Shear Centre (2) ---THIN-WALLED OPEN CROSS SECTIONS (Ref: Gere Gere & Timoshenko pp316-328)
1. Introduction In the last lecture we looked at the shear centre of a singly symmetric beam. We found that the resultant shear force in the cross section acts through a point, S, which generally does not coincide with the centroid of the cross section. The shear centre always lies on an axis of symmetry. In this lecture we investigate the position of the shear centre of a channel cross section. To do this we must first derive an extension to the shear formula, τ=QS z/(Ib) for thin walled sections.
a b
s
z
d
c
m
y x
x
dx
n
P Fig.1 Thin-walled open sections
2. Shear stress in thin walled sections Many steel and aluminium structural forms (cold formed sections, channels, zeds, sigma etc.) can be classed as thin walled open sections. The wall thickness is small compared to the other dimensions of the beam. It must be also be open (i.e. not closed like a hollow box or cylinder), see Fig.1. The beam element shown in Figure 2 is loaded with a shear force, P, which acts through the shear centre, S , parallel to the y axis. The y and z axes shown are principal axes and the x axis is along the centroid of the beam. The beam has a uniform cross section in the x direction. The force P induces a bending moment about the z axis and the normal stress is given by Eq(1): σ x
=−
M z y
(1)
I z
s
S P
a
F1
t C
z
b dx
d
c
F2
y
Fig.2 Shear in thin walled beam (element)
Taking the element abcd , dx wide we can examine the equilibrium. The difference between the force, F1 and F2 due to direct stress variations (as a consequence of d /dx) must be balanced by the shear force along the beam M /dx) at the face cd . (2) τ tdx = F − F 1
2
where t is is the thickness at distance s from the free edge. The The force, F1, is integral of d irect stress over the area of the face, ad . s s M z1 (3) F 1 = σ x dA = − ydA I z
∫
∫
0
0
where Mz1is the bending moment at face ad . Similarly the force F2 is: s
F 2 =
∫
x dA
σ
=−
0
M z 2 I z
s
∫ ydA
(4)
0
The shear stress is then got from Eq(2) as τ
=
M z 2 − M z1 1 dx
s
I t ∫ z
0
ydA =
QS z
(5)
I z t
The last part of Eq (5) (5) is true if the section has a constant thickness. If the thickness thickness varies with s then the integral version given in Eq (5) would have to be used. This formula is very similar to the original shear formula and it is valid for any shape of thin walled open cross section. The shear flow can be calculated from QS z (6) q = τ t = I z
3. Shear distribution in an I section Figure 3 shows the pattern of shear flow and the stress distributions for an I beam. The shear stress patterns are derived in G&T p321. This derivation is not part of the syllabus of Structural/Stress Analysis 2. The shear in the web varies from τ2 to τmax in a parabola, as we previously showed. We can also demonstrate that the shear stress distribution in the flanges is triangular.
τ1 τ2 S
C
z
τmax
τ2
e Q
τ1 Fig.3 Shear in I beams
y
Fig.4 Shear centre of Channel
4. Shear centre in a channel section The channel section in Figure 4 is bent about the z axis. Examining the shear in the flange, Sz is a function of s. At s from the free end the first moment is h S z ( s ) = st (7) 2 It is simply the area of t hat part of the flange times the distance to the N-A. From Eq (7) we can see that Sz varies linearly with s . When s equals b we are at the end of the flange. The maximum flange shear acts here and is given by S z ( s ) =
h
2
bt
(8)
The stress varies linearly with s from 0 at the free end to a maximum τ1 at the connected end of the flange. This gives the triangle shape shown in figure 4. The peak flange shear stress is: τ 1
=
bhQ
(9)
2 I z If the channel is of constant thickness, then τ1 will also be the shear at the top of the web (i.e. τ2=τ1 ). If the web had a different thickness tw , to the flange, tf then τ2 =τ1 (tf/tw) The maximum shear in the web is at the neutral axis and can be shown to be
⎛ h ⎞ Qh = ⎜b + ⎟ ⎝ 4 ⎠ 2 I z
τ
max
(10)
The horizontal shear force in either flange can be obtained by multiplying the average flange shear by the area of the flange:
⎛ bτ ⎞ Qhb 2 t Q1 = ⎜⎜ 1 ⎟⎟t = 4 I z ⎝ 2 ⎠ The web shear is made up of a rectangle, τ2h and a parabola of area 2 3
(
τ max
− τ 2 ) h
(11)
(12)
Thus the vertical force in the web is Q2 = ht τ 2 +
2 3
(
τ
max
− τ 2 ) ht =
ht
3
(τ 2
Q bth 2 th 3 ( ) + 2τ max ) = + 2 12 I z
(13)
But the second moment of inertia, Iz, is made up of the web part th3/12 and two flanges, bh2/4, i.e. I z =
bth
2
2
th
+
3
(14)
12
Thus we find that the shear force in the web equals the applied shear Q2=Q which could be deduced from the vertical equilibrium of the cross section. If the applied shear force, Q, is positioned at the shear centre then the sum of moment are also equal to zero. Hence taking moment about point S: Q1 h − Q2 e = 0
(15)
Thus, the distance, e, from the shear centre to the centroid of the web is Q1h/Q2 which gives e=
2
2
b h t
4 I z
=
3b 2 h + 6b
(16)
when Eq(14) is used for Iz. Eq(16) is valid for channels of constant thickness. If the web has thickness, tw, and the flange has a different thickness, tf , G&T p325 gives the formula for e as e=
3b 2 t f
(17)
h t w + 6b t f
Note that in the above derivation the distance b is from the free end to the centroid of the web and h is the distance from the centroid of one flange to that of the other. Question 1: Calculate the distance , e , from the back of a 203×76×23.8 kg channel section. The flange thickness of this section is 11.2 mm, web thickness 7.2mm. the overall depth is 203.2mm and the overall width of the flange is 76.2mm.
h/2
S
z
C
Question 2: Figure 5 shows a slit rectangular thin walled tube of constant thickness t . Derive the following formula for t he distance e from the centreline of the wall of the tube to S.
e=
h/2
b(2h + 3b)
2(h + 3b)
e
Fig.5 Slit tube
b/2
b/2
y
