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The Design of Modern Steel Bridges - TEDI

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146 <strong>The</strong> <strong>Design</strong> <strong>of</strong> <strong>Modern</strong> <strong>Steel</strong> <strong>Bridges</strong><br />

ðstb cos yttwÞsin yt<br />

<strong>The</strong> shear force resisted by tension field action is<br />

ðt t1Þbtw<br />

Equating<br />

st ¼<br />

ðt t1Þ<br />

sin yt cos yt<br />

Due to st, the pull on the flanges per unit length is<br />

and its vertical component is<br />

sttw sin yt<br />

sttw sin 2 yt<br />

ð5:43Þ<br />

<strong>The</strong> compressive force P tf on a vertical stiffener is equal to the vertical component<br />

<strong>of</strong> the total pull over length a on the flanges. Thus<br />

Ptf ¼ statw sin 2 yt<br />

t t1<br />

¼<br />

sin yt cos yt<br />

¼ðt t1Þatw tan yt<br />

atw sin 2 yt, using equation (5:43)<br />

ð5:44Þ<br />

<strong>The</strong> inclination <strong>of</strong> the membrane forces y t for the maximum shear resistance<br />

due to tension-field action has been found from parametric studies never to<br />

exceed p/4, nor does it exceed the angle <strong>of</strong> the diagonal <strong>of</strong> the web panel with<br />

the horizontal for aspect ratio a/b <strong>of</strong> the panel up to 3. Hence P tf can be taken as<br />

Ptf ¼ðt t1Þatw, or ðt t1Þbtw<br />

whichever is smaller.<br />

Due to st, the pull on an end vertical stiffener per unit height is<br />

and its horizontal component is<br />

sttw cos yt<br />

sttw cos 2 yt<br />

Assuming some end fixity, the bending moment on an end post is<br />

sttw cos 2 b<br />

yt<br />

2<br />

10<br />

¼ 1<br />

10 ðt t1Þtwb 2 cot yt<br />

from equation (5.43).<br />

From parametric studies for the maximum shear resistance due to tensionfield<br />

action, an upper band for cot y t was found to be 80/y d, where y d is the<br />

inclination <strong>of</strong> the diagonal <strong>of</strong> the web panel with the horizontal in degrees. <strong>The</strong><br />

design bending moment M y on an end stiffener can thus be expressed as<br />

My ¼ 8ðt t1Þtwb 2 =yd

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