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

Shear of Open and Closed Section BeamsMegson Ch. 17Open Section BeamsConsider only shear loads applied through shear center (no twisting)Torsion loads must be considered separately

AssumptionsAxial constraints are negligible

Shear stresses normal to beam surface are negligibleNear surface shear stress = 0Walls are thin

Direct and shear stresses on planes normal to the beam surface are const through the thickness

Beam is of uniform sectionThickness may vary around c/s but not along the beam

Thin-WalledNeglect higher order terms of t (t2, t3, )

Closed Section BeamsConsider both shear and torsion loading

Force equilibrium: General stress, Strain, and Displacement RelationshipsS the distance measured around the c/s from some convenient origin

z Direct stress (due to bending moments or bending action of shear loads)

Shear stresses due to shear loads or torsion loads (for closed section)

s Hoop stress, usually zero (non-zero due to internal pressure in closed section beams)

zs = sz =

shear flow; shear force per unit lengthq = t (positive in the direction of s)

Force equilibrium (contd)From force equilibrium considerations in z-direction:

Force equilibrium in s-direction gives

Stress Strain RelationshipsDirect stress: z and s strains z and sShear stress: strains (= zs = zs)

Express strains in terms of displacements of a point on the c/s wallvt and vn: tangential and normal displacements in xy plane Not used(1/r: curvature of wall in x-y plane)

Stress Strain RelationshipsTo obtain the shear strain, consider the element below:

Shear strain:

Center of TwistEquivalent to pure rotation about some pt. R (center of twist [for loading such as pure torsion])For the point N

Origin O of axes chosen arbitrarily, and axes undergo disp. u, v,

Center of Twist (contd)

ButEquivalent to pure rotation about some pt. R (center of twist [for loading such as pure torsion])

Center of twist contAlso from

Comparing Coefficients with:

Position of Center of Twist

Shear of Open Section BeamsThe 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 CENTERNot 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 contFrom:

Shear of Open Section Beams contIntegrating 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 problemDetermine 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?tyxhh/21243Shear Flow Distribution (Sx = 0):

Shear sample problemBottom Flange: 1-2, y = -h/2, x =-h/2 + S10 S1 h/2Show this: EXAM TYPE PROBLEM

Shear sample problem

Shear sample problemIn web 2-3:

y =-h/2 + S2x = 0 for 0 S2 hSymmetric distribution about Cx with max value at S2 = h/2 (y = 0) and positive shear flow along the webShear Flow S2 = 0

Shear sample problemIn web 3-4:

y =h/2 x = S3 for 0 S3 hShear 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 sidesSample ProblemCalculate the shear center of the thin-walled channel shown here:

Sample problem shear centerThe shear center (point S) lies on the horizontal (Cx) axisat some distance s from the web. If a shear load Sy passes through the shear center it will produce no twist.

Lets look at the shear flow distribution due to Sy:Since Ixy = 0 and Sx = 0Further:Then:

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

Along the web 2-3 , y = -h/2 + S2Sample problem shear centerAt point 3: S1 = h

Along the top flange 3 - 4 , y = h/2Sample problem shear centerAt point 4: S3 = bGood Check!

Sample problem shear center

Shear Flow Distribution due to SyThe moments due to this shear flow distribution should be equal to zero about the shear centerSsSySolve for s to find the shear center location: