Week 4 Challenge

Design of Beams

1. Section classification problem for listed beam shapes

Calculate section classification of following beam shapes

• ISMB 500
• ISMC 200
  • PG 1300 x 450 x 20 x 12

• Double angle (Back to back) – 2 x ISA 75x75x6
• RHS - 150 x 100 x10
• Builtup box – 1000 x 450 x 20 x 12

ANSS

PROPERTIES OF SECTION -

ISMB 500

h= 500 mm

bf= 180 mm

tf= 17.2mm

tw= 10.2 mm

FROM IS TABLE 2

FLANGE CRITRION = bf/tf = 180 / 17.2 = 10.46 >9 .4 `epsilon`

web critrion = d/ tw = h-2(tf -r) /tw

=500 - 2 (17.4-17)/ 10.2 = 48.94 < 84 `epsilon`

so scection is compact

ISMC 200

h=200

bf=75 mm

tf =11.4 mm

tw=6.1 mm

root radius = 11 mm

fFLANGE CRITRION = bf/tf = 75/11.4 =6.5 <9.4 `epsi`

web critrion = df/tw =h-2(tf+r) =200-2(11.4+11) /tw = 25< 42 `epsi` so s

setion is plastic

fange critrion = b/tf= 450/50 = 9 <9.4 `epsi`

web critrion = d/tw = 1300/12 = 108.33 < 126 `epsi`

so section is compact

• Double angle (Back to back) – 2 x ISA 75x75x6

tf= 6 mm

bf =150 mm

tw = 12mm

d = 75 +6 =81mm

flange critrion = bf/tf = 150/ 6 =25 >15 `epsilon`

web critrion = d/tw = 81/ 12 =6.75 > 15 `epsilon`

section is semi compact

• Builtup box – 1000 x 450 x 20 x 12

bf = 1000-40=960 mm

2. Effective length problems

Calculate effective length of rafter madeup of Builtup I section about major (x-x) and minor axis (y-y) having purlin spaced at 1.2 m.

ANS

let the rafter is BC

COS15 = ( 40/2) /BC

BC= 20/ COS 15 = 20.705 M

EFFECTIVE LENGTH ABOUT MINIOR AXIS = 1.2 M

ABOUT MAJOR AXIS = 0.8 *(20.705-2.4) =14.644 M

3. Deflection of a beam problem

• Calculate deflection at the centre of the beam shown below. EI = 20 x 10⁶ Nm²

tke moment about R2

R1 x 7 = 100*5+100*2 +6*7*3.5

R1 = 121 KN

R2 = 200+42- 121 =121 KN

TAKE SECTION X-X , at a distance x FROM R1 at last span

AND APPLY MACALUS PRINCIPLE,

EI d^2y/dx^2 = Mx

EI d^2y/dx^2 = 121 *x - 100*(x-2) -100*(x-5) - 6*x*x/2 +6*(7-x)*(7-x)/2 -------1

integrate this one

EI dy/dx = 121 *x^2/2 +c1 - 100*(x-2)^2/2 -100 *(x-5)^2/2 - 6*x^3/6 - 6* (7-x)3/6 --------2

integrate aain

EIY = 121*X^3/6 +C1 X + C2- 100*(X-2)^3/6 - 100* (X-5)^3/6 -X^4/4 +(7-X)^4/4 ------------------3

SOPE AT R1 = X=0 , Y = 0

SLOPE AT R2 , X = 7 ,Y =0

PUT VALUE IN 3

0= 0+0+C2 = 0

C2 =0

PUT X =7 ,Y = 0

0= 121*7*7*7/6 + 7C1 +0 -100*5*5*5/6 -100 *2*2*2 /6

CALCULATE THE VALUE OF CI AND C2 , PUT IN EQN , WE GET DELECTION

• Calculate deflection at the centre of the beam shown below. EI = 25 x 10⁶ Nm²

PUT THE VAUE OF X= 3M IN EQN E

WE WILL GET THE MAXIMUM DEFLECTION.AT THE CENTER.

• A simply supported beam is 10 m long with udl of 10 kN/m Calculate the Flexural stiffness (EI) of the beam which limits the deflection to 5 mm at the middle.

AS WE KNOW THE DEFLECTION OF UDL OF SIMPLY SUPPORTED BEAM IS `delta`= 5WL^4/384 *EI

5 mm = 5 *10 *10 /384 *EI

EI = 5*10*10*10*10*10*1000/384*0.005 = 26 X 10^7 N MM2

4. Moment capacity determination problem

Select a standard hot rolled section for the beam which is simply supported of length 6 m spaced at 3 m c/c, carrying a floor Dead load of 5 kN/m² (including Self weight) & Live load of 6 kN/m² and a concentrated load of 100 kN at mid span of beam. The beam is laterally restrained and assumes bearing length of 50 mm

Tip steps:

• Calculate maximum bending moment & required plastic modulus
• Select suitable trial section
• Calculate design shear force and design shear strength
• Check design capacity of the section
• Check for deflection, web buckling, bearing

FACTORED MOMENT Mo = 1.5* 447 = 670.5 KN M

ZP= M/ ( fY/Ym0)

= 670.5*1000 *1000 *1.1 / 250 =2950 cm3

FROM IS 800 TABLE H

CHOOSE SECTION

ISMB 600 , ZP = 3510.63 CM3

h = 600 mm

bf= 210mm

tf= 20.8 mm

tw= 12mm

r = 20mm

FLANGE CRITRIN0= (bf/2)/tf = (210/2)/ 20.8 =5.04 < 9.4 , section plastic

web critrion = d/tw = 600- 81.6 = 43 >84 ,section plastic

section is plastic

check for shear

Vd = fy*Av/ `sqrt3` *ym0

= 250* h*tw / `sqrt3` *ym0

= 250* 600* 12 / `sqrt3` *1.1 = 944 kn

0.6 Vd = 0.6*944 =566 kn > 243 kn ,so safe

design bendig strengt

Md = `beta zp fy`/ ymo

=section is plastic so `beta` =1

Md = 1* 3510.63 *250 /1.1 =797 kn m

Mo< Md so section is safe.

check for deflecton

maximum deflection = span /360 = 6000/360 = 16.67 mm

`del` = 5 WL 4/384 Ei + PL^3/48 EI

= 5*18 *6000^4/384 *2*10^5 *98193 + 100 *6000^3/ 48 *2*10^5 *98193

5. Purlin design problem

Design channel purlin, considering sag rod at mid span for the roof truss having following data

• Span of truss : 20 m
• Spacing of Truss : 4 m c/c
• Roof slope : 20 degrees
• Spacing of purlin : 2 m
• Wind suction : 1 kN/m²
• Live Load : 0.75 kN/m² ,Dead Load 1.4 kN/m²

DESIGN OF PURLIN ;

DEAD LOAD = 1.4*2= 2.8 KN/M

LIVE LOAD = 0.75*2= 1.5 KN/M

WND LOAD= 1*2 =2 KN/M

COMPONENT OF DEAD LOAD NORMAL TO THE SHEETIG= 2.8*COS 20 =2.63 KN/M

COMPONENT OF LIVE LOAD NORMAL TO THE SHEETING =1.5 *COS20= 1.40KN/M

WIND LOAD NORMAL TO THE CHANNEL = 4 KN/M

COMPONENT OF DEAD LOAD ARALLEL TO THE SHEETIG =2.8*SIN20 =0.957 KN/M

COMPONENT OF LIVE LOAD PARAELL TO THE SHEETING = 1.5 S1N 20= 0.513KN/M

WIND LOAD PARALLEL TO THE CHANNEL =0

COMBINATION IN Z DIRECTION OR NORMAL TO THE CHAANNEL

=1.5 DL +1..5LL

=1.5 *2.63+ 1.5*1.40 =6.045 KN/M

Y DIRECTION

1.5 DL+1.5 LL =1.5(0.9570+0.513) =2.205 KN/M

WE DESIGN FOR LOAD 6.045 KN/M

BENDING MOMENT MZ=WL2/8 =6.045*4*4/8 =12.09knm

BENDING MOMENT FOR My =wl2/8 =2.205*4*4/8=4.41 knm

SHEAR FORCE

Vz= WL/2 =6.045*4/2 =12.09 KN

Vy = WL/2 =2.205*4/2 = 4.41 KN

REQUIRED SECTION MODOLUS

Zp = Mz*ymo/ fy +2.5 (d/b) My *ymo/fy

= 12.09 *10^6 *1.1/250 +2.5(150/75) 4.41 *10^6 *1.1/250

=53196 +97020 =150216 = 150.2166*10^3 mm3

choose section ISMC 175

d =175mm

bf= 75mm

tf=10.2mm

tw= 5.7mm

flange critrion b/tf= 75/10.2 =7.32 < 9.4

web critrion d/tw= 175- (10.2+7) /tw= 27 <84

section is plastic

shearcapacity-

Vd=250*175*5.7/1.1*1.723 =130kn

0.6 Vd = 0.6*130 = 78kn>12.09 kn so safe

moment capacity =

Mdz= Zpz*fy/ymo =

=161.65*10^3 *250 /1.1 *10^6 == 36.37knm <1.2*ze*fy/ymo

,<1.2* 139.8 *250/1.1 = 61.0kn

hence safe