14 Pl Conectora en Arrisotres Excentricos

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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14 Pl Conectora en Arrisotres Excentricos

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