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8/12/2019 14 Pl Conectora en Arrisotres Excentricos
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PROYECTO
N
REV. B. MEMORIA DE CALCULO DE ESTRUCTURAS
HSS Chevron Brace Connection-1
Material Properties:
Es= 29000.0ksi
Fy Fu
Beam: W14X26 36.00ksi 58.00ksi
Brace: HSS4X4X3/16 36.00ksi 58.00ksi
Gusset Plate: 36.00ksi 58.00ksi
Geometric Properties
Beam: d: 13.70in tw= 0.23in kdes:
Brace: H: 4.00in B: 4.00in Ag:
X1: 1 Y1: 1 55/84
FEXX= 70.00ksi
Pu= 56.4kips compression
Tu= 50.3kips tension
Ww= 3/8 , in gusset plate thickness
Determine required brace-to-gusset weld size
Since the brace loads are axial, the angle between the longitudinal brace axis and line of force
is: w: 0.00Fw=0.6FEXX(1+0.5SIN1.5w)= 42.0 ksim3 4.72in
N= 4.00und
Wreq'd=Pu/(NFw(0.707)m3)+1/16 1/16in. added to the weld size is to account for the sl
Wreq'd= 0.20inThe minimum weld size for this connection is:
Wreq'd= 1/4 in 0.250 in
Determine required gusset plate thickness:
We=Ww-Wreq'd-1/16= 1/16 in
t1 req'd=(0.6FEXXWe)(0.707)x(2)/(0.6Fy)t1 req'd= 0.13in
t1 req'd= 3/8 in 0.375 in
Check gusset plate buckling (compression brace)
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r=t1/12= 0.11in
From the figure, the distance:
m1= 9.00in
Limiting slenderness ratio:
4.71E/Fy= 133.7
Since the gusset is attached by one edge only, the buckling mode could be a sidesway type as
shown in Commentary Table C-C2.2. In this case use:
K= 1.2
K m1/r= 99.77
4.71(E/Fy)= 133.68Fe=2E/(K m1/r)2= 28.76 ksi
Fcr=(0.658 Fy/Fe)Fy 21.32 ksi
Fcr=0.877Fe= 25.22 ksi
Fcr= 21.32 ksi
m2= 4.00in
lw=m2+2xm3xtan30= 9.45in
Aw=lw t1 reqd= 3.54in
Pn=FcrxAw= 75.55 kips
FPn=FFcrA= 68.00 kips > Pu , Ok
Check tension yielding of gusset plate (tension brace)
FRn=FFyAw= 114.83 kips > Tu, Ok
Check shear strength at brace-to-gusset welds
Ae=N m3 t= 3.29in
Nominal Shear Strength
FRn=F0.6FyAe= 70.97 kips > Pu , Ok
Check shear lag fracture in HSS brace
/x=(B2+2BH)/(4(B+H))= 1.500inU=1-(/x/m3)= 0.682
An=Ag-2txt1= 2.42in2
Ae=UAn= 1.65in2
FRn=FFuAe= 71.92 kips > Pu , Ok
Calculate interface forces
Design the gusset-to-beam connection as if each brace were the only brace and locate each
braces connection centroid at the ideal centroid locations to avoid inducing a moment on thegusset-beam interface, similarly to uniform force method special case 3.
eb=d/2= 6.85in
=atan(Y1/X1)= 31.14Let /==eb tan= 4.14in
= 4.20in=ec 0.0r = (( + ec)2 + ( + eb)2) = 8.04inHub=Pu/r 29.46 kipsVub=ebPu/r 48.05 kips
Determine required gusset-to-beam weld size
The weld length is twice the horizontal distance from the work point to the centroid of the
gusset-to-beam connection, , for each brace. Therefore,l=2= 8.40in
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Since the gusset straddles the work line of each brace, the weld is uniformly loaded. Therefore,
the available strength is the average required strength and the fillet weld should be designed
for 1.25 times the average strength.
u= 2 number of gusset
D req'd=1.25Pu/(0.75x0.6xFEXXx0.707x1/16xlxu)
D req'd= 3.01 /16
D req'd= 1/4 in
Check gusset thickness (against weld size required for strength)
Weld fracture plate fracture
0.75x0.6xFEXXDx2/(16x21/2) 0.75x0.6xFuxWw
Ww0.0884xFEXXD/Fu= 0.321in t1 req'd= 0.375in
Check local web yielding of the beam
N= 14.67in supporting length
FRn=F(N+5kdes)Fytw= 151.86 kips cos PuVertical reaction:
cos Pu= 48.23
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FECHA:
4/1/2014
0.735in
2.58in2 t: 0.174in
t in HSS
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Ok
Ok
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