Chapter 26 Closed section beams 679 Chapter 27 Open section beams 718
26.2 Shear stress distribution at a built-in end of a closed section beam
This special case of structural constraint is of interest due to the fact that the shear stress distribution at the built-in end of a closed section beam is statically determinate.
Figure 26.1 represents the cross-section of a thin-walled closed section beam at its built-in end. It is immaterial for this analysis whether or not the section is idealized since the expression for shear flow in Eq. (17.19), on which the solution is based, is applicable to either case. The beam supports shear loads Sx and Sy which generally will produce torsion in addition to shear. We again assume that the cross-section of the beam remains undistorted by the applied loads so that the displacement of the beam cross-section is completely defined by the displacements u, v, w and the rotation θ referred to an arbitrary system of axes Oxy. The shear flow q at any section of the beam is then given by Eq. (17.20), that is
q=Gt
pdθ dz +du
dz cosψ+ dv
dz sinψ+ ∂w
∂s
At the built-in end,∂w/∂s is zero and hence q=Gt
pdθ
dz + du
dz cosψ+dv dzsinψ
(26.1)
Fig. 26.1Cross-section of a thin-walled beam at the built-in end.
in which dθ/dz, du/dz and dv/dz are the unknowns, the remaining terms being functions of the section geometry.
The resultants of the internal shear flows q must be statically equivalent to the applied loading, so that
q cosψds=Sx
q sinψds=Sy
qp ds=Syξ0−Sxη0
⎫⎪
⎪⎪
⎪⎪
⎪⎬
⎪⎪
⎪⎪
⎪⎪
⎭
(26.2)
Substitution for q from Eq. (26.1) in Eqs (26.2) yields dθ
dz
tp cosψds+ du dz
t cos2ψds+dv dz
t cosψsinψds= Sx G dθ
dz
tp sinψds+ du dz
t sinψcosψds+dv dz
t sin2ψds= Sy G dθ
dz
tp2ds+du dz
tp cosψds+ dv dz
tp sinψds= (Syξ0−Sxη0) G
⎫⎪
⎪⎪
⎪⎪
⎪⎬
⎪⎪
⎪⎪
⎪⎪
⎭
(26.3)
Equations (26.3) are solved simultaneously for dθ/dz, du/dz and dv/dz. These values are then substituted in Eq. (26.1) to obtain the shear flow, and hence the shear stress distribution.
Attention must be paid to the signs ofψ, p and q in Eqs (26.3). Positive directions for each parameter are suggested in Fig. 26.1 although alternative conventions may be adopted. In general, however, there are rules which must be obeyed, these having special importance in the solution of multicell beams. Briefly, these are as follows. The positive directions of q and s are the same but may be assigned arbitrarily in each wall. Then p is positive if movement of the foot of the perpendicular along the positive direction of the tangent leads to an anticlockwise rotation of p about O.ψis the clockwise rotation of the tangent vector necessary to bring it into coincidence with the positive direction of the x axis.
Example 26.1
Calculate the shear stress distribution at the built-in end of the beam shown in Fig.
26.2(a) when, at this section, it carries a shear load of 22 000 N acting at a distance of 100 mm from and parallel to side 12. The modulus of rigidity G is constant throughout the section:
Wall 12 34 23
Length (mm) 375 125 500
It is helpful at the start of the problem to sketch the notation and sign convention as shown in Fig. 26.2(b). The walls of the beam are flat and therefore p andψare constant along each wall. Also the thickness of each wall is constant so that the shear flow q is independent of s in each wall. Let point 1 be the origin of the axes, then, writing
26.2 Shear stress distribution 683
Fig. 26.2(a) Beam cross-section at built-in end; (b) notation and sign convention.
θ=dθ/dz, u=du/dz andv=dv/dz, we obtain from Eq. (26.1)
q12=1.6Gv (i)
q23=1.0G(375×0.886θ−0.886u−0.5v) (ii)
q34=1.2G(500×0.866θ−v) (iii)
q41=1.0Gu (iv)
For horizontal equilibrium
500×0.886q41−500×0.886q23=0 giving
q41=q23 (v)
For vertical equilibrium
375q12−125q34−250q23=22 000 (vi) For moment equilibrium about point 1
500×375×0.886q23+125×500×0.886q34=22 000×100 or
3q23+q34=40.6 (vii)
Substituting for q12, etc. from Eqs (i), (ii), (iii) and (iv) into Eqs (v), (vi) and (vii), and solving forθ, uandv, givesθ=0.122/G, u =9.71/G,v=42.9/G. The values of θ, u andv are now inserted in Eqs (i), (ii), (iii) and (iv), giving q12=68.5 N/mm, q23=9.8 N/mm, q34=11.9 N/mm, q41=9.8 N/mm from which
τ12=42.8 N/mm2 τ23=τ41=9.8 N/mm2 τ34=9.9 N/mm2
Fig. 26.3Built-in end of a beam section having a curved wall.
We note in Example 26.1 that there is a discontinuity of shear flow at each of the corners of the beam. This implies the existence of axial loads at the corners which would, in practice, be resisted by booms, if stress concentrations are to be avoided. We see also that in a beam having straight walls the shear flows are constant along each wall so that, from Eq. (17.2), the direct stress gradient ∂σz/∂z=0 in the walls at the built-in end although not necessarily in the booms. Finally, the centre of twist of the beam section at the built-in end may be found using Eq. (17.11), i.e.
xR= −v
θ yR= u θ
which, from the results of Example 26.1, give xR= −351.6 mm, yR=79.6 mm. Thus, the centre of twist is 351.6 mm to the left of and 79.6 mm above corner 1 of the section and will not, as we noted in Section 26.1, coincide with the shear centre of the section.
The method of analysis of beam sections having curved walls is similar to that of Example 26.1 except that in the curved walls the shear flow will not be constant since both p andψin Eq. (26.1) will generally vary. Consider the beam section shown in Fig.
26.3 in which the curved wall 23 is semicircular and of radius r. In the wall 23, p=r andψ=180+φ, so that Eq. (26.1) gives
q23=Gt(rθ−ucosφ−vsinφ) The resultants of q23are then
Horizontally : π
0
q23cosφr dφ Vertically :
π
0
q23sinφr dφ Moment (about 0) :
π
0
q23r2dφ
The shear flows in the remaining walls are constant and the solution proceeds as before.
26.2 Shear stress distribution 685
1
2R
2R q41
q34 q12
q23 20 kN
2 R O
φ
3 4
Fig. 26.4Beam section of Example 26.2.
Example 26.2
Determine the shear flow distribution at the built-in end of a beam whose cross-section is shown in Fig. 26.4. All walls have the same thickness t and shear modulus G;
R=200 mm.
In general at a built-in end (see Eq (26.1)) q=Gt
pdθ
dz + du
dz cosψ+dv dzsinψ
Therefore, taking O as the origin and writingθ=dθ/dz, u=du/dz andv=dv/dz
q41=Gt(−2Rθ+v) (i)
q12=Gt(−Rθ+u) (ii)
q34=Gt(−Rθ−u) (iii)
q23=Gt(−Rθ+ucosφ−vsinφ) (iv) From symmetry
q12=q34
i.e.
Gt(−Rθ+u)=Gt(−Rθ−u) Therefore
u=0 Resolving vertically
q412R− π
0
q23sinφR dφ=20×103
i.e.
q41−1 2
π
0
q23sinφdφ= 10 000 R Substituting from Eqs (i) and (iv) gives
−Rθ+1.79v= 10 000
GtR (v)
Now taking moments about O
q412R 2R+q122R R+q342R R+ π
0
q23R2dφ=20 000×2R which gives
2q41+q12+q34+ 1 2
π
0
q23dφ= 20 000 R Substituting from Eqs (i), (ii), (iii) and (iv)
2Gt(−2Rθ+v)−2GtRθ+ Gt 2
π
0
(−Rθ−vsinφ) dφ= 20 000 R from which
Rθ−0.13v= −2641.7
GtR (vi)
Solving Eqs (v) and (vi)
v= 4432.7
GtR , Rθ= −2065.4 GtR Therefore
q41=Gt
2×2065.4
200Gt + 4432.7 200Gt
=42.8 N/mm Similarly
q12=q34=10.3 N/mm Finally
q23=10.3−22.2 sinφN/mm
26.3 Thin-walled rectangular section beam 687