Portal_frame2_2007-2008

Portal Frame II Side and End grits Both side and end girts are used to mount the side or end cladding, the difference between the two is that the side girts are supported on the main column and the end girts are supported on the end gables column.

Design Procedure: 1- The acting loads: • Dead Load: - Own weight (10 – 20) kg/m. - Weight of the steel sheets wc = (5-8) kg/m2 for single layer. •

Live Load : -

•

For case of maintenance there is a concentrated load of 100 kg

Wind load: - In this case the wind load will be a main load so it will be a case A - Wwind = ( (Ce + Ci ) x K x q ) x a Ce = 0.8 , Ci = 0.3 K= 1.0 for h ≤ 10 m = 1.1 for h ≤ 20 m q = 70 kg/m2 in Cairo

2- The straining actions: • Wx = due to wind load only Wx = Wwind MX =

•

WX × S 2 W ×S , QX = X 8 2

Wy = due to dead loads and live load Wy = wc x a + o.w. Py = 100 kg W × S 2 PY × S W × S PY MY = Y + , QY = Y + 8 4 2 2 1/12-P.F.II 2007-2008

Portal Frame II Where S is: The span of side girts = spacing between main frames columns. The span of end girts = spacing between end gables columns. And a is the spacing between side or end girts. 3- Choice of Section: fb =

Mx My + ≤ Fbcx Sx Sy

As the channels are non-compact or slender sections so Fbcx = 1.4 t/cm2 Assume Sx = 7 Sy for hot rolled C sec. or Sx = 6 Sy for cold formed sec. M x + (6or 7)M y cm3 1.4 And from the tables we choose the appropriate channel.

By solving the first equation we get the Sx required =

4- Checks 1- Bending stress: My M fb = x + ≤ Fbcx = 1.4t / cm 2 Sx Sy 2- Shear stress: Q q x = x ≤ 0.35Fy A web Qy qy = ≤ 0.35Fy A flanges 2- Deflection "due to live load only": δ act =

P ×S 3 span < 48E × IY 300

N.B.: In case of not satisfied we increase the channel section or use tie rod to reduce S in the Y direction, and so we decrease My and the deflection. MY with 1 tie = 1 / 4 MY without tie rod MY with 2 tie = 1 / 9 MY without tie rod For case of using cold formed section we must use tie rod, to safeguard the deflection N.B. We have to check another case of wind, if wind is suction, Ce = 0.5 Wx = Wwind = 0.5 x K x q x a W ×S2 MX = X And Luact = span 8 2/12-P.F.II 2007-2008

Portal Frame II Rafter Splices -

What is the splice? It is a connection joining 2 parts of the rafter together.

-

Why and where we put the splice? The splice is put to divide the rafter in parts so the max part length is less than 12 m to be able to transport it to site. It is preferred to be as near as possible from the point of zero moment.

-

Design straining actions : As the rafter section is subjected to M , N , Q so the splice will be subjected to the same straining actions. • • •

N is small and can be neglected Q is the actual shear force at the position of the splice. Qact = Y -WT . X of splice M is the max moment capacity of the section. Mmax = Sx x Fbcx "at splice section"

Splice

WT

?

x tan? < 12 m

< 12 m

x of splice h

L / 2 > 12 m span = L

X Y

3/12-P.F.II 2007-2008

Portal Frame II 2 head plate splice. " using Pretensioned bolts "

M max

Q act

B

fb

2 tf X1 e p

X

2

f1 h

H I

2

Pretension bolts: “as before page 9 to 16 E.C.2” Arrange bolts As the connection is designed on max moment, so it must be symmetric. By using pretensioned bolts the head plate is fully effective, so Ix = B H3 / 12, Y = H / 2 fb = Mmax Y / Ix = 6 Mmax / B H2 f1 = Mmax X2 / Ix

X1 = 2 + tf + e + P /2 X2 = H / 2 – X 1

Check bolts : " Always N is neglected for bolts as it decreases tension on bolts" Text,b1,M = 1 / 2 ( ( fb + f1 ) /2 x ( B x X1 ) ) 0.8 T Text,b2,M = 1 / 2 ( ( f1 + 0 ) /2 x ( B x X2 ) )

0.8 T

Check shear stress on bolts : Qact / n Ps n is the total number of bolts.

4/12-P.F.II 2007-2008

Portal Frame II Check on weld between head plate and rafter section: Properties of the weld :

b f -2S

Awhz = ( bf – 2S ) S x 2 + 4 x 0.4 x bf x S Awvl = 2 x 0.8 x h x S Awtot = Awhz + Awvl

0.4 b f

Ix = 2 ( S x (0.8h)3 / 12 ) + 2 ( b f – 2S ) S x (( h + S ) / 2)2 + 4 ( 0.4 bf x S ) x ( h / 2 – S / 2 – tf )2

0.8h

Checks: At point 1q1 = 0 f1 = N / Awtot + Mmax x ( h /2 + S ) / Ix

0.72 t/cm2

At point 2q2 = Qact / Awvl f2 = N / Awtot + Mmax x ( 0.8 h / 2 ) / Ix R2 = ( f22 + 3q22 ) 0.72 t/cm2 x 1.1 N.B.: if N is neglected put N=0

5/12-P.F.II 2007-2008

Portal Frame II Example: It is required for the previous example:

a. Design and draw the rafter splice at 4m from column Use M24 grade (10.9) bolts. b. Design Side girts as : 1 – Hot rolled C section 2 – Cold formed C section

Solution: a. Rafter splice: At splice sec:Qact = Y -WT . X of splice = 6 – 0.5 x 4 = 4 ton Mmax = Sx x Fbcx = 1.536 x 904 = 1388 cmt Splice Type I 17.0 X1

1.27 4.0 8.0

2

fb f1

X2

M

36 40

2

H = 36 + 2 + 2 = 40 cm Ix = B H3 / 12 = 17 x 403 / 12 = 90666.6 cm4 Y = H / 2 = 20 cm X1 = 2 + tf + e + P /2 = 2 + 1.27 + 4 + 4 = 11.27 cm X2 = H / 2 – X1 = 20 – 11.27 = 8.73 cm fb = Mmax Y / Ix = 1388 x 20 / 90666.66 = 0.3 t/cm2 f1 = Mmax X2 / Ix = 1388 x 8.73 / 90666.66 = 0.13 t/cm2

Q act

Check bending stress on bolts : Text,b1,M = 1 / 2 ( ( 0.3 + 0.13 ) /2 x ( 17 x 11.27 ) ) = 20.6 > 0.8 T = 17.84

unsafe

Text,b2,M = 1 / 2 ( ( 0.13 + 0 ) /2 x ( 17 x 8.73 ) ) = 4.82 < 0.8 T safe Try to safeguard the first row by putting 4 bolts / row , but B / 2 = 17 / 2 = 8.5 cm < tw / 2 + 2 e + p = 0.4 + 2 x 4 + 8 = 16.4 we cant use 4 bolts / row. Increase H and move second row of bolts up the flange 6/12-P.F.II 2007-2008

Portal Frame II fb

4.0

H = 36 + 8 + 8 = 52 cm

8.0 4.0 1.2 7 4.0

Ix = B H3 / 12 = 17 x 523 / 12 = 199194.66 cm4 Y = H / 2 = 26 cm X1 = tf / 2 + 2e = 1.27 / 2 + 2 x 4 = 8.63 cm X2 = H / 2 – X1 = 26 – 8.63 = 17.37 cm fb = Mmax Y / Ix = 1388 x 26 / 199194.66 = 0.181 t/cm2 f1 = Mmax X2 / Ix = 1388 x 17.37 / 199194.66= 0.12 t/cm2

52.0

f1

36.0

8.0

S T . P L 1 0m m

Check bolts : Text,b1,M = 1 / 2 ( ( 0.181 + 0.12 ) /2 x ( 17 x 8.63 ) ) = 11 < 0.8 T = 17.84

safe

Text,b2,M = 1 / 2 ( ( 0.12 + 0 ) /2 x ( 17 x 17.37 ) ) = 8.8 < 0.8 T safe Check shear on bolts : Qact / n = 4 / 8 = 0.5 ton < Ps = 7.11 ton

safe

Thickness of head plate: We take a strip of width b/2 = 17/2 = 8.5cm, we calculate the moment from this strip Mp = force in bolt x e = 11 x 4 = 44cmt

44 MP t t ×( P ) = × ( P ) = Fb = 0.72Fy t/cm2 3 3 h ×t 8 ×t P 2 2 ( 1 P) ( ) 12 12 so tp = 4.4 cm very big fb =

Check on weld between head plate and rafter section: Properties of the weld: Assume size of weld S = 8 mm for web and S flange and stiff = 12 mm Awhz = 8 x 0.4 x 17 x 1.2 = 65.28 cm2 Awvl = 2 x 0.8 x 36 x 0.8 + 0.8 x 8 x 4 x 1.2 = 76.8 cm2 Awtot = Awhz + Awvl = 142.08 cm2

0.8x8

0.8x36

Ix = 2 (0.8 x (0.8 x 36)3 / 12) + 4 (0.4 x 17 x 1.2) x ((36 + 1.2) / 2)2 + 4 (0.4 x 17 x 1.2) x ( 36 / 2 – 1.2 / 2 – 1.27)2 + 4 (1.2 x (0.8 x 8)3 / 12) + 4 (0.8 x 8 x 1.2) x (36 / 2 + 0.5 x 8)2 = 37942.7 cm4 Checks: At point 1-

0.4x17

q1 = 4 / 76.8 = 0.052 t/cm2 f1 = 1388 x (18 + 8) / 37942.7 = 0.95 t/cm2 R1 = (0.952 + 3x0.0522) = 0.96 t/cm2 > 0.72 t/cm2 x 1.1 try to increase stiff length and recheck 7/12-P.F.II 2007-2008

Portal Frame II b. Design of Side girts 1- As hot rolled C section 1- The acting loads: • Dead Load: - Own weight 20 kg/m. - Weight of the steel sheets wc = 6 kg/m2 . •

Live Load : -

•

Py = 100 kg

Wind load: - Wwind = ( (Ce + Ci) x K x q ) x a Ce = 0.8, Ci = 0.3 K= 1.0 for h ≤ 10 m q = 70 kg/m2 in Cairo

2- The straining actions: •

Wx = due to wind load only Wx = Wwind = 1.1 x 1 x 70 x 1.5 = 115.5 kg /m = 0.115 t/m'

WX × S 2 0.115 × 62 = 0.517 mt = 8 8 W ×S 0.115 × 6 QX = X = = 0.345 t 2 2 MX =

•

Wy = due to dead loads and live load Wy = wc x a + o.w. = 6 x 1.5 + 20 = 29 kg/m = 0.029 t/m' Py = 100 kg = 0.1 t W × S 2 PY × S 0.029 × 62 0.1× 6 = = 0.28 mt MY = Y + + 8 4 8 4 W × S Py 0.029 × 6 0.1 = = 0.137 t Qy = y + + 2 2 2 2

3- Choice of Section:

fb =

Mx My + ≤ Fbcx Sx Sy

As the channels are non compact sections so Fbcx = 1.4 t/cm2 Assume Sx = 7 Sy for hot rolled sec.

8/12-P.F.II 2007-2008

Portal Frame II Sx required =

M x + 7M y 1.4

= (51.7 + 7 x 28) / 1.4 = 176.9 cm3

Choose C 200 4- Check 1- Bending stress: Luact = zero as the compression flange is fully laterally supported by the corrugated sheets. Section is non-compact as we are using channels Fbcx = 1.4 t/cm2 M M 51.7 28 fb = x + y = + = 1.3 < Fbcx = 1.4t / cm 2 Sx Sy 191 27 2- Shear stress: Q 0.345 qx = x = = 0.02 ≤ 0.35Fy A web 20 × 0.85 Qy 0.137 = = 0.008 ≤ 0.35Fy qy = A flanges 2 × 7.5 × 1.15

3- Deflection "due to live load only": δ act =

0.1× 6003 P ×S 3 span 600 = = 1.44cm < = = 2cm 48E × IY 48 × 2100 ×148 300 300

9/12-P.F.II 2007-2008

Portal Frame II 2- As cold formed C section 1- The acting loads: • Dead Load: - Own weight 10 kg/m. - Weight of the steel sheets wc = 6 kg/m2. •

Live Load : -

•

Py = 100 kg

Wind load: - Wwind = (( (Ce + Ci) x K x q ) x a Ce = 0.8, Ci = 0.3 K= 1.0 for h ≤ 10 m q = 70 kg/m2 in Cairo

2- The straining actions: We will use 2 tie rods • Wx = due to wind load only Wx = Wwind = 1.1 x 1 x 70 x 1.5 = 115.5 kg /m = 0.115 t/m'

WX × S 2 0.115 × 62 = 0.517 mt = 8 8 W ×S 0.115 × 6 QX = X = = 0.345 t 2 2 MX =

•

Wy = due to dead loads and live load Wy = wc x a + o.w. = 6 x 1.5 + 10 = 19 kg/m = 0.019 t/m' Py = 100 kg = 0.1 t As there is 2 tie rods S = 6/3 = 2.0 m W × S 2 PY × S 0.019 × 22 0.1× 2 MY = Y + = + = 0.0595 mt 8 4 8 4 W × S Py 0.019 × 2 0.1 Qy = y + = + = 0.07 t 2 2 2 2 3- Choice of Section:

fb =

Mx My + ≤ Fbcx Sx Sy

As the channels are non compact sections so Fbcx = 1.4 t/cm2 Assume Sx = 6 Sy for cold formed sec. M x + 6 M y 51.7 + 6 × 5.95 Sx required = = = 62.5 cm3 1.4 1.4 Choose C180x75x4 10/12-P.F.II 2007-2008

Portal Frame II 4- Checks:a- Code limits for slender sections. For web subjected to moment h = H – 2r – 2 t = 180 – 2(6)-2(4) = 160 mm h 160 = = 40 < 200 Ok t 4 For Unstiff. flange b = B – r – t = 75 – 6 – 4 = 65 mm b 65 = = 16.25 < 40 Ok t 4 Section satisfies code limits. b- Determine the effective parts of the section. 1- Flange Unstiffened flange subjected to compression

= 1 , K = 0.43

b = 65 mm

b t Fy 65 4 2.4 = 0.87 = 44 K σ 44 0.43 λ − 0.15 − 0.05ψ 0.87 − 0.15 − 0.05 ×1 ρ= P = = 0.885 < 1 λP2 0.87 2

λP =

be =

b = 0.885 x 65 = 57.53 mm

2- Web Stiff. Web subjected to moment

= -1 , K = 23.9

h = 160 mm b t Fy 160 4 2.4 λP = = = 0.29 44 K σ 44 23.9

ρ=

λP − 0.15 − 0.05ψ 0.29 − 0.15 − 0.05 × −1 = 2.25 > 1 = λP2 0.292

The web is fully effective. As the flange is not fully effective so we must calculate the new properties of the section. IX eff = IX table – IX of the reduced part = 606.25 – 0.747x0.4 (9-0.2)2 = 583.11 cm4 IY eff = IY table – IY of the reduced part = 67.2 – (

0.4×0.7473 0.747 2 ) ) = 59cm4 + 0.4×0.747×(7.5 −1.9 − 12 2

11/12-P.F.II 2007-2008

Portal Frame II

a- Check bending stress. Luact = zero as the compression flange is fully laterally supported by the corrugated sheets. Fbcx = 1.4 t/cm2

51.7 5.95 MX MY y + x = 9+ (7.5 − 1.9) =1.36 < 1.4 t/cm2 IX IY 583.11 59 b- Check shear stress. fbc =

Very small shear force, shear stress will be safe Q 0.345 qx = x = = 0.05 ≤ 0.35Fy A web 18 × 0.4 Qy 0.07 qy = = = 0.012 ≤ 0.35Fy A flanges 2 × 7.5 × 0.4 − 0.7 × 0.4 c- Check deflection. "due to live load only" δ act =

0.1× 2003 P ×S 3 span 200 = = 0.13cm < = = 0.66cm 48E × IY 48 × 2100 × 59 300 300

12/12-P.F.II 2007-2008