roof truss sample manual calculation.doc

A Howe roof truss design with 29 members using angle bars supporting a load of 10 kips on the upper joints and 5kips on the lower joints. The truss has a span of 72 ft .The angle  bar has 1 hole per leg with bolt diameter of  in. !"sing A#$ steel and the A%&' &pe(ifi(ation).

T*"&& +etails ,embers A-   A'  4 -'  4 -+  , '+  4, '  /4 +  ,/ +  6,   /6 8  / 8  6 H  6H 8H  H 8%  % H%

 Note: For A36 steel  :  #$ u  5

-ar or(es !kips) Tension 'ompression 73.25 52.5 32.5 5 2. 5 5#.0# 90 27.5 90 #1.2 112.5 12.5 112.5 10.$1 120 5

/ength !ft) 12.7# 9 9 9 12.7# 9 9 9 12.7# 9 9 9 12.7# 9 9

TOP CHORD (COMPRESSION)

or A-; T  73.25 < /12.7# Assume kl=r  1#0 < a .3 A  73.25 = .3  .3 Try ∟6 x 6 x ¾

A  .33 r 1.17

kl=r  !12.72>12) = 1.17  1#0.53 ? kl=r  1#0.53 < a  .7$

(by interpolation)

  !.7$)!.33)  7#.9# @ T

 ⁄⁄unsafe!

Try ∟6 x 6 x ⅞

A  9.7# r  1.17

kl=r  !12.72>12) = 1.17  1#0.53 ? kl=r  1#0.53 < a  .7$   !.7$)!9.7#)  5.2#  T

 ⁄⁄safe!

 ⁄⁄Use ∟6 x 6 x ⅞

B&in(e -+; +; H are members of the (hord whi(h run (ontinuousl:; we design the se(tion with the ma>imum for(e. or H; T 112.5 Assume kl=r  30 <  a  19.19 A  112.5 = 19.19  5.$ Try ∟6 x 6 x ¾

A  .33 < r  1.17

kl=r  !9>12) = 1.17  92.#1 ? kl=r  92.#1 < a  1#.9#

(by interpolation)

  !1#.9#)!.33)  117.57  T

 ⁄   ⁄safe!

 ⁄⁄Use ∟6 x 6 x ¾

BOTTOM CHORD (TENSION)

B&in(e A; '; 8; and 8% are members of the (hord whi(h run (ontinuousl:; we design the se(tion with the ma>imum for(e. or 8%; T  120 Ag  T = !0.$:) Ag  T = !0.$>#$) Ag  120 = !0.$>#$)  5.5$ Ae  T = !0.5u )!0.9) Ae  T = !0.5>5)!0.9) Ae  120 = !0.5>5>0.9)  3.$0 Try ∟6 x 6 x ½

A  5.75

T  0.$!#$)!5.75)  123.2  T

 ⁄⁄safe!

Ae  0.9C 5.75D 2!E)!)F  3.$125 T  !3.$125)!0.5)!5)  1##.7$  T

 ⁄⁄safe!

 ⁄⁄Use ∟6 x 6 x ½

WEB (Tension Members BC, DE, FG, HI)

or -'; T  32.5 Ag  T = !0.$:) Ag  !32.5) = !0.$>#$)  1.97 Ae  T = !0.5u )!0.9) Ae  32.5 = !0.5>5)!0.9)  1.$# Try ∟3½ x 3½ x 5⁄16 

A  2.09

T  !0.$)!#$)!2.09)  35.13  T

 ⁄⁄safe!

Ae  0.9C2.09 G 2!E)!5=1$)F  1.5# T  1.5#!0.5)!5)  33.#7  T

 ⁄⁄safe!

 ⁄⁄Use ∟3½ x 3½ x 5⁄16 

or +; T  27.50 Ag  T = !0.$:) Ag  27.5 =!0.$>#$)  1.27 Ae  T = !0.5u )!0.9) Ae  27.5 = !0.5)!5)!0.9)  1.05 Try ∟2½ x 2½ x 5⁄16 

A  1.3$

T  !0.$)!#$)!1.3$)  #1.53  T

 ⁄⁄safe!

Ae  !0.9)C1.3$D2!E)!5=1$)F  1.07 T  1.07!0.5)!5)  #1.0#  T

 ⁄⁄safe!

 ⁄Use ∟2½ x 2½ x 5⁄16   = 

or 8; T  12.5 Ag  T = !0.$:) Ag  12.5= !0.$)!#$)  0.5 Ae  T = !0.5u )!0.9) Ae  12.5 =!0.9)!0.5)!5)  0.3 Try ∟2½ x 2½ x 3⁄16 

A  0.92

T  0.$ !#$)!0.92)  19.7  T

 ⁄⁄safe!

Ae  0.9C0.92D2!E)!#=1$)F  0.$2 T  0.$2!0.5)!5)  17.9  T

 ⁄⁄safe!

 ⁄⁄Use ∟2½ x 2½ x 3⁄16 

or H%; T  5 Ag  T = !0.$:) Ag  5=!0.$>#$)  0.2# Ae  T = !0.5u )!0.9) Ae  5=!0.5)!5)!0.9)  0.19 Try ∟1½ x 1½ x ⅛

A  0.#59

T  0.$!#$)!0.#59)  7.75  T

 ⁄⁄safe!

Ae  0.9C0.#59D2!E)!)F  0.1 T  0.1!0.5)!5)  5.22  T

 ⁄⁄safe!

 ⁄⁄Use∟1½ x 1½ x ⅛

WEB (Compression Members)

or '+; T  5#.0# Assume kl=r  1#0 < a  .3 A  5#.0# = .3  $ Try ∟6 x 6 x 9⁄16 

A  $.3# < r  1.1

kl=r  !12.7#>12) = 1.1  129.3$ ? kl=r  129.3$ < a  .91

(by interpolation)

  !.91)!$.3#)  57.29  T

 ⁄⁄safe!

 ⁄⁄Use ∟6 x 6 x 9⁄16 

or ; T  #1.2

Assume kl=r  1#0 < a  .3 A  #1.2 = .3  #.$0 Try ∟6 x 6 x 5⁄16 

A  #.$5 < r  1.2

kl=r  !12.7#>12) = 1.20  127.# ? kl=r  127.# < a  9.2$

(by interpolation)

  !9.2$)!#.$5)  ##.  T

 ⁄⁄safe!

 ⁄⁄Use ∟6 x 6 x 5⁄16 

or 8H; T  10.$1 Assume kl=r  1#0 < a  .3 A  10.$1 = .3  1.2 Try ∟4 x 4 x ⅜

A  2.$ < r  0.7

kl=r  !12.7#>12) = 0.7  19#.# ? kl=r  19#.# < a  #.9

(by interpolation)

  !#.9)!2.$)  11.#  T

 ⁄⁄safe!

 ⁄⁄Use ∟4 x 4 x ⅜