Steel Diaphragms

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Design of steel diaphragms Chris Hendy

25/1/06

Design of steel diaphragms

Typical external forces acting on the diaphragm in a rectangular box are shon in

!igure 1" These une#en forces can be split into a torsional case and symmetric shear

case as shon in !igure 2 and $ respecti#ely" !igures 2 and $ also sho the shear

flos acting on the edges of the diaphragm" These result from the difference in the

shear flos acting in the box alls each side of the diaphragm" The shear flos from

#ertical shear in the box alls in !igure $ can be calculated using the usual

I / yVAq = approach"

The diaphragm can% in principle% be designed from the shear stresses and direct

stresses resulting from the combination of actions applied by the to cases in !igure 2

and $" &n reality% the boundary shear stresses are unli'ely to be distributed according

to this elastic theory of the box cross(section because of the sudden discontinuity

hich occurs at a diaphragm" )*ccurate+ determination of stresses hoe#er cannot

be predicted by simple hand methods such as these% hich for example on,t predict

the high concentration of shear stress that can occur at the bottom of the eb and

diaphragm ith hard bearings" The difficulty of calculating stresses is the reason for

the extensi#e guidance gi#en in -.500 art $ 2000"

Despite the abo#e% the calculation method in -.500 art $ 2000 still uses !igures 2

and $ as a basis" Direct stresses can be calculated by first determining the moment

acting on a #ertical section" Considering such a section at distance x from a eb% but

outside a bearing% the moment from the torsional case in !igure 2 is

x B

T Dqx Dqx xqD M T =×=×+×= 2 13

The moment from the shear case in !igure $ is

Rx D xq D xq x Dq M R ,av R ,av R ,av R =×=×+×= 2 23

The factor of )2+ in e4uation 23 corresponds to ) d 2+ in -.500 art $ clause

7"18"5"2"$% hich accounts for non(uniform boundary stresses hen not assessed

accurately" * factor of 2"0 is conser#ati#e as% in this case% it conser#ati#ely assumes

that the hori9ontal shear stress acting on the diaphragm is e4ual to the a#erage #ertical

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25/1/06

shear stress hen it is actually less3 and that it remains constant at this #alue along

the top flange hen !igure $ shos it reduces3" !or a more normal box% ha#ing dec'

cantile#ers% the shear flo due to #ertical shear is as shon in !igure and the #alue

of )2+ therefore becomes e#en more conser#ati#e because of the re#erse shear flo in

the cantile#er"

The total design moment is then% from 13 and 23

x B

T R M M M

RT

+=+= $3

This moment corresponds to that in the the first part of clause 7"18"6"2"$ and is

applied to an )&+ section comprising the diaphragm plate forming the eb3 and a

idth of attached box top and bottom flange forming the flanges3" The direct

stresses are calculated from this moment and cross(section"

Figure 1 – Rectangular box with uneven reactions

!or #ertical sections inside the bearings% the moment ill contain a component from

the bearing reaction and this needs to be added into e4uation $3" *lso the effect of

any loadings from traffic applied directly to the diaphragm should be considered in

the moment e4uilibrium" These lead to other terms in clause 7"18"6"2"$"

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D

B

R1 R2




25/1/06

.hear stresses beteen a bearing and the ad:acent eb can be calculated from the

shear flo

({ }

d e

av , RT

t D

Dqq +

here e D is the net depth of the diaphragm alloing for

holes and cut outs and d t is the diaphragm thic'ness" ;lsehere% the shear can% in

theory% be calculated from #ertical e4uilibrium using !igures 2 and $" !ormulae in

-.500 art $ clause 7"18"6"2" hoe#er ma'e alloance for une#en seating on the

bearings so the results there cannot be obtained solely from the statics of figures 2 and

$"

&f there are access holes in the diaphragm% as in !igure 1% the 9ones underneath and

abo#e the holes may be need to be designed for secondary bending arising from the

shear passing through the reduced section depth" <uidance on hen to consider this

is gi#en in -.500 art $ clause 7"18"2"= i"e" if the clear distance abo#e or belo the

hole measured to either the flange or a cut out in the diaphragm plate if less3 exceeds

0"8 times the hole idth"

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Figure 2 – Shear flow due to torsional component

Figure 3 – Shear flow due to uniform shear component

of 5

R/2 R/2

D / Rq R ,av 2=

*pproximation to pea' shear flo is

R/2 R/2

x

D

BD / T A / T qT 220 ==

T

x BD / T A / T qT

220

==

D




25/1/06

Figure 4 – Shear flow due to uniform shear component in a box with cantilevers

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R/2 R/2

D / Rq R ,av 2=

ea' shear flo >

R/2 R/2

x

D

Steel Diaphragms

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