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

2. 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

3. 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)

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

5. 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)

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

7. 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])

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

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

10. 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)

11. Shear of Open Section Beams cont…
From:

12. 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:

13. 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):

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

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

16. 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

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

18. 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:

19. 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:

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

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

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

23. Sample problem shear centerSy
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: