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Thin Walled Sections

The effect of shear deformation on the results of a structural analysis is usually negligible. Where it is more significant, it will usually suffice to make a simple approximation to the shear deformation area of members with a cross-section such as those shown in Fig 1. The usual approximation is, by analogy with a simple rectangular beam, to take 56\frac{5}{6} or 1112\frac{11}{12} of the total web area (overall depth × web thickness).

For those rare structures where the shear deformation is very important it may be necessary to use a more exact value for the area. This Note gives formulae for FF for the cross-sections of Fig.1 where

As=Fdt6A_{s} = F\frac{dt}{6}


These were derived from the virtual work formula, shear deflection per unit length = q2Gtds\int_{}^{}{\frac{q^{2}}{Gt}ds}. Here qq is the shear flow at any point at the middle of the wall thickness, the shear stress qt\frac{q}{t} is assumed constant across the wall thickness tt, and the integration extends over the whole cross-section.

To see what the formulae mean in practice, they were applied to steel sections taken from the handbook with the results shown in the table below. In a web or flange with varying thickness, tt was assumed constant at its average value. The smaller values of FF in a range correspond to cross-sections with squarer aspect ratios. Samples only of UBs, UCs and channels were taken.

It can be seen that, for sections with top and bottom flanges bending, as nature intended, in their strong direction, the usual approximation is satisfactory (although it should be noted that d is the distance between flange centres, not overall depth). For the more bizarre sections used in bending, values of FF are seen to be lower than expected, especially when they are perversely bent in their weakest direction.

Bending in strong direction
UB5.72 to 5.81
Joist5.17 to 5.78
UBP5.25 to 5.28
UC5.28 to 5.53
RHS5.0 to 5.49 db=[1:2]\quad\frac{d}{b} = \lbrack 1:2\rbrack
Channel5.06 to 5.60
Angle3.68 to 4.62 db=[1:2]\quad\frac{d}{b} = \lbrack 1:2\rbrack
Tee from UB4.78 to 4.97
Tee from UC4.11 to 4.35
Bending in weak direction
Channel2.12 to 3.78
RHS4.02 to 5.0 db=[1:2]\quad\frac{d}{b} = \lbrack 1:2\rbrack)
Angle2.55 to 3.68 (for db\frac{d}{b} = 0.5 to 1)
I’s, T’s, & cruciform5.0

With regard to calculating shear stresses, the exact distribution is not normally required, or even usable, because Codes of Practice base the shear strength on an allowable average shear stress calculated on the total net area DtDt. However, the shear distribution is sometimes required to design welds or concrete stitches, and, since it was found in the process of deriving the formulae for FF, formulae for the stress factors kik_{i} and kmk_{m} are given. Here,

qi=kiVdq_{i} = k_{i}\frac{V}{d}

is the shear flow in the web at the junction with the flange, and

qmi=kmVdq_{mi} = k_{m}\frac{V}{d}

is the maximum shear flow in the web, in which VV is the shear force at the section.

For circular annuli, assuming that the stress is constant across the wall thickness tt, both the deflection and maximum stress can be obtained using a shear area of half the actual area, that is πrt\pi rt where rr is the mean radius.


λdtbTβbtdTφdb\lambda \equiv \frac{dt}{bT} \qquad \beta \equiv \frac{bt}{dT}\qquad \varphi \equiv \frac{d}{b}

Shear deformation area

As=Fdt6A_{s} = F\frac{dt}{6}

Type A1

F=(6+λ)24β+6+2λ+0.2λ2F = \frac{(6 + \lambda)^{2}}{4\beta + 6 + 2\lambda + 0.2\lambda^{2}}

Types A2 & A3

F=(6+λ)2β+6+2λ+0.2λ2F = \frac{(6 + \lambda)^{2}}{\beta + 6 + 2\lambda + 0.2\lambda^{2}}

Special case of A3 with constant wall thickness so that T=t,λ=2φ,β=2φT = t,\lambda = 2\varphi,\beta = \frac{2}{\varphi}

F=10φ(3+φ)25+15φ+10φ2+2φ3F = \frac{10\varphi(3 + \varphi)^{2}}{5 + 15\varphi + 10\varphi^{2} + 2\varphi^{3}}

Type B1

F=(4+λ)22β+3.2+1.4λ+0.2λ2F = \frac{(4 + \lambda)^{2}}{2\beta + 3.2 + 1.4\lambda + 0.2\lambda^{2}}

Special case of B1 with constant T=t,λ=φ,β=1φT = t,\lambda = \varphi,\beta = \frac{1}{\varphi}

F=5φ(4+φ)2(1+φ)(10+6φ+φ2)F = \frac{5\varphi(4 + \varphi)^{2}}{(1 + \varphi)\left( 10 + 6\varphi + \varphi^{2} \right)}

Types B2 & B3

F=(4+λ)20.5β+3.2+1.4λ+0.2λ2F = \frac{(4 + \lambda)^{2}}{0.5\beta + 3.2 + 1.4\lambda + 0.2\lambda^{2}}

Special case of B3 with constant thickness T=t2,λ=2φ,β=2φT = \frac{t}{2},\lambda = 2\varphi,\beta = \frac{2}{\varphi}

F=20φ(2+φ)25+16φ+14φ2+4φ3F = \frac{20\varphi(2 + \varphi)^{2}}{5 + 16\varphi + 14\varphi^{2} + 4\varphi^{3}}

Type C

F=5F = 5

Stress Factors

Type A1, A2 & A3

kj=66+λkm=32[4+λ6+λ]k_{j} = \frac{6}{6 + \lambda}\quad k_{m} = \frac{3}{2}\left\lbrack \frac{4 + \lambda}{6 + \lambda} \right\rbrack

Type B1, B2 & B3

kj=64+λkm=32[(2+λ)2(1+λ)(4+λ)]k_{j} = \frac{6}{4 + \lambda}\quad k_{m} = \frac{3}{2}\left\lbrack \frac{(2 + \lambda)^{2}}{(1 + \lambda)(4 + \lambda)} \right\rbrack