Pin Design AISC

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  • 8/18/2019 Pin Design AISC

    1/3

    PIN CONNECTION

    Connecting Plates

    t= 0.63 in thickness of the plate

    a= 1.378 in

    a) Bearing d= 1.26 in Pin diameter

    b= 4 in

    0.7936 35 ksi

    25.0 kip 50 ksi

    b) Yielding

    2.5

    52.8 ip

    Pin

    a) !"ear

    1.25 in2

    #$.% kip 35 ksi

    50 ksi

    &llo'able !trengt" ( 25.0 ip *82.8 ip  +e,-iered !tregt" E++O+  

    pb = in2 !"=

    + n 

    t( !

    #=

    $= in2

    Pn 

    t(

    p=πd2 %4

    p=

    Pn 

    s/ ( !

    "=

    !#=

     Rn=1 .8 F 

     y A

     pb

    Ω=2.00

     Pn= F 

     y A

    g

    Ωt =1.67

     Pn=0 .6 F u A pΩsf =2.00

    d

         a

    b

  • 8/18/2019 Pin Design AISC

    2/3

    PIN CONNECTION

    Connecting Plates

    t= 3.15 in thickness of the plate

    a) Tensile r-pt-re a= 4.375 in increase a &section '5.2(

    d= 5 in Pin diameter

    b= 16.25 in ok)) &section '5.2(

    5.594 in 70 ksi

    85 ksi

    %*$.$ kip

    b) !"ear r-pt-re

    43.313 in2

    0%.5 kip

    c) Bearing

    15.75

    **2.# kip

    d) Yielding

    51.2

    2%5.1 ip

    Pin

    a) !"ear

    70 ksi

    85 ksi

    19.63 in2

    00.% kip

    b) le3-ral 4ielding

    *= 4.79 in

    1458.3 +ip.in

    1218.5 +ip

    $2*.1 ip

    &llo'able !trengt" ( $2*.1 ip ip  +e,-iered !tregt" O 66

    beff =2t,0.63 &in(

    beff 

    = !"=

    !#=

    Pn 

    t(

    sf =2t&a,d%2(

    sf =

    Pn 

    s/ (

    pb

     = in2

    + n 

    t(

    $= in2

    Pn 

    t(

    p=πd2 %4 !

    "=

    !#=

    p=

    Pn 

    s/ (

    -p=

    Pn=

    Pn 

    t(

     Pn=2 tbeff  F uΩ

    t =2.00

     Pn=0 .6 F 

    u A

    sf 

    Ωsf =2.00

     Rn=1 .8 F  y A  pbΩ=2.00

     Pn= F 

     y A

    g

    Ωt =1.67

     Pn=0 .6 F 

    u A

     p

    Ωsf =2.00

     Pn=4  M 

    u/  L

    Ωt =1.67

    d

         a

    b

  • 8/18/2019 Pin Design AISC

    3/3

    PIN CONNECTION

    Connecting Plates

    t= 3.75 in thickness of the plate

    a= 3.5 in

    a) Bearing d= 5 in Pin diameter

    b= 14 in

    18.75 70 ksi

    8.# kip 85 ksi

    b) Yielding

    52.5

    2200.1 ip

    Pin

    a) !"ear

    19.63 in2

    $8. kip 75 ksi

    100 ksi

    &llo'able !trengt" ( $8. ip 01.0 ip  +e,-iered !tregt" O 66

    pb = in2 !"=

    + n 

    t( !

    #=

    $= in2

    Pn 

    t(

    p=π

    d

    2

     %4

    p=

    Pn 

    s/ ( !

    "=

    !#=

     Rn=1 . 8 F 

     y A

     pb

    Ω=2.00

     Pn= F 

     y A

    g

    Ωt =1.67

     Pn=0 .6 F u A pΩsf =2.00

    d

         a

    b