Steel book

Eelo University
Steel structure design
At
Senior Engineering

EELO
UNIVERCITY
Department of Civil engineering
Steel Structure Design
Project Report
2017-2018
GROUP TWO
Group Names ID:No
1. Yasin Said Mohamed 1469
2.Mohamed Ahmed Mohamed 1471
3. Abdirahman Farah Ainab 1467
4.Mohamed Abdi nour Mohamud 488

Table of contacts
1. Roofing system……….………………………………………..1
 Iron sheet
 Purlines
 Truss analysis
2. Slab………………………………………………………………….9
3. Beams…………………………………………………...............19
 Beam under the truss
 Beam under the slab
4. Columns………………………………………………………….37
 Colum A
 Column B
 Column C
5. Connections…………………………………………………….52
6. Footing Design…………………………………………………56
 Footing A
 Footing B
 Footing C
7. Stair Case…………………………………………………………66

Introduction
The aim of this book is to provide students and practicing engineers with a guide of
structural steel design to meet the requirement of BS 5950:Part 1: 2000 Structural
Use of Steelwork in Building. The emphasis has been to illustrate the clauses in the
code rather than to match practical cases exactly. The first part of the book gives
basic design concepts of structural elements comprising beam, column, connection,
roof truss, and plate girder. In the second part, it presents worked examples of design
of structural steel elements which are of commonly used in building frame
structures. The examples have different design problem, which require different
approach of loading analysis and design formula.
Steel structure is a metal structure which is made of structural steel* components connect
with each other to carry loads and provide full rigidity. Because of the high strength grade of
steel, this structure is reliable and requires less raw materials than other types of structure like
concrete structure and timber structure.
In modern construction, steel structure is used for almost every type of structure including heavy
industrial buildings, multi-story buildings, equipment support systems, infrastructure, bridges,
towers, airport terminals
*Structural steel is steel construction material which fabricated with a specific shape and
chemical composition to suit a project’s applicable specifications.
Depending on each project’s applicable specifications, the steel sections might have various
shapes, sizes and gauges made by hot or cold rolling, others are made by welding together flat or
bent plates. Common shapes include the I-beam, HSS, Channels, Angles and Plate.
4 reasons why steel structure is the best choice?
1. Cost savings
Steel structure is the cost leader for most projects in materials and design. It is inexpensive to
manufacture and erection, requires less maintenance than other traditional building methods.
2. Creativity
Steel has a natural beauty that most architects can’t wait to take advantage of. Steel allows for
long column-free spans and you can have a lot of natural light if you want it in any shape of
structure.

3. Control and Management
Steel structures is fabricated at factory and rapidly erected at construction site by skilled
personnel that make safe construction process. Industry surveys consistently demonstrate that
steel structure is the optimal solution in management.
4. Durability
It can withstand extreme forces or harsh weather conditions, such as strong winds, earthquakes,
hurricanes and heavy snow. They are also unreceptive to rust and, unlike wood frames, they are
not affected by termites, bugs, mildew, mold and fungi.
What are pre-engineered buildings?
Pre-engineered buildings are built over three members connected to each other:
 Primary members (columns, rafters, bracing…)
 Secondary members (Z or C purlins, girts and eave struts)
 Roof and wall sheeting connected to each other
 Other building components

Design Of
Roofing System
Designed By: Sheet1/74
Date:
18/1/2018
Abdirahman Farah Ainab
Yasin Said Mohamed
Mohamed Ahmed Mohamed
Mohamed Abdinour Mohamud
Checked By: Eng. Sabaax
Reference Purlines Design Remark
RHS:Table
Given Data
Spacing of truss = 6m
Nod to nod spacing =3m
Roof sheet and purline = 0.3KN/m2
( on slope)
Self-weight of truss = 0.2KN/m2
( on plan)
Imposed load = 1kn/m2
(on plan)
Design of purline
king post = tan16*9 = 2.58m
un factored load on purlin = = 0.312 KN/m2
un factored imposed load = 1 KN/m2
on plan
Total un factored load = (0.31+1)*6m*3m = 23.61 KN/m2
On each Node
Slope of Roof = 16o
< 30o
For Rectangular Hallow Section (RHS)
Zlimit = = = 78.7cm3…………..Elastic
Dlimit = = = 85.7mm
Blimit = = = 40cm3
Try 120*80*8 (Zx=87.5) form code
3188975

Design Of
Roofing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of truss members Remark
Design load = 1.4 (0.312+0.2) +1.6(1)
=0.761+1.6 = 2.31 KN/m2
Total point load on Each node =2.31*6*3 = 41.65 KN
Truss Analysis
Determinate Or In-Determinate
 Determinate trusses
M+R=2J m =21 r =3 j =12
21+3=2*12
24=24
Stability Checking
 Truss internally unstable (M < 2J-3)
 Truss Internally Stable (M 2J -3)
2*12-3 =21
Section truss

Design Of
Roofing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Fy
∑ y
∑ x
∑
∑ x
∑ y
∑ x
∑
Af = = 62.47 KN
Ay -20.82 -41.65 -41.56 20.82 + 62.47 = 0
Ay = 62.47KN
Joint A
∑ y
62.47 -20.82 + Sin160 (FAC) =0
FAC = = 151.1KN (c)
FAB- Cos160 (151.1) = 0
FAB = 145.2KN (C)
Joint B
FBC = 0
-145.1 + FBD = 0
FBD = 145.2KN (T)
Joint C
-41.65 +sin160 (FCE) - Sin160 (FCD) +Sin160 (151.1) = 0
-41.65 + 41.65 + 0.2756 FCD = 0
- 7 FCD………………………………………… ………………Equation
Cos160 (151.1) + Cos160 (FCE) + Cos160 (FCD) = 0
145.2 + 0.961FCE FCD …………………………..Equation (2)

Design Of
Roofing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
∑ y
∑ y
∑ y
Joint C …..cont”
0.2765FCE – 0.2756 FCD = 0
0.961FCE +0.961FCD = - 145.2
FCE = – 75.5KN (C)
FCD = – 75.5KN (C)
Joint D
∑ x
-145.5 + Cos160 (75.5) +FDF = 0
-145.2 + 72.57 + FDF = 0
FDF = 72.63KN (T)
FDE – Sin160 (75.5) = 0
FDE = 20.8KN (T)
Joint E
-20.8 - 41.65 - Sin160 (FEF) + Sin160 (75.5) + Sin160 (FEG) = 0
- 20.8 – 41.65 - 0.256FEF + 20.8 + 0.256FEG = 0
-41.65 – 0.256FEF + 0.256FEG = 0
0.256FEF -0.256FEG = 41.65 ……………………… Equation
75.5 Cos160 + FEG Cos160 + Cos160 (FEF) = 0
0.961FEG + 0.96FEF = - 7 7……………………………Equation )

Design Of
Roofing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
∑ y
∑
Joint E… …Cont
0.256FEF -0.256FEG = 41.65
0.961FEG + 0.96FEF = - 72.57
FEG = 37.8 (T)
FEF = 113.3 (C)
Joint G
-41.65-FGE-sin16FGI-sin16(37.8)=0
-41.65-FGE-0.275FGI-10.41=0
-FGE-0.275FGI= ……………… Equation
-FGE-10.38-52.06 =0
FGE=62.44(c)
Cos16(37.8)+cos16FGI = 0
-36.3+0.96FGI=0
=
FGI=37.76(T)

Design Of
Roofing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
∑ y
∑ x
Joint D
124.94 -62.44- Sin160 (113.3) + Sin160 (FDI) = 0
-31.27 + 0.275FDI = 0
FFI = = - 113.44 (C)
FFH – 72.63 + Cos16(113.3) - Cos16(113.3) = 0
FFH 72.63 (T)
MEMBER FORCE
FAC -151.1KN
FAB -145.2 KN
FBD 145.2 KN
FCE -75.5 KN
FCD -75.5 KN
FDF 72.63 KN
FDE 20.8 KN
FEG 37.8 KN
FEF -113.3 KN
FGE -62.44 KN
FGI 37.76 KN
FFI -113.44 KN
FFH 72.63 KN

Design Of
Roofing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Code page
29
Design of top members (tension members)
Tension Force Fv = 145.2KN
Design strength Py = 275KN/m2
Area of section required = 145.2 x = 5.28cm2
Try square hallow section = 50 x50 x4 SHS ( Ag = 7.19cm2
)
Members are welded at the connection, therefore no deduction for bolt holes
at the cross sectional area. Members are symmetrical section, there is no
reduction in cross section, hence, Ae = Ag
Tension capacity, Pt = Ae*Py = (7.19 x 100 x275/1000)
= 197.7KN
Ft < Pt = 145.2 KN < 197.7 KN……………………… ok
Design of bottom members ( compression member)
For the bottom chord, lateral and vertical restraints are provided at 4.5m
Spacing respectively. From the analysis results, the compression force in
member(JG) is greater than (GE), therefore check the capacity of member JG
Only, which has higher axial load
Design the member using rectangular hallow section in Grade S 275
Compression member, Fc = 151.1KN
Try section = 90 x90 x8( Ag = 25.6 cm2 )
r = 3.32
d 90 -3(8) = 66mm
d/t = 66/8 = 8.25 < 40…… Section Is Not Slender
Cl : 3.4.3
Effective net
area

Design Of
Roofing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Le = 0.85 x 4500 = 3825mm
= = = = 115.2
Pc = 123.5 N/mm2
Pc = Ag x Pc = 2560 x 123.5/1000 = 316.15KN……………..Ok
Try square hallow section = 50 x50 x4 SHS ( Ag = 7.19cm2
)
Table 24(a)

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
R.C.C. Design of Slab

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
R.C.C. Design of Slab Remark
Fcu = 30N/mm2 Designing
Fy = 460N/mm2 a) slab 1
Thickness of slab = 150mm b) slab 2
Screed thickness = 20mm
Concrete cover for slab (c) = 25mm
Live load =5 KN/m2
Concrete density (pc) = 25Kg/m3
Second Floor
2.1) Loading System (S1)
(Two edge continuous slab)
= = 1.4 < 2 Two way slab
Self-weight of slab = c * h=0.15x25 = 3.75KN/m2
Finishing is Assume to = 1KN/m2
Total dead load on the slab (n) =3.75+1 = 4.75 KN/m2
Live load = 5KN/ m2
Ultimate Design load = 1.4(DL) + 1.6(LL) = 1.4(4.75) +
1.6(5)
= 14.65 KN/ m2

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
At the mid span short side
Assume = 12mm
D = h – cover - /2
D = 281– 25 - = 250mm
Msx1 = SX * nLx2 = 0.0522* 14.65* 6.362 = 30.16KN.m/m
K = = = 0.016 < 0.156
Z = d(0.5+ √ = d(0.5+ √ = 0.97d > 0.95d
Asmin = = = 325mm2
Area of steel req = = = 290 mm2 > Asmin
Therefore provide T10 – 200 mm c/c (Apr = 395mm2/m)
Single
reinforcement
use 0.95d

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Mid span of longer side
D = 250 – 25 -12 -12/2 = 207mm
Msy1 = SX * nLx2 =0.034* 14.475*6.362 = 19.9 KN.m/m
K = = = 0.015 < 0.156
Z = d(0.5+ √ = d(0.5+ √ = 0.97 > 0.95d
Area of steel = = = 226.8 mm2
Provide T 10 – 275c/c (As = 287mm2)
Continuous edge short side
d = 207 – 30 – 12/2 = 171mm
Msy1 = SX * nLx2 = 0.0752 * 14.475* 6.362 = 38.6KN.m/m
K = = = 0.044 < 0.156
Z = d(0.5+ √ = d(0.5+ √
= 0.94d > 0.95d (use 0.95d)
As = = = 550. mm2 > As min
ProvideT12 – 125c/c (As pro = 632 mm2
)
OK
Use 0.95d

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Continuous edge of longer side
d = 281 – 30 - = 250mm
Msy1 = SX * nLx2
=0.045* 14.475*6.362
= 40.45KN.m/m
K = = = 0.02 < 0.156 (Satisfactory)
Z = d(0.5+ √ = d(0.5+ √ = 0.969d
= 0.096d > 0.95d ( use 0.95d)
As req = = = 99.3 mm2
Provide T10 – 300 ( As pro = 263 mm2
)
Distribution Reinforcement
Apply as minimum = = = 260 mm2
Therefore provide T 10 – 287 c/c (As pro = 287 mm2
)

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Shear checking
Vsx1= VX * nLx= 0.526 X 14.65 X 6.36 = 49 KN
V = = = 0.196
= = 0.22 < 3
= = 1.6 > 1
Vc = ( ) ( ) ( ) = 0.632 x 0.6 x 1.12 x 1.062
= 0.81 > V=0.196
Shear link is not required

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Deflection Checking
( )actual = 25.44( )basic = 26
= = 0.61
FS = = = 250
Modification factor = 0.55 + = 0.55 +
= 1.25 < 2
( )actual 26 x 1.25 = 32.5 > 25.44 “satisfactory”
Crack Checking
Maximum clearance distance 3d @ 750mm
3 x 119 = 357mm
357 – 10 = 347

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
LOADING SYSTEM
Self-weight of slab = c * h=0.15x25 =3.75KN/m2
Finishing is Assume to be = 1KN/m2
Total dead load on the slab (n) =3.75+1 = 4.75 KN/m2
Live load = 5KN/ m2
Ultimate Design load = 1.4(DL) + 1.6(LL) = 1.4(4.75) + 1.6(5)
= 14.65 KN/ m2
Middle of short span
Msx1 = SX * nLx2
= 0.0416x 14.65 x 6.362
= 24.65KN.m
D = 281– 25- = 250mm
K = = = 0.013 < 0.156
Z = d(0.5+ √ = d(0.5+ √ )
= 0.98d > 0.95d use 0.95d=
As req = = = 237.5mm2
Asmin = = 365.3mm2
As <Asmin therefore use Asmin
Provide T10- 200 c/c ( Aspro = 395 mm2
)

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Middle of long span
Msy1 = SX * nLx2
=0.028* 14.65*6.362
= 1 6.592 KN.m/m
D = 281-25- = 250mm
K = = = 0.0088< 0.156
Z = d(0.5+√ = d(0.5+ √ = 0.99d>0.95d use 0.95d
As req = = As req = = 160mm2
Asmin = = 365.3mm2
As req< Asmin therefore use Asmin
Provide T 12 – 200 c/c ( Aspro = 395 mm2
)
Long Continuous Side
Msy1 = SX * nLx2
=0.037* 14.65*6.362
= 21.92 KN.m/m
d = 281 – 25 - = 250 mm
K = = = 0.0116 < 0.156
Z = d(0.5+ √ = d(0.5+ √ = 0.986d >0.95d use 0.95d
As = = = 211.2mm2
Asmin = = 365.3mm2
< Asmin therefore Asmin
Provide T 10 – 200 c/c ( Aspro = 395 mm2
)

Design Of
Slab System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Check for Shear
Vsx1 = VX * nLx2
= 0.456 X 14.65 X 6.36 = 42.4KN
V = = = 0.168
= = 0.14 < 3
= = 1.6 > 1 ……………………2
Vc = ( ) ( ) ( ) = 0.63 x 0.526 x 1.12 x 1.062
= 0.4 > V =0.37
Shear link is not required
Checking for deflection of short span
= = 0.529
FS = = = 265.8
Modification factor = 0.55 +
( )
=0.55 + = 1.78< 2
( )actual = 25.2 ( )basic= 26
( )allow 26 x 1.78= 46.3>25.2 oky

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel beam Remark
= = 2.25 <2 Two way slab
Self-weight of slab = Ƥc h =0.281x25 .
= 7.025KN/m2
Self-weight of screed = 0.02x25
=0.5KN/m2
Total dead load on the slab (n) =8.525KN/m2
Dead load of beam from slab = WD = * +
= [
( )
]
= 18.07 x 1.255 = 22.67 x 2 side KN/m
= 45.36 KN/m
Self -weight of beam = 0.98KN/m
Total dead load on the beam = 46.30KN/m
(LL) on Beam from slab (n) WL = [
( )
] = [
( )
]
= 13.4KN/m x 2 sides
= 22.88 KN/m

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Ultimate load design = 1.4(DL) + 1.6(LL) = 1.4( 46.36) + 1.6(26.6)
= 107.5KN.m
M = = = 1112.75 KN.m
Sx = = 4046.3cm3
Try UB = 914x419x388 ( Sx = 17700)

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Section geometrical property
Plastic modulus sx =17700mm3
Elastic modulus Zx =15800mm3
Depth D = 921mm
Width B = 420.5mm
Web thickness t =21.4mm
Flange T =36.5mm
Depth between fillet d =799.6mm
Roof radius r =24.1mm
Flange slenderness b/T =5.74
Web slenderness d/t =37.4
Moment of inertia I =720000cm4
Buckling parameter u =0.885
Torsional index x =26.7
Check for design strength ( Py) and section classification for flange thickness
of T = 18.2, Py = 275 N/mm2
ɛ = * + = * + = 1.037
Compacting limiting of volume of b/T = 9ɛ = 9.3 x 1.03 = 9.3 > 5.74
Compacting limiting of volume of d/t = 80ɛ = 80 x 1.03 = 81 > 37.4
Therefore the section is Plastic
Table 9
Table 11 Node(b)

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
check for shear capacity at the support (max.shear)
FV = 490.3KN
1) Shear capacity pv=0.6 PYAv
Pv= 0.6 PYDt =0.6 x 21.4 x 921 x 265 x 10-3
= 3133.8kN > 490.3--------------------------OK
2) Check section for moment capacity
Moment max = 1112.75KN.m
0.6Pv = 0.6 x 3133.8
= 1880.3 KN.m
Fv < Pv = 490.3 < 1880.3 KN.m…it is low shear ………… OK
3) Moment capacity with low shear load for plastic section
Mcx = Py Sx 1.2PyZx
= 265 x 17700 x 10-3
1.2 x 265 x 15800 x 10-3
= 4690.5 5024.4
Mcx = 4690.5 KN.m
M < Mx 1112.75 < 4690.5 KN.m………… ……………….. OK
Cl:4.2.3
Cl:4.2.5

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
4) Check for lateral torsional buckling capacity
M
M = 950kn
Mlt = 0.85
Mb = SxPb =
Pb = uv w
5) Checking bearing and buckling at support
I. Bearing
Fv < Pwb
Local capacity of the web: Pbw = ( b1 +nk)tpw
The section properties of angle : 160 x160 x 18
. t = 18
.r = 16
Stiff bearing length b1 = is obtained by taking a tan at 450 through the
. bearing i.e along the tangent to root angle
b1 = 2t + 0.8r – c
b1 = 2(18) +0.8(16) – 8.76 = 40.04
k = T+r = 36.5 + 24.1 = 60.6
n = 2
Pwb = ( b1 +nk)tpyw
= (40.04+2(60.6) x 21.4 x 265 x 10-3
= 914.39KN
Forced applied through the flange
914.39 > 490.3 KN ………………………………….OK
Therefore bearing stiffener is not required.
Cl:3.1.3
Cl:4.5.2.1

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Bucking
ae = 40.04/2 = 20.02 mm ( ae = 0.7d)
Buckling resistance off the un-stiffened web.
Px = x
√
= x
√
x 914.39
= 053x 1.8
= 14,086.5
Fv = 490.3 KN < 858 KN …… ……………………………. OK
Check for deflection under servicebility loads
[
( )
] = 10.6 x 1.255 = 13.3 x 2 = 26.6 KN
E= 205KN/mm2
I= 720000cm4
=* + x104
=* + x104
= 0.16 mm
Limiting lim=span/360 = 9100/360 = 25.27 > 0.16 …………Ok
Cl: 4.5.3.1
Table 8

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel beam two Remark
= = 2.25 <2 Two way slab
Self-weight of slab = Ƥc x h = 0.281x25 .
= 7.025KN/m2
Self-weight of screed = 0.02x25
= 0.5KN/m2
Self-weight of brick wall 18Kgm/m3 x 0.15 x 4
= 10.8KN/m
Total dead load on the slab (n) = 19.325KN/m
Dead load of beam from slab = WD = * +
= [
( )
]
= 40.9 x 1.08 = 51.4 KN/m
Self -weight of beam = 0.98KN/m
Total dead load on the beam = 52.4 KN/m
Live load on Beam from slab (n) WL = [
( )
]
= [
( )
]
= 13.3 KN/m
Ultimate load design = 1.4(DL) + 1.6(LL) = 1.4( 52.4) + 1.6(13.3)
= 94.64 KNm2

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
M = = = 979.6 KN.m
Sx = = = 3562cm3
Try UB = 914x419x388 ( Sx = 17,700)

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Depth D = 921mm
Width B = 420.5mm
Web thickness t = 21.4mm
Flange T = 36.5mm
Depth between fillet d = 799.6mm
Roof radius r = 24.1mm
Flange slenderness b/T = 5.74
Web slenderness d/t = 37.4
Moment of inertia I = 720000cm4
Buckling parameter u = 0.885
Torsional index x = 26.7
of T = 18.2, Py = 275 N/mm2
ɛ = * + = * + = 1.037
Compacting limiting of volume of b/T = 9ɛ = 9.3 x 1.03 = 9.3 > 5.74
Compacting limiting of volume of d/t = 80ɛ = 80 x 1.03 = 81 > 37.4
Therefore the section is Plastic
Table 9
Table 11
Node(b)

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
FV = 430.6 KN
pv=0.6 PYDt =0.6 x 21.4 x 921 x 265 x 10-3
= 3133.8kN > 430.6--------------------------OK
Check section for moment capacity
Moment max = 979.6 KN.m
0.6Pv = 0.6 x 3133.8
= 1880.3 KN.m
Fv < Pv = 430.6 < 1880.3 KN.m…it is low shear ………… OK
Moment capacity with low shear load for plastic section
Mcx = Py Sx 1.2PyZx
= 265 x 17700 x 10-3 1.2 x 265 x 15800 x 10-3
= 4690.5 5024.4
Mcx = 4690.5 KN.m
M < Mx 979.6 KN.m < 4690.5 KN.m………………………….. OK

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
1) Check for deflection under servicebility loads
[
( )
] = 10.6 x 1.255 = 13.3 x 2 = 26.6 KN
E = 205KN/mm2
I = 720000cm4
=* + x104
=* + x104
= 0.16 mm
M
M=950kN
Mlt=0.85
Mb= SxPb =
Pb = uv w
U=0.885
Le= 1.0xL = 1.0x9100 = 9100
Ry= 95.9
𝛌y = 9100/95.9 = 94.8
X = 26.7
𝛌/x = 94.8/26.7 = 3.35
V= 0.89
𝛌 uv w = x x x = 79.5
Py =265 =79.5
Mb = 161x10-3
x17700 = 2849.7
Mb/mlt > M 2849.7/.85 = 3352.5KN > 979.6 KN.m

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
II. Bearing
Fv < Pwb
. t = 18
.r = 16
b1 = 2t + 0.8r – c
b1 = 2(18) +0.8(16) – 8.76 = 40.04
k = T+r = 36.5 + 24.1 = 60.6
n = 2
Pwb = ( b1 +nk)tpyw
= (40.04+2(60.6) x 21.4 x 265 x 10-3
= 914.39KN
914.39 > 490.3 KN ………………………………….OK
Bucking
ae = = 20.02 mm ( ae = 0.7d)
Px = x
√
= x
√
x 914.39
= 053x 1.8 x 914.39 = 858
Fv = 430.6 KN < 858 KN ………………………………………. OK

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel beam three Remark
Self-weight of slab = Ƥc x h = 0.281x25 .
= 7.025KN/m2
Self-weight of screed = 0.02x25 = 0.5
Total dead load = 8.8KN/m2
= 17.61 x 2 sides = 35.22 KN/m
Self-weight of beam = 0.98 KN/m
Total dead load = 36.22
Live load = x 2 sides = 20 sides KN/m
Ultimate load = 1.4(DL) + 1.6(LL) = 1.4(36.2) + 1.6(20) = 86.68 KN/m

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
M = = = 371.8 KN.m
Shear = = 247.8 KN
Sx = = = 1352 cm3
Try UB = 610x 229 x 125(sx = 3880cm3)
Depth D = 612.2
Width B = 229
Web thickness t = 11.9
Flange T = 19.6
Depth between fillet d = 547.6
Roof radius r = 12.7
Flange slenderness b/T = 5.84
Web slenderness d/t = 48
Moment of inertia I = 98800cm4
Buckling parameter u = 0.874
Torsional index x = 34.1
of T = 19.6, Py = 275 N/mm
ɛ = * + = * + = 1.0
Compacting limiting of volume of b/T = 9ɛ = 9x 1 = 9 > 5.84
Compacting limiting of volume of d/t = 80ɛ = 80 x 1 = 80 > 48
section is class one (plastic section)

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
FV = 247.8 KN
pv=0.6 PYDt =0.6x 612.2 x 19.6 x 275 x 10-3
= 1979.8 kN > 247.8--------------------------OK
1) Moment capacity with low shear load for plastic section
Mcx = Py Sx 1.2PyZx
= 275 x 3880 x 10-3 1.2 x 275 x3220 x 10-3
= 1067 1062.6 KN.m
Mcx = 1062.6 KN.m
M < Mx 371.8 KN.m < 1062.6 KN.m………………………….. OK

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
1) Check for deflection under servicebility loads
= 10 x 2 sides = 20 KN
E = 205KN/mm2
I = 98800cm4
=* + x104
=* + x104
= 7.7 mm
M
M=371.8kN
Mlt=0.85
Mb= SxPb
Pb = uv w
U = 0.874
Le = 1.0xL = 1.0x 6000 =6000
Ry = 49.7
𝛌y = 6000/49.7 = 120.7
X = 41.8
𝛌/x = 120.7/34.1 = 3.5
V= 0.89
uv w = x 7 x 7x = 93.8
Py =275 =145.4
Mb = SxPb =3880 x145.4 x 10-3
= 564 KN
Mb/mlt = 564/.85 = 663.7 KN > 371.8 KN…………………………. OK
Table 19

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
III. Bearing
Fv < Pwb
. t = 18
.r = 16
b1 = 2t + 0.8r – c
b1 = 2(18) +0.8(16) – 8.76 = 40.04
k = T+r = 36.5 + 24.1 = 60.6
n = 2
Pwb = ( b1 +nk)tpyw
= (40.04+2(60.6) x 21.4 x 275 x 10-3
= 948.89KN
948.8 > 247.8KN ………………………………….OK
IV. Bucking
ae = b1/2 = 40.04/2 = 20.02 mm ( ae = 0.7d)
Px = x
√
= x
√
x 948.8
= 0.53x 2.1 39 x 948.8 =1075.7
Fv = 247.6 KN < 1075.7KN ……………………………………. OK

Design Of
Beam System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel Column Remark

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel column Remark
Dead load (kn) Imposed load Total
Un-
factored
Factored x
1.4
Un-factored Factored
x 1.6
Level3(roof)
R1-3
R2-3
62.4
0.98
87.5
1.37
……….
……….
----------
----------
R2-1
R2-2
Self-weight of
column
64.8
64.8
.98
90.7
90.7
1.37
67.4
67.4
108
108
R1-3
R2-3
Self-weight of
column
64.8
64.8
90.7
90.7
0.98
67.4
67.4
108
108
Total 453.6 432 885.63
1) To determine nominal moment
Geometrical properties of section
356 x 406 x 467
D = 436.6 r = 15.2 ry = 107
B = 412.4 d = 290.1 rx = 107
T = 35.9 = 3.56 Ag = 595.5
T = 58 = 8.08 u = 0.839
I = 183118 x = 6.86 Zy = 3293
Zx = 8388 Sx = 10009 Sy = 5038
Total design axial load : Fc = 885
Design strength Py
Thickness of the thickest element of the steel. Section
T = 58 > 16
Py = 255

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Normal moment Mx
ex = 100 + = 100 + = 138.3
Mx = 318.3 x 198.7 x 10-3
= 63.24 KN.m
Normal moment My
ey = 100 + = 100 + = 117.95 mm
My = 198.7 x 117.9 x 10-3
= 23.43 KN.m
The ratio of large to lower column stiffness
= = X = 1.5
Since the ratio of column stiffness is equal 1.5, then the nominal moment can
be divided equally.
Mx = 63.24/2 = 31.62 KN.m
My = 23.34/2 = 11.72KN.m
To determine the capacity of column design strength Py
Thickness of the thickest element of the steel section
T = 58 < 63, Py = 255
Section classification
ɛ = * + = * + = 1.04
flange classification
= 3.56
b/T = 9ɛ = 9.3 x 1.04 = 9.3>3.56
Flange is class 1
Web classification
= 8.08
Check weather web is class 1

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Fc = is positive because axial is compression
R1 = = = 0.22 < -1 r1 1 ……………….OK
class is plastic section
Compression capacity for class 1 section
Pc = Ag*Pe
To determine Pc
Effective length = Le = 1 x 6000 = 6000mm
Slender ratio
x = = = 34.
y = = = 56
Rolled H section greater than 40mm
For buckling about x-x axis use struc curve c
For buckling y-y axis use struc curve d
x = 34.3 Py { } = 231.4 KN/mm2
y = 56 Py { } = 179 KN/mm2
Therefore Pcy = 179 KN/mm2
Compression Capacity Pc = Ag*Pc
= 595.5 x 102
x 179 x 10-3
= 10659.45 KN
Buckling resistance moment for column in simple construction
lt = 0.5(l/ry)
= 0.5(6000/107)
= 28
Py = 255
Pb = 255
Mbs = Pb* Sx = 255 x 10,009 x 10-3
= 2552.3 KN.m

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Moment capacity about minor axis
Py Zy = ( 255N/mm2
) ( 8388 x 10-3
)
= 2138.9 KN/m
Solving the intersection equation
+ + = 1
= + + = 1
= 0.08302 + 0.012389 + 0.00548 1
= 0.1 1
Try UC = 356 x 406 x 467

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Dead load (KN) Imposed load Total
Un-
factored
Factored x
1.4
x 1.6
Level3(roof)
R1-3
R2-3
R3-3
124.8
0.98
……..
174.72
1.37
………
……….
……….
……….
----------
----------
…………
R1-2
R2-2
R3-2
Self-weight of
column
64.8
64.8
129.6
0.98
90.7
90.7
181.4
1.37
67.4
67.4
134.8
108
108
215.68
R1-1
R2-1
R3-1
Self-weight of
column
64.8
64.8
129.6
0.98
90.7
90.7
181.4
1.37
67.4
67.4
134.8
108
108
215.68
Total 904.43 863.36 1767.8
356 x 406 x 467
D = 436.6 r = 15.2 ry = 107
B = 412.4 d = 290.1 rx = 107
t = 35.6 = 3.56 Ag = 595.5
T = 58 = 8.08 u = 0.839
I = 183118 x = 6.86 Zy = 3293
Zx = 8388 Sx = 10009 Sy = 5038
Total design axial load : Fc = 1767.8
Design strength Py
T = 58 > 16
Py = 255

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Normal moment Mx
ex = 100 + = 100 + = 138.3
Mx = Pc x xe = 318.3 x 968.92 x 10-3
= 308.4 KN.m
= = X = 0.66 < 1
be divided equally.
Mx = 63.24/2 = 31.62 KN.m
T = 58 < 63, Py = 255
ɛ = * + = * + = 1.04
= 3.56
b/T = 9ɛ = 9.3 x 1.04 = 9.3 > 3.56
Flange is class 1
Web classification
= 8.08

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
R1 = = = 0.22 < -1 r1 1 ……………….OK
Pc = Ag. Pe
To determine Pc
Slender ratio
x = = = 34.
y = = = 56
x = 34.3 Py { } = 231.4 KN/mm2
y = 56 Py { } = 179 KN/mm2
= 595.5 x 102 x 179 x 10-3
= 10659.45 KN
Buckling resistance moment for column in simple construction
lt = 0.5(l/ry) = 0.5(6000/107) = 28
Py = 255
Pb = 255
Mbs = Pb* Sx = 255 x 10,009 x 10-3 = 2552.3 KN.m

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Py Zy = ( 255N/mm2) ( 8388 x 10-3)
= 2138.9 KN/m
+ 1
= + = 1
= 0.16 + 0.12 = 0.28 1
= 0.28 1
Try UC = 356 x 406 x 467

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Dead load (KN) Imposed load Total
Un-
factored
Factored x
1.4
x 1.6
Level3(roof)
R1-3
R2-3
R3-3
R4-3
124.8
0.98
……..
……..
174.72
1.37
………
………
……….
……….
……….
----------
----------
…………
R1-2
R2-2
R3-2
R4-2
Self-weight of
column
64.8
64.8
64.8
64.8
0.98
90.7
90.7
90.7
90.7
1.37
67.4
67.4
67.4
67.4
108
108
108
108
R1-1
R2-1
R3-1
R4-1
Self-weight of
column
64.8
64.8
64.8
64.8
0.98
90.7
90.7
90.7
90.7
1.37
67.4
67.4
67.4
67.4
108
108
108
108
Total 904.43 863.36 1767.8
356 x 406 x 467
D = 436.6 r = 15.2 ry = 107
B = 412.4 d = 290.1 rx = 107
t = 35.6 = 3.56 Ag = 595.5
T = 58 = 8.08 u = 0.839
I = 183118 x = 6.86 Zy = 3293
Zx = 8388 Sx = 10009 Sy = 5038
Total design axial load : Fc = 1591.7
Design strength Py
T = 58 > 16
Py = 255

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Normal moment Mx
ex = 100 + = 100 + = 138.3
Mx = Pc x xe = 318.3 x 968.92 x 10-3 = 308.4 KN.m
= = X = 0.66 < 1
be divided equally.
Mx = 63.24/2 = 31.62 KN.m
T = 58 < 63, Py = 255
ɛ = * + = * + = 1.04
= 3.56
b/T = 9ɛ = 9.3 x 1.04 = 9.3 >
3.56
Flange is class 1
Web classification
= 8.08

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
R1 = = = 0.22 < -1 r1 1 ……………….OK
Pc = Ag. Pe
To determine Pc
Slender ratio
x = = = 34.
y = = = 56
x = 34.3 Py { } = 231.4 KN/mm2
y = 56 Py { } = 179 KN/mm2
= 595.5 x 102
x 179 x 10-3
= 10659.45 KN
Buckling resistance moment for column in simple
construction
lt = 0.5(l/ry)
= 0.5(6000/107)
= 28
Py = 255
Pb = 255
Mbs = Pb* Sx = 255 x 10,009 x 10-3
= 2552.3 KN.m

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Py Zy = ( 255N/mm2
) ( 8388 x 10-3
)
= 2138.9 KN/m
+ 1
= + = 1
= 0.16 + 0.12 = 0.28 1
= 0.28 1
Try UC = 356 x 406 x 467

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel Base plate for column Remark
7
Axial load of column = 1767.8 KN for column = 7
Bearing strength of concrete = 0.6Fcu …………………………………………………..
Take Fcu =
Area required ( ) = = =
{ } =
√
= = 15.33
Find the thickness of the base plate
tp = * + ………………………………………………………………………………………
where
Pyp = design strength of the of the base plate 7
tp = 15.33 * + = 7.84mm
7 ……………………………………………………………….. OK
Cl: 4.13.1
4.13.2.2

Design Of
Column System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel footing Remark

Design Of
Connection
System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel connection Remark
1) Assume the bottom row of bolts does not carry shear force.
Fs = = 7 7
Ft ∑
Assume center of rotation is about the bottom bolts:
Ymax = 400mm
∑ y
Ft
For simple of method
+
Shear capacity of connection;
Shear capacity of bolt:
Ps =
As = At 7 (Assume failure at thread of bolt)
Ps = 400N/mm2
Shear force, Ps = 7 7 7 7 … …
Shear capacity of bolt
Pbb = d.t.pbb FS 7 7 … …
(t is taken as the smaller of thickness offend plate and flange. The
thickness of column flange for = 356 x 406 x 467 UC is 58mm)
Table 30
Table 31

Design Of
Connection
System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Design of Steel Connection Remark
Bearing capacity of plate;
Pbs = Pbs 0.5.e.t. Pbs = =
=
Pbs = 7 … … … … … … … … … … … … … …
Pnon = 7 7 7
Check interaction formula
+ = +
7
……………Ok
Therefore the connection is Safe
Check block shear capacity:
Pr = 0.6Py.t{ }
Lv =
Pr = 0.6 x 275 x 58{ }
Pr … … … … … … … … … … …
Table 32
Table 34
Cl: 6.2.4

Design Of
Connection
System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Solution
2) To determine the size of weld which is safe and economic for the
beam and plate connection;
Fs
Ft
* + =
Ft
Resultant force, FR =
FR = 0.3 N/mm
Use weid size 5mm, electrode class E35, Py =220N/mm2
Pw = 0.7.s.Pw 7 77 0.3KN

Design Of
Connection
System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Connection beam to beam
1) Assume the bottom row of bolts does not carry shear force.
Fs = =
Ft ∑
Assume center of rotation is about the bottom bolts:
Ymax = 300mm
∑ y
Ft
For simple of method
+
Shear capacity of connection;
Shear capacity of bolt:
Ps =
As = At (Assume failure at thread of bolt)
Ps = 400N/mm2
Shear force, Ps = … …
Shear capacity of bolt
Pbb = d.t.pbb FS … …
(t is taken as the smaller of thickness offend plate and flange. The
thickness of column flange for = 914 x 419x 388 UB is 21.4mm
Table 30
Table 31
Table 32

Design Of
Connection
System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Bearing capacity of plate;
Pbs = Pbs 0.5.e.t. Pbs = =
=
Pbs = … … … … … … … … … … … … … …
Pnon = 7
Check interaction formula
+ = + ……………Ok
Therefore the connection is Safe
Check block shear capacity:
Pr = 0.6Py.t{ }
Lv =
Pr = 0.6 x 275 x 21.4{ }
Pr … … … … … … … … … … …
Cl:6.2.4

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Solution
3) To determine the size of weld which is safe and economic for the
beam and plate connection;
Fs
Ft
* + =
Ft
Resultant force, FR =
FR = 0.3N/mm
Use weid size 5mm, electrode class E35, Py =220N/mm2
Pw = 0.7.s.Pw 7 77
Table 37

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
FOUNDATION DESIGN Remark
ZERO MOMENT
Given Data
Total axial load = 1767 KN.m
Fcu = 40 KN/mm2
Fy = 460 KN/mm2
Solution
 Total live load and dead load.
Third floor loading
 Dead load (DL) = 904.43KN/m
 Live load (LL) = 863.36KN/m
Assume that = 25 > 20
Cover = 45 mm
a) Determine foundation thickness, h.
Deform types = La = 27
La = 27 x 26 = 675 mm
Assume that bar is bend = 200 mm
L = 675 – 200
= 475 mm
H = 475 + 45 + 2 x 25 = 570 mm = 600 mm

Design Of
Footing System
Date:
18/1/2018017
Yasin Said Mohamed
Reference
b) Service limit state
1.0 GK + 1.0 QK = 1.0 X 904.43X 1.0 X 631.4
=1767KN + self weight of the footing
= 1767 + 10% ( 1767)
= 1944.7 KN
D = 600- 45 – 25 - 25/2 = 517.5 518 mm
Assume Bearing capacity = 200KN/m2
Required base Area = =
= 9.72 m2
Provide a base area of square = 9.7 m2
Footing self-weight = height footing x base Area xUnit-weight of
concrete
= 0.6 x 9.7 x 25 = 145.5 KN
Column design axial load = 1(DL) + 1(LL) = 1767KN
Earth pressure = = = 182.16KN/m2

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Try UC = 356 x 406 x 467
Moment = 182.1KN/m2
x 2.51 x x 3.1 =1778.2 KN
d = 600 – 45 – 25 – 25/2 = 517.5mm = 518mm
K = = = 0.053 < 0.156
Z = d(0.5 + √ ) = 0.937d ok
Ar req = = 5554.79 mm2
Check Armin= 0.13%bh = =2418 mm2
T25-200c/c (AS pro = 2450mm2
)

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Shear Check
Checking maximum shear
Vmax= = = 0.275 < 0.8 = 5.05
Normal Shear / Vertical shear
= = = 0.15N/ mm
VC= x ( ) x x ( )
= 0.632 x 0.531 x 1 x 1.17
= 0.4
Punching Shear
Critical perimeter = ( column perimeter + 8 x 1.5d)
= + 8 x 1.5 x 518
= 2461.6 + 6216
= 8677.6 mmVc
Area within perimeter =[ ]
= [ ]
= 3.81 x mm2
Punching shear force = 182( 3.12
– 3.81) = 1055.6KN
Punching shear stress = =
= 0.234
0.234 < 0.4
Check Crack = 3d @ 750mm
= 3 x 518 = 1554 mm > 750mm
Actual distance between =
– –
=
– –
= 353.75mm < 750 mm
H > 200mm

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Referenc
e
FOUNDATION DESIGN Remar
k
Check = = = 0.15 < 0.3 …………………………ok
Detailing

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Maximum moment footing
load(N) = 1767.6 KN
Moment = 308KN.m
PMAX = + = +
=282.7 + 118.4
= 401.1
PMIN= - = -
=282.7 -118.4
= 164.3
Assume the area is =2.5x2.5 = 6.25m2
Central load = 1.047 /2.5 x 241.8 = 101.26
401.1 – 101.26 = 299.9 kNm
Moment maximum
= (300x 2.5 x 1.047 x0.5235)+* 7 +
= 411.07 + 79.38
=490.45

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
d = 600 – 45 – 25 – 25/2 = 517.5mm
K = = = 0.018 < 0.156
Z = d(0.5 + √ = 0.98d > 0.95d
Use 0.95d
Z = 0.95 x 517.5 = 491 mm
ASreq = = = 2283.6 mm2
Check Asmin= 0.13%bh = =234
T25 – 200c/c (AS prov=2450 mm2)
Maximum shear checking
Average: 164.3 + = 282.7
Vmax = = = 0.341< 0.8 = 5.05
Normal Shear / Vertical shear
= = = 0.189N/ mm
= = 0.77
VC = x ( ) x x ( )
= x x 77 x
= 0.632 x 0.961 x 0.94x 1.69
= 0.964

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
N = 282.7 X 2.5 (1.047 - 0.518) = 373.87 KN
= = 0.288 < VC = 0.9……..ok
Punching Shear
Critical perimeter = ( column perimeter + 8 x 1.5d)
= 2461.6 + 8 x 1.5 x 518
= 2461.6 + 621.6
= 3083.2 mm
Area within perimeter = [ ]
= [ ]
= 3.81 x mm2
Punching shear force = 282.2 (2.52
– 3.81)
= 688.56KN
Punching shear stress = =
= 0.43
0.43< 0.964
Check Crack = 3d @ 750mm
= 3 x 518 = 1554 mm > 750mm
Actual distance between =
– –
=
– –
= 277.5mm > 750 mm
H > 200mm
Check = = = 0.189< 0.3…………………….OK

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Detailing

Design Of
Stair Case
Date:
18/1/2018
Yasin Said Mohamed
Reference
Stair case DESIGN Remark
Design platform members for stair case show in figure:
Using A36steel F =22 k
Stringer are to plate
Headers are to be Channel section
DL = 97.1spf
LL = 105.3 spf
Solution
1. Face stringers (see Note below)
W = = = 6.8 k
From table 5.28 on Page 5.1
Select (3/16” x 10” plate* w = 6.8k)
R = w/2 = 6.8/2 = 3.4k
Selection is…………………………………………………. ok
Figth header
W = ksf x ksf x = 0.378k/ft
P -= 2 x 3.4 = 6.88k
M = + pl/4 = + 6.8 x 11.4/4 = 25.52kip.ft
Sreq = = = 13.9 = 14 find S
From the table 5.32; on page 5-14
C10x 20; S = 15.8; I = 78.8
Limit defilection
I > 0.00228(1.6P +wl) = 0.00228(1.6 x 6.8 + 0.378 x 11.4) (11.4)2
6 x 104
16in < 36.6………………. OK
R w + p/2 = 0.378 x 11.4/2 + 6.3 5.5K

Design Of
Footing System
Date:
18/1/2018
Yasin Said Mohamed
Reference
Platform header
W = 0.378 x 11.4 5 /2 = 11.7
M = wl/8 = 0.378x 11.4/8 = 16.6kips
S = = = 9in
5” from table 5.32 on page 5-14
Section
C9 x 13.4; S = 10.6; I = 47.9
Check deflection
= 0.0007633 = 0.000763 x11.7 = 0.27 in
Allowble = p x 12 / 360 = 0.38 in…………………………………… ok
R = w/2 = 11.7/2 = 5.8k
Support point A
Laod R platform = 5.8k
Select support member from table 5.28 and 5.31 on page 5-12 & 5- 13
Support B
Load R of flight header + R of wall stringer = 11.7+5.8

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