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 50 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

Aim & Introduction- Structural steel sections are to be clssified into Plastic ( Class-1) , Compact ( Class-2) , Semi compact ( Class-3) & Slender ( Class-4)

Given section- ISMB 500 -

Result :- Thus it is concluded that Flange is Plastic, Web is Plastic but the beam section ISMB 500 as a whole is Plastic

Given section ISMC 200

Result -Hence it is concluded that the given channel section is Plastic

Given section PG 1300 x 450 x 50 x 12

Result :Thus it is concluded that the section of the plate girder is compct

Double angle (Back to back) – 2 x ISA 75x75x

Builtup box – 1000 x 450 x 20 x 12

Result :Thus it is concluded that the built up box section is Plastic

Q-2. Effective length problems

Calculate effective length of following Top chord beams (double angle) having length L1, L2, Vertical member (single angle) having length L3 and inclined member (single angle) having length L4.

Aim & Introduction- Effective length of various members in given truss are to be found out.

Truss is given in which top chord, vertical member and inclined members have been with their length have been marked with effective length are to be determined

Procedure( Solution)

Top chords with length `L_1 & L_2` are in compression which is continous so Effective length will be as follows-

`L_1` (Effective) = 0.85 `L_1`

`L_2`(Effective) = 0.85 `L_2` ( 0.7 L to 1.0 L )

`L_3`(Effective) = 1.0 L3 ( Vertical member in tension)

`L_4`(Effective) = 0.85 `L_4` ( Discontinous compression member)

Result :- It is concluded that the Member having length`L_1` , `L_2` is under compression are of effctive length 0.85 `L_1`, and 0.85 `L_2` , effective length of `L_3` is 1.0`L_3`and that of inclined member with length `L_4` is 0.85`L_4`

Q-2 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.

Aim & Introduction: A truss is given with following data-

Span= 40m , Rise= 5.36 m , Angle of truss = 15 degree , Base of column are pinned joint

Procedure (Method) : Inclined Length of rafter AB in half truss = (40/2)/COS 15 =26.35 M

Let us assume that on half of the rafter is divided into 14 panels

Inclined Length of each panel = Inclined Length of PR/ No. of pannel= 26.35/14 = 1.882 m = 1822 mm

Effective length `L_(eff)` = 0.8 L= 0.8x 1822 = 1457.6 mm

Result : It is concluded that the Efeective length of the top chord of truss is 1457.6 mm

Q-3. Deflection of a beam problem

Q (i) Calculate deflection at the centre of the beam shown below. EI = 20 x 106 Nm2

Aim & Introduction- Deflection at the center of the beam is to be found out

Given- EI = 20x106N-m2">106N−m2106�-�2, Beam witt UDL of 6 N/m and two point loads each of 100 N symmetrically placed

Procedure- (Refer figure given below)

`R_1+R_2` =100+100+6x7 = 242 --------------(1`

Taking Moment about B `R_2`x7 = 100x5+100x2+42x3.5 = 847\

`R_2` =121 N , From eqn (1) `R_1` = 121 N

Considering Moment about X-X

`M_(XX)`- 121 (x) + 100(x-2) + 6(x)(x/2)+100(x-5) =0

`M_(XX)`= 121 (x) - 100(x-2) - 6(x)(x/2) - 100(x-5)

`EI (d^2y//dx^2)` = 121 (x) - 100(x-2) - 6(x)(x/2) - 100(x-5)

EI (dy/dx) = 121(`x^2/2`) +`C_1` - 100 `(x-2)^2/2` -6(`x^3/6`) -100 `(x-5)^2/2`

EIy = 121 (`x^3/6`) + `(C_1x)`+`C_2` - 100 `(x-2)^3/6` -6 (`x^4/24`) -100 `(x-5)^3/6`

0 = 6917.66 +7`C_1`- 2083.33 -600.25 -133.33

7`C_1` = 2083.33+600.25+133.33 - 6917.66

`C_1` = - 4100.42 /7 - 585.77

Applying boundary conditions- x=0, y=0 then `C_2`=0

x=7, y=0 this gives `C_1`= -585.77

EIy = `121 (x^3/6)`- 585.77(x)- 100 `(x-2)^3/6` -6`(x^4/24)`-100 `(x-5)^3/6`

For max. deflection x= 3.5 ( due symmetrical loading and SF =0 at x = 3.5 m

EIy = 864.64 - 2050.19 - 56.25 - 37.51 = - 1279.31

EIy = -1279.3

y = -1279.31/EI = -1279.31x1000/20x`10^6` = 0.0639 mm

Result - It is concluded that the deflection at the mid point of the beam shall be maximum due to the symmetry of loads which is = 0.0625 mm

A beam with EI = 25x `10^6 Nm^2` having UDL @ 1.5 KN/m and point load of 2 KN & 5 KN is given

`R_1+ R_2` = 2+5+ 1 .5x5 = 14.5 --------(1)

Taking moment about A

`R_2`x5 =5(2.5) + 2 (1.5) +7.5(2.5)

`R_2` = 34.25/5 = 6.85 KN

From equation (1) we get `R_1` = 14.5 - 6.85 = 7.65 KN

Taking moment about X-X

`M_x` - 7.65 (x) +2( x-1.5) + 1.5(x) (x/2) + 5(x-2.5) = 0

`M_x` = 7.65 (x) - 2( x- 1.5) -1.5 (x)( x/2) - 5( x- 2.5)

EI (`(d^2y)/ (dx^2)`= 7.65 (x)-2 ( x-1.5) - 1.5 (x)(x/2)-5 ( x- 2)

Integrating-

EI (dy/dx) = 7.65 `(x^2/2)` +`C_1`- 2`(x-1.5)^2/2`-1.5`(x^3/6)`-5`(x-2.5)^2/2` -------------------------(2)

Now differenciating equation (2)-

EI y = 7.65 `(x^3/6)`+ `C_1(x)`+`C_2`-2`(x-1.5)^3/6`- 1.5`(x^4/24)`- 5`(x-2.5)^3/6` ---------------(4)

Applying boundary conditions x=0 , y=0 in equation (4) which gives `C_2`=0

Now using x= 5 , y =0 in equation -4)

159.75 + 5`C_1`- 42.875- 39.0625 - 13.02

`5C_1` = 42.875 +39.0625+13.02- 159.75 = - 64.7925 `C_1` = - 64.7925/5 = - 12.96

Substituting the values of `C_1 & C_2` in equation (4), we get

EI y= 7.65 `(x^3/6)` -12.96 (x) -`(x-1.5)^3`-1.5`(x^4/24)`-5`(x-2.50)^3/6`

Deflection at the middle where 5KN force is acting,

substituting x= 2.5 m in equation (4)-

EI y= 19.92 -32.4 -1.0 -2.44 = - 15.92 KN `m^3`

EI y at x = 2.5 m = - 15.92`xx10^9` N `mm^3`

y at centre = - 15.92`xx10^9`/ 25`xx10^6xx10^3` = - 0.6368 mm

Result - It is concluded that deflection at the centre due to given loading arrangement of beam is 0.6368 mm ( -ve sign indicate that the deflection is below the X axis

Q-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 unrestrained and assumes bearing length of 50 mm

Tip steps:

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

Aim & Introduction- Determination of moment capacity of a simply supported beam.

• Hot rolled section for beam is to be choosen
• Beam is simply supported
• Length of Beam = 6m
• Spacing of Beams = 3m c/c
• Dead load on beam = 5 KN/m2">m2�2 ( including self weight)
• Live load on beam= 6 KN/m2">m2�2
• Concentrated load on beam = 100 KN
• Beam is laterally unsupported/unrestrained
• Bearing length = 50 mm
• Assume grade of steel Fe 410 , Partial safety factor = 1.1

Result- Thus it is conclued that the beam ISWB 600 @ 133.7 Kg/m, is capable to resist Bending moment of 548.7 KN-m safely

Q-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

Aim & Introduction- Design of channel purlin for roof truss

Span=20 m , spacing of truss= 4 m c/c , roof slope=20 degree , spacing of purlin = 2 m , wind suction = 1 KN/m2">m2�2 , LL= 0.75 KN/m2">m2�2 ,DL= 1.4KN/m2">m2�2

roof slop = 20 degree

Result :- It is concluded that for purlin of truss ISMC 200 section is safe in shear, moment , over all strength and deflection.