1. Unique empaistic design,
insure combined effect.
Unique
empaistic shape and smooth convexity and concave
can prevent bringing air bubble when moulding concrete
to affect hold-wrap effect. |
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| Close arranging A-type knaggy empaistic
on WD-914 structure building-bear board is in the
center of steel-bear board, and reaching the depth
of 4.5 mm, greatly strengthen hold-wrap force between
steel-bear board and concrete; there is long-round
protrudent empaistic at the two sides of wave crest
which more strengthens hold-wrap force. |
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| 2. Both strength-adding ribs + shearing
force nail central welding, insuring scissors-nail
bring force into play. |
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| The two pieces of concave symmetrical
ribs can make scissors-nail in the center of flute
penetrate steel-bear board to weld on the girder. |
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| The calculation formula of anti-shearing
coefficient k of shearing force nail is: |
K＝0.85/nr0.5(wr/ha)[(hs/ha)]-1.0<
=1.0
( from YB9238-92《Structure Design and Construction
Regulations of Steel-Concrete Combined Building》)
<BS5950>.
But said formula
is not precise enough, among which there is much
favorable assumption, even not including the influence
of weld location of shearing force nail on anti-shearing
power of shearing force nail.
|
| According
to the research about anti-shearing power of shearing
force nail in structure building-bearing board from
the two professors Mottram and Johnson of Britain
Warwich University, the formula is put forward as
the result of the test of the two professors: |
K=(0.75r/nr0.5)[hs/(hs+ha)]<
=1.0
Among: nr is the quantity of shearing force nails
of one board-rib of combined girder-section. ( Nr<=3
)
Wris the
average breadth of press typed armor plate-rib.
hsis the height
after shearing force nail is weld.（hs<
=ha+75mm)
ha
is the height of rib of press typed armor plate.
r is the minimum of {Wr/ha;(e/ha)+1;2.0}
e is the distance between
the center of shearing force nail and the wall of
board-rib. |
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| According to the
test research of Mottram and Johnson, the suggesting formula
to put forward is: |
K=(0.75r/nr0.5)[hs/(hs+ha)]<
=1.0
Among, r is the minimum of{Wr/ha;(e/ha)+1;2.0}
Take WD-914 for example: |
| |
a.a. Weld shearing
force nail in the central location ( e=76.25 ),
Wr/ha=152.5/76=2
(e/ha)+1=(76.25/76)+1=2
r=Min{Wr/ha;(e/ha)+1;2.0}=2
K=(0.75*2/10.5)[120/(120+76)]
=1.5*0.612245
=0.918368
b.b. Weld
shearing force nail in the central location excursion
30 mm(e=152.5/2-30=46.25)
Wr/ha=152.5/76=2
(e/ha)+1=(46.25/76)+1=1.61
r=Min{Wr/ha;(e/ha)+1;2.0}=1.61
K=(0.75*1.61/10.5)[120/(120+76)]
=1.2075*0.612245
=0.739286
0.739286/0.918368=0.805 |
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