitrarily, and axes undergo disp. u, v,  Center of Twist For the point N Origin O of axes chosen arbitrarily, and axes undergo disp. u, v,  Equivalent to pure rotation about some pt. R (center of twist [for loading such as pure torsion])

Center of Twist (cont’d) Equivalent to pure rotation about some pt. R (center of twist [for loading such as pure torsion]) But

Comparing Coefficients with: Center of twist cont… Also from Comparing Coefficients with: Position of Center of Twist

Shear of Open Section Beams The open section beam supports shear loads Sx and Sy such that there is no twisting of the c/s (i.e. no torsion loads) For this, shear loads must pass through a point in the c/s called the SHEAR CENTER Not necessarily on a c/s member Use the equilibrium eqn. And obtaining z from basic bending theory (no hoop stresses, s = 0)

Shear of Open Section Beams cont… From:

Shear of Open Section Beams cont… Integrating with respect to s starting from an origin at an open edge (q = 0 at s = 0) gives: For a c/s having an axis of symmetry, Ixy = 0. Then eq. for qs simplifies to:

Shear sample problem y Determine the shear flow distribution in the thin-walled z-section shown due to shear load Sy applied through its shear center (no torsion). Where is the shear center? And the centroid? 4 3 h x t 1 h/2 2 Shear Flow Distribution (Sx = 0):

Shear sample problem Show this: EXAM TYPE PROBLEM Bottom Flange: 1-2, y = -h/2, x =-h/2 + S1 0 ≤ S1 ≤ h/2

Shear sample problem t y x h h/2 1 2 4 3

Shear sample problem In web 2-3: y =-h/2 + S2 x = 0 for 0 ≤ S2 ≤ h Shear Flow S2 = 0 Symmetric distribution about Cx with max value at S2 = h/2 (y = 0) and positive shear flow along the web

Shear sample problem In web 3-4: y =h/2 x = S3 for 0 ≤ S3 ≤ h Shear Flow Distribution in z-section

Calculation of Shear Center If a shear load passes through the shear center, it will produce NO TWIST  M = 0 If c/s has an axis of symmetry, the shear center lies on this axis For cruciform or angle sections, the shear center is located at the intersection of the sides Sample Problem Calculate the shear center of the thin-walled channel shown here:

Sample problem shear center The shear center (point S) lies on the horizontal (Cx) axis at some distance ξs from the web. If a shear load Sy passes through the shear center it will produce no twist. Let’s look at the shear flow distribution due to Sy: Since Ixy = 0 and Sx = 0 Further: Then:

Sample problem shear center Along the bottom flange 1-2, y = -h/2 At point 2: S1 = b

Sample problem shear center Along the web 2-3 , y = -h/2 + S2 At point 3: S1 = h

Sample problem shear center Along the top flange 3 - 4 , y = h/2 At point 4: S3 = b Good Check!

Sample problem shear center Sy Shear Flow Distribution due to Sy ξs S The moments due to this shear flow distribution should be equal to zero about the shear center Solve for ξs to find the shear center location: