9.7Tension Field Beams 317
Fig.9.17
Determination of plastic hinge position.
Then,
Sxccφ= 2 Mpc′φ+σtctwsin^2 θ
c^2 c
2
φ
TheminimumvalueofSxisobtainedbydifferentiatingwithrespecttocc,thatis,
dSx
dcc
=− 2
Mpc′
c^2 c
+σtctw
sin^2 θ
2
= 0
whichgives
c^2 c=
4 Mpc′
σtctwsin^2 θ
(9.56)
Similarly,inthetensionflange,
c^2 t=
4 Mpt′
σtttwsin^2 θ
(9.57)
Clearly,fortheplastichingestooccurwithinaflange,bothccandctmustbelessthanb.Therefore,
fromEq.(9.56),
Mpc′ <
twb^2 sin^2 θ
4
σtc (9.58)
whereσtcisfoundfromEqs.(9.52)and(9.53)atthemidpointofWX.
TheaverageaxialstressinthecompressionflangebetweenWandXisobtainedbyconsideringthe
equilibriumofhalfofthelengthofWX(Fig.9.18).
Then,
Fc=σcfAcf+σtctw
cc
2
sinθcosθ+τmtw
cc
2