PROJECT : PAGE : CLIENT : DESIGN BY :

JOB NO. : DATE : REVIEW BY :

INPUT DATA & DESIGN SUMMARYMEMBER SIZE RT 1 3/4 4 1/8 d b A Ix Iy E Wt (lbs/ft)

4 1.75 1.38 2.7 0.7 10100 1.62F tu = 30 ksiF ty = 25 ksiF cy = 25 ksi

AXIAL COMPRESSION FORCE P = 3 kips, ASDSTRONG GEOMETRIC AXIS EFFECTIVE LENGTH kL x = 6.4 ftWEAK GEOMETRIC AXIS EFFECTIVE LENGTH kL y = 3.2 ft

M rx = 1.2 ft-kips, ASDSTRONG GEOMETRIC AXIS BENDING UNBRACED LENGTH L bx = 2 ftSTRONG DIRECTION SHEAR LOAD, ASD V strong = 30 kips

M ry = 0.35 ft-kips, ASDWEAK DIRECTION SHEAR LOAD, ASD V weak = 20 kips

THE DESIGN IS ADEQUATE.

ANALYSISCHECK COMPRESSION STRESS IN AXIAL FORCE

10.13 ksi, (compression in column, AA ADM-IA 3.4.7)

Where r x = 1.410 in, (Page VI-34 to VI-37)

r y = 0.730 in, (Page VI-34 to VI-37)

kL / r = Max( kL x / r x , kL y / r y ) = 54.47E = 10100 ksi, (Table 3.3-1)B c = F cy [1 + (F cy / 2250)

0.5 ] = 27.64 , (Table 3.3-4)

D c = (B c / 10) (B c / E)0.5 = 0.14 , (Table 3.3-4)

C c = 0.41 (B c / D c ) = 78.38 , (Table 3.3-4)

n u = 1.95 , (Table 3.4-1)

n y = 1.65 , (Table 3.4-1)

S 1 = (B c - n u F cy / n y ) / D c = 0.00 , (Eq. 3.4.7-4)

S 2 = C c = 78.38 , (Eq. 3.4.7-5)

11.80 ksi, (compression in web, AA ADM-IA 3.4.9)

Where b / t = 30.000 , (Figure 3.4.9-1)k 1 = 0.35 , (Table 3.3-4)k 2 = 2.27 , (Table 3.3-4)B p = F cy [1 + F cy

(1/3) / 11.4] = 31.41 , (Table 3.3-4)

STRONG GEOMETRIC AXIS BENDING MOMENT

WEAK GEOMETRIC AXIS BENDING MOMENT

Aluminum RT Member Capacity Based on Aluminum Design Manual 2005 (AA ADM-IA)

TENSILE YIELD STRESS (T5, T6, T7, T8, or T9)

COMPRESSIVE YIELD STRESS (T5 to T9)

TENSILE ULTIMATE STRESS (T5 to T9, Tab 3.3-1)

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(cont'd)D p = (B p / 10) (B p / E)

0.5 = 0.18 , (Table 3.3-4)

S 1 = (B p - n u F cy / n y ) / (1.6 D p ) = 6.66 , (Eq. 3.4.9-4)

S 2 = k 1 B p / (1.6 D p ) = 39.22 , (Eq. 3.4.9-5)

f a = P / A = 2.17 ksi < F a = Min (F a 1 , F a 2 ) = 10.13 ksi[Satisfactory]

CHECK TENSION STRESS IN BENDING MOMENTSF = 15.15 ksi, (tension in bending, AA ADM-IA 3.4.2 & 3.4.4)

Where k t = 1.00 , (Table 3.4-2)

f = Max (M rx / S x , M ry / S y ) = 10.51 ksi < F = 15.15 ksi[Satisfactory]

Where S x = 1.37 in, (Page VI-34 to VI-37)

S y = 0.84 in, (Page VI-34 to VI-37)

CHECK COMPRESSION STRESS IN STRONG GEOMETRIC AXIS, x - x, BENDING MOMENT

15.15 ksi, (compression in beam, AA ADM-IA 3.4.14)

Where S 1 = (B c - F cy )2 / (1.6 D c )

2 = 129.82 , (Eq. 3.4.14-4)

S 2 = (C c /1.6)2 = 2399.9 , (Eq. 3.4.14-5)

J = 1.8 , (Page VI-34 to VI-37)S c = S x = 1.37 in, (Page VI-34 to VI-37)

C b = 1.00 , (AA ADM-IA 4.9.4)

L b S c / [C b (I y J)0.5 / 2] = 57.29

15.15 ksi, (compression in flanges, AA ADM-IA 3.4.16)

Where b / t = 12.000 , (Figure 3.4.8-1)k 1 = 0.5 , (Table 3.3-4)k 2 = 2.04 , (Table 3.3-4)

S 1 = (B p - F cy ) / (1.6 D p ) = 22.88 , (Eq. 3.4.16-4)

S 2 = k 1 B p / (1.6 D p ) = 56.04 , (Eq. 3.4.16-5)

19.72 ksi, (compression in web, AA ADM-IA 3.4.18)

Where h / t = 30.000 , (Figure 3.4.18-1)c c = c o = 2 in

Min (F ty /n y , F tu / k t n u , 1.3F ty /n y , 1.42F tu / k t n u ) =

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F hfor StnhmDB ht forF S S

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(cont'd)m = 0.65 , (Table 3.3-4)B br = 1.3 F cy [1 + F cy

(1/3) / 7] = 46.08 , (Table 3.3-4)

D br = (B br / 20) (6 B br / E)(1/3) = 0.69 , (Table 3.3-4)

S 1 = (B br - 1.3 F cy ) / (m D br ) = 0.45 , (Eq. 3.4.18-4)

S 2 = k 1 B br / (m D br ) = 51.05 , (Eq. 3.4.18-5)

f bx = M rx / S x = 10.51 ksi < F bx = Min (F bx 1 , F bx 2 , F bx 3 ) = 15.15 ksi[Satisfactory]

CHECK COMPRESSION STRESS IN WEAK GEOMETRIC AXIS, y - y, BENDING MOMENT

13.94 ksi, (compression in flanges, AA ADM-IA 3.4.16)

Where b / t = 30.000 , (Figure 3.4.8-1)

f by = M ry / S y = 5.02 ksi < F by = 13.94 ksi[Satisfactory]

CHECK COMBINED COMPRESSION AND BENDING (AA ADM-IA 4.1.1)

1.27 < 1.33 , (1.33 if IBC/CBC 1605.3.2 apply)[Satisfactory]

Where C mx = 0.85

C my = 0.85

F ex = 2 E / n u (kL x /r x ) 2 = 17.23 ksiF ey = 2 E / n u (kL y /r y ) 2 = 18.47 ksi

CHECK SHEAR STRESS (AA ADM-IA 3.4.20)

8.75 ksi, (for strong shear)8.75 ksi, (for weak shear)

Where h / t = 30.000 , (for strong shear, Figure 3.4.18-1)h / t = 12.000 , (for weak shear, Figure 3.4.18-1)B s = (F cy / 3

0.5 ) [1 + (F cy / 30.5 ) (1/3) / 9.3] = 18.21 , (Table 3.3-4)

D s = (B s / 10) (B s / E)0.5 = 0.08 , (Table 3.3-4)

C s = 0.41 B s / D s = 96.55 , (Table 3.3-4)

S 1 = (B s - F cy / 30.5 ) / (1.25 D s ) = 39.09 , (Eq. 3.4.20-4)

S 2 = C s / 1.25 = 77.24 , (Eq. 3.4.20-5)

f s = V strong / A w = 3.75 ksi < F s = 8.75 ksi, (for strong shear)f s = V weak / A w = 5.71 ksi < F s = 8.75 ksi, (for weak shear)

[Satisfactory]

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