Project 2

 

1. Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows:

    

Width	30m
Length	50m
Eave Height	9m
Bay spacing	6m
Soil type	Medium
Safe Bearing Capacity	200 kN/m2
Roof slope	1 in 12

Assume suitable sections for structural elements. Follow IS800:2007, IS1893 and IS 875

1. Prepare a Design Basis Report for the project
2. Create structural model using STAAD Pro Connect Edition
3. Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
4. Design using MS – Excel
   1. Steel column
   2. Rafter
   3. Base plate
   4. Pedestal
   5. Z- purlin

1. Design Foundations using STAAD Foundation.
2. Attach necessary sketches and drawings wherever required.
3. All stipulations, assumptions and design parameter
4. must adhere to Indian Standards.

 

 

 

ANS-

 

1. ROOF SLOPE ASSUME =  X 

 

 

GIVEN SLOPE  =1/12 

 

B= 30M

 

     X/[b/2]    =  1/12

 

   X/15       1/12

 

 

X        =   15/12  = 1. 25   m

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  3-D RENDERED MODEL 

 

 

 

 

 

 

 

2. material   

 

 STEEL 

 

3.SPECIFICATIONS 

 

-MEMBER TENSION IN BRACING SYSTEM 

BETWEEN RAFTERS

-REALEASE MOMENTS  END AND START 

 

 

 

 

 

 

 

 

 

 4 . SUPPORT    - PIN SUPPORT 

 

5. LOADING 

 

LOAD CASES DETAILS -

 

1. EQ+X

2.EQ+Z 

 

3.DEAD LOAD   -

 

-self weight -1.15 KN/M

-Main frame rafter   GY-1.856  KN/M

-GABLE FRAME RAFTER -.928 KN/M 

- MAIN FRAME COLUMN GY-1.31  KN/M 

 -GABLE  FRAME COLUMN GY -0.655  KN/M ,

-GABLE / WIND COLUMN GY -1.04  KN/M ,

-WIND FRAME  POINT LOAD FY -1 KN/M ,

-GABLE FRAME POINT LOAD FY-0.58  KN/M

 

 

4.LIVE LOAD -

-Main frame rafter   GY-5  KN/M

-GABLE FRAME RAFTER -2.625 KN/M 

-WIND FRAME  POINT LOAD FY -4.375 KN/M ,

-GABLE FRAME POINT LOAD FY-2.187  KN/M

 

 

SEISMIC DEFINATION:

Zone Parameters:

MEMBER WEIGHT :

ASSIGN to edit list from followed by dead load 

-

 

-self weight  Y -1.15 KN/M

-Main frame rafter   GY-1.856  KN/M

-GABLE FRAME RAFTER -0.928 KN/M 

- MAIN FRAME COLUMN GY-1.31  KN/M 

 -GABLE  FRAME COLUMN GY -0.655  KN/M ,

-GABLE / WIND COLUMN GY -1.04  KN/M ,

 

 

 

 

 JOINT WEIGHT -

 

LOAD CASES DETAILS -

5. WIND LOAD [O+ CPi]

 

 

6.WIND LOAD [O-CPi]

 

7 WIND LOAD [9O+ CPi] 

 8.WIND LOAD [9O-CPi]

 

 

 

 

 

 

 

 

DESIGN OF RAFTER -

 

 

 

 

 

 

2. Design a simply supported gantry girder to carry electric overhead travelling crane

Given: 

Span of gantry girder = 7 m

Span of crane girder = 9 m 

Crane capacity = 250 kN 

Self-weight of trolley, hook, electric motor etc. = 40 kN 

Self-weight of crane girder excluding trolley = 100 kN 

Minimum hook approach = 1.0 m 

Distance between wheels = 3 m 

Self-weight of rails = 0.2 kN/m

 

 

 

DESIGN OF STEEL COLUMN:

Solution:

Flange thickness = T = 12.7 mm.

Overall height of Column ISMB400 = h = 400 mm.

Clear depth between flanges = d = 400 – (12.7 x 2)
= 374.6 mm.

Thickness of web = t = 10.6mm.

Flange width = 2b = bf = 250 mm.

Hence, half Flange Width = b = 125 mm.

Self –weight = w = 0.822 kN/m.

Area of cross-section = A = 10466 mm2.

Radius of gyration about x = rx = 166.1 mm.

Radius of gyration about y = ry = 51.6 mm.

Type of section:

b/T = 125/12.7= 9.8 < 10.5

d/t =374.6/10.6 =35.3 < 42

(Table 3.1 of IS: 800)

Hence, cross-section can be classified as “COMPACT”.

Effective Sectional Area, Ae = 10,466 mm2

(Since there is no hole, (Clause 7.3.2 of IS: 800)
no reduction has been considered)

Effective Length:
As, both ends are pin-jointed effective length, KLx = KLy = 3m

Slenderness ratios:

KLx/rx = 9000/166.1 =54.1

KLy/ry = 9000/51.6 =174.41

Non-dimensional Effective Slenderness ratio, $\lambda$:

$\mathrm{\lambda \; =}$

    =$\frac{\sqrt{2.5}X{(174)^}^{\mathrm{2/}}}{{\mathrm{\pi ^}}^{2}X2X{\mathrm{10^}}^5}$= 3.83

Value of $ts=\surd 2.5w(a2-0.3b2)\gamma m0fy>tf$ from equation: 

`ence,  = 0.34 for buckling class ‘b’ will be considered.
Hence,  = 0.5 x [1+0.34 x (0.654-0.2)+0.6542] = 0.79`

Calculation of x from equation x:

 = 0.809

Calculation of fcd from the following equation:

= 0.809 x 250/1.1 = 183.86 N/mm2

Factored axial load in kN.
pd = Ag fcd
    = 10466 x 183.86/1000 = 1924.28 kN.

 

 

DESIGN OF RAFTER:

Solution: 

Span of Rafter = 6 m

Dead load = 18KN/m

Imposed load = 40KN/m

support bearing = 100mm

yield strength = 250N/mm2

Design load calculation:

Factores load 1.5(DL+LL) =87 KN/m

Factores Bending moment = $\frac{W{\mathrm{l^}}^{\mathrm{2/8}}}{}$= 65.25 KN/m

Section modulus Required:

Z reqd = (65.25 x x1.1) /250 =287100 mm3 = 287.1 cm3

Section classification:

ISMB-200

A = 323.3 mm2

D = 200mm

B = 100mm

t = 5.7 mm

T = 10.8 mm

Ixx = 2235.4 cm4

Iyy = 150 cm4

Zp = 375.35 cm3

Moment of resistence of the cross-section:

Md = $\frac{\beta bZpf\mathrm{y/}}{\gamma m0}$

      = (1 x 375.35 x 250) / ( 1.1)

Md = 85.306 > 65.25 KN/m

 

DESIGN OF BASE PLATE:

Strength of concrete, Fcu = 40 N/mm2

Yield strength of steel, fy = 250 N/mm2

Material factor, $\gamma m$= 1.1 KN

Factoresl oad = 1500 KN

Steel column section:

Thickness of flange, T = 12.7 mm

Area required:

Bearing strength of concrete = 0.4fcu = 0.4 x 40 = 16 N/mm2

                                          = (1500x1000) /( 16)

                                          = 93750 mm2

Let size of plate , Bplate = 450 mm

                          Dplate = 300 mm

                 Area of plate= 135000 mm2

projection on each side = a=b =25mm

W = (1500x1000) /( 450 x 300)

     = 11.11 N/mm2

Therefore, Thickness of Base Plate, clause 7.4.3.1

= 7.3 mm < 12.7 mm

Size of Base plate 450 x 350 x 16 mm

 

DESIGN OF PEDESTAL:

Grade of concrete = 40 N/mm2

Load                    = 200 KN

Moment               = 120 KN

Horizontal shear   = 20 KN

Yield strength       = 250 N/mm2

Length of base plate = 450 mm

Width of base plate = 350 mm

C/C distance of bolt in group-Z = 300 mm

C/C distance of bolt in group-X = 180 mm

Bearing strength of concrete Fc = 16 N/mm2

Depth of Column = 300 mm

Width of Column = 250 mm

Anchor Bolt Details

Dia of anchor bolts =24 mm

No:of anchor bolts in each side = 4

Total no:of anchor bolts, n = 8

Gross area of the bolt ;Asb' = 452.16 mm2

Net area of bolt 'Anb' = 352 mm2

Ultimate tensile strength of bolt 'fub' = 400 N/mm2

Fyb (anchor bolts) = 240 N/mm2

Base plate Details

Ultimate tensile strength of plate 'fu' = 490 N/mm2

Thickness of plate = 16 mm

Yield stress of plate = 330 N/mm2

Anchor bolt design

Area of the plate =  157500 mm2

Stress,   Maximum pressure =$\frac{\mathrm{F/}}{A}+\frac{6XM\mathrm{z/}}{B{\mathrm{L^}}^2}$

                                         = 10.6 N/mm2 < 16 Hence OK

             Minimum pressure =

                                         = -9.04 N/mm2 < 16 Hence OK

 Centroid = = 242.89 mm

a =L/2-C/3 = 144.04 mm

e = (L-Ld) / 2 =75 mm

y = (L - C/3-e) = 294.04 mm

Tension in anchor bolt along the length of plate, FT = (Mz - Fa)/Y

                                                                          = 364.38

Tension per bolt                                                   = 91 KN

Shear per bolt                                                     = 1.63 KN

Shear Check

Factored shear force = Vsb = 1.63 KN

                                 Vd,sb = 81290.9 N

                                          = 81 KN

Factored                      = 65.03 KN

Tensile Check

Factored tensile force in bolt, Tb = 91.1 KN

Tensile strength of bolt        Ts,b = Tn,b /$\gamma mb$= 126 KN

Td,b = 98.65 KN

Combined Unity Check

Vsb/Vdb = 0.025

Tb/Tdb = 0.92

Unity check = 0.85 <  1, Hence OK

Anchor Bolt Length

 Bond strength in tension, $\tau bd$= 1.4 N/mm2

Anchor length required = Tb(3.14*$\tau bd$)

                                  = 863.44 mm

Let Anchor Bolt Length = 900 mm

 

 

 

 

 

Width	30m
Length	50m
Eave Height	9m
Bay spacing	6m
Soil type	Medium
Safe Bearing Capacity	200 kN/m2
Roof slope	1 in 12

 

DESIGN OF -Z PURLIN-

 

 

 

STEP -1                                                WIND LOAD ON MEMBERS
		WIND LOAD ON MEMBERS				F     =	[Cpe -Cpi] .A. Pd
		INTERNAL PRESSURE COFFICIENT				Cpi	-0.5
		external pressure cofficient				Cpe
		length of building				L	L	=70M	80
		WIDTH OF BUILDING				B		=35M	40m
		HEIGHT OF COLUMN				H		=	10
		RATIO				L/B		=	2
						H/W		=	0.26
		ROOF ANGLE				1/12		=	4.76 DEGREE
							TAKE	=	5 degree
		DESIGN WIND PRESSURE                                          Pd							1.48 kn/m2
		governing spacing				for rafter			7m
step-2	Cpe values -Roof
	TABLE 6
					0 DEGREE		90 DEGREE
				EF	GH	EG	GH
				-0.9	-0.4	-0.8	-0.4
STEP-3	WIND LOAD ON RAFTER
angle	face	Cpe	Cpi +	Cpi -	Cpe - Cpi	Cpe - Cpi	Pd for +	Pd for   -
					for +	for -
0 DEGREE	EF	-0.9	0.5	-0.5	-1.4	-0.4	-14.52	-4.15
	GH	-0.4	0.5	-0.5	-0.9	0.1	-9.34	1.04
90  DEGREE	EG	-0.8	0.5	-0.5	-1.3	-0.3	-13.49	-3.11
	FH	-0.4	0.5	-0.5	-0.9	0.1	-9.34	1.04

 

 

 

2. Design a simply supported gantry

girder to carry electric overhead travelling crane

Given: 

Span of gantry girder = 7 m

Span of crane girder = 9 m 

Crane capacity = 250 kN 

Self-weight of trolley, hook, electric motor etc. = 40 kN 

Self-weight of crane girder excluding trolley = 100 kN 

Minimum hook approach = 1.0 m 

Distance between wheels = 3 m 

Self-weight of rails = 0.2 kN/m

 

 

0	SPAN OF GANTRY GIRDER				7M
	SPAN OF CRANE GIRDER				9M
	CRANE CAPACITY				250KN
	SELF WEIGHT of trolley ,hook,motor				40kn
	self weight of crane girder exciluding trolley				100KN
	Minimum hook approch				1m
	distance between wheels				3m
	self weight of rails				0.2 kn/m
	Maximum moment due to vertical load
	weight of trolley + weight of load lifted [CRANE CAPCITY]				=[250+40]	290 kn
	self weight of crane girder exciluding trolley   100KN				ly	100 kn
	for maximum reaction on girder ,the moving load should be
	as close to the gantry as possible
	Reaction  at  A		RA		For maximum reaction on gantry girder, the moving load should be as close the gantry girder as possible. Ra=$\frac{290X8}{9}+\frac{100}{2}=307.77KN$	=307.77 KN
	LOAD ON GANTRY GIRDER FROM EACH WHEEL				307.77/2	=158.88 KN
	FACTORED WHEEL LOAD				1.5 X158.88	=230.83KN
	-MAXIMUM MOMENT = Wheel are equl distance from center of grder                   Me                                    -moment due to impact =[0.25x333.90]=83.47 kn-m				Me  RD= 230.83 x (3-1.5-0.75) + 230.83 x (3+0.75)    = 148.40 KN                                 7 Max. moment, Me = 148.40 x 2.25 = 333.90 KN-m	=333.90kn
	let self weight of girder					= 2kn/m
	dead load [ self wt. + rails ]					2.2 kn/m
	FACTORED dead  LOAD				2.2 x 1.5	3.3 kn/m
	M due to dead L.					20.21 kn/m
	M due to impact load				[0.25x333.90]=83.47 kn-m	83.47 kn/m
	factored mom. Due to all  vertical loads  [Mz= 333.90 +83.47+20.21]				M=	482.17 kn/m
STEP- 3	MAXIMUM MOM. DUE TO LATERAL LOADS
	Horizontal force transfered to rails                                        kn				[10/100]x[250+40]	29kn
	horizontal force distributed over 4 wheels on each wheels    loads				29/4	7.25 kn
	factored horizontal force				1.5 x 7.25	10.875kn
maximum mom.	MAX .MOM .  IN gantry  girder the position of loads is same as earlier expect that it is horizontal so by proportioning we get				[10.875/230.83]x333.90	15.73 kn -m
SHEAR FORCE	vertcal shear due to wheel loads				230.83+[230.83 X3]/7     For maximum shear force on the girder, the trailing wheel should be just on the girder as shown below.	329.75 kn
	impact					82.13
	self weight					9.9 kn
	total shear					420.5kn
	total shear due to surge					17.28kn
step 4    section determination
	minimum economic depth				D=7000/12	L/12
						583.33 mm
	WIDTH OF COMPRESSION FLANGE				L/40 TO L/30=7000/30 TO  7000/40
					175     to	233
	REQUIRED PLASTIC MODULUS			Zp	1.4   X  M/Fy
					1.4 X437.58X10^6/250	=2450X10^3 MM3
	let us try   ISMB 550 WITH ISMB 250 ON COMPRESSION FLANGE
STEP -5	PROPERTIES OF SELECTED SECTION TAKE AS PER CODE