My First Challenge through My First Project

Note : This is my first challenge and being answered through my First Project BECAUSE by mistake I clicked submitt button after writing answer to only one question i.e. Q-5 then I came to know that my opportunity ihas come to an end for first challenge . I contacted the support team and apprised about the facts and requested to resolve my problem as I wanted to take up the challenge in the right sprit. I am very greatfull that the technical support team was kind enough to listened my problem and advised me to answwer ypour first challenge through my first Project and publish it in Public domain. In this way your First challenge shall be evaluated considering my mistake .

Mean while I found that my challenge has bee evaluated in which I secured 20/150. I also found that answers to the q-3 was also provide by your team whch I did not attempt as answering this question had no sense. It is therefore very kindly requested to reevaluate my first challege leaving the marks of q-3 so that I can assess myself as well as clear picture may be reflected before every one who are associated.

I was finding difficulty in copying the images thats why Answers do not carry Fig./sketches or images although these might have facilitated while checking.

Q-5 : Expansion Joint size determination

Solution:

given : Original length of steel Bridge at -20 deg centigrde = 2000 m

Temp. coefficient of steel `alpha`= 12x 10^-6 /oC

L_T = L_0 + alpha `times` delta T

2000+12 x10 ^-6 `times` 60

Elongated length = 20001.44 m----- Ans

length difference = 1.44 m -----Ans

Q-1:

Aim: Which out of the two through truss system shown is most efficient in carrying gravity load ?

Solution : Truss A is the most efficient in arrying gravity load because in this truss system diagnol members are in tension for gravity load and most of the vertical members are in Compression. Such type of truss is called Pratt Truss

Q-4

Target: Load calculatipon problem

Given: Floor Framing diagram indicating various beams and Girders

Floor Load = 10 KN/m²

Load on Interior Beam, Interior Girder and Exterior Girder to be calculated

Solution-

(i) Load on interior Beam-

L_y/l_{x =} 8/2 =4 > 2, One way slab

Load on marked interior beam will come from half the slab area from two adjacent slab = wl_x/2+wl_x/2 + Self wt of beam

= 10x2/2+10x2/2 + Self wt of beam

= 20 KN + Self wt of beam- Ans

(ii) Load on interior girder-

L_y/l_{x =} 8/6 =.33 < 2, Two way slab

Load on marked interior Girder will come from two triangular portion of djacent slab = wl_x/3+wl_x/3 + Self wt of beam

= 10x6/3+10x6/3 + Self wt of beam

= 40 KN+ Self wt of beam—Ans

(iii) Load on Exterior girder-

L_y/l_{x =} 8/6 =.33 < 2, Two way slab

Load on marked Exterior Girder will come from one triangular portion slab

= wl_x/3 Self wt of beam

= 10x6/3 + Self wt of beam

= 20 KN+ Self wt of beam—Ans

Q-3

Target- Determine MOI,

(i) Channel section

I_XX = I_XXI + I_XX2 + I_XX3

I_{XXI =} I_G1 + Ah² ( Using Parallel axis theorem)

= bd³/12 + Ah²

= 80X10³/12 + 80X10 x 95²

( h= Distance between the CG of Channel section & top flange=95 mm)

= 6666.67 +7220000= 7226666.67 mm⁴ = 722.67 cm⁴

Similarly

I_{XX3 =} I_G3 + Ah²

Same as I_XXI due to symmetry

Therefore I_XX2 = 722.67 cm⁴

I_{XX2 =} I_G2 + Ah² ( Here h=0)

Therefore I_{XX2 =} I_G2

= bd³/12 = 8x180³/12 = 3888000mm⁴ = 388.8 cm⁴

I_XX = I_XXI + I_XX2 + I_XX3

= 722.67 + 722.67 + 388.8

= 1834.14 cm⁴ ---Ans

(ii) Plate Girder

I_XX = I_XXI + I_XX2 + I_XX3

I_{XXI =} I_G1 + Ah² ( Using Parallel axis theorem)

= bd³/12 + Ah²

= 300X30³/12 + 30 x 300 x 515²

( h= distance between CG of I beam and top flange = 515 mm)

= 675000 +2387025000 = 2387700000 mm^{4 =} 238770 cm⁴

Similarly

I_{XX3 =} I_G3 + Ah²

Same as I_XXI due to symmetry

Therefore I_XX2 = 238770 cm⁴

I_{XX2 =} I_G2 + Ah² ( Here h=0)

Therefore I_{XX2 =} I_G2

= bd³/12 = 8x1000³/12 = 666666666.67 mm⁴ = 66666.67 cm⁴

I_XX = I_XXI + I_XX2 + I_XX3

= 238770 + 238770 + 66666.67

= 544206.67 cm⁴ ---Ans

(iii) MI of ISMB 300 with top and bottom plate 12x160 mm about XX (major axix)

Dimensions of ISMB 300 as per standard

Depth= 300 mm, Width = 140 mm, Web Thickness = 7.7 mm,

Flange thickness= 13.1 mm

I_XX = I_XXI + I_XX2 + I_XX3

Top flange : I_{XXI =} I_G1 + Ah² ( Using Parallel axis theorem)

= bd³/12 + Ah²

= 140x13.1³/12 + 140x13.1x (150-13.1/2)²

= 26227.13 +37739873.19 = 37766100.92 mm³

MI of 12 mmx160mm top plate along with flange about XX-

= MI of top flange + MI of plate

= 37766100.92 + bd³/12+ Ah²

= 37766100.92 + 160x12³/12 +12x160x (150+6)²

= 37766100.92 + 23040 +46725120

= 84514260.92 mm⁴ = 8451.43 cm⁴

Similarly MI of bottom flange and plate due to similarity

I_{XX3 =} 8451.43 cm⁴

MI of Web portion about XX₂ : I_XX2 = Idb³/12 = 7.1x (300-13.1-13.1)³/12

( Here h=0)

= 12144438.34 mm⁴ = 1214.44 cm⁴

Total MI of Reinforced section about XX = 8451.43 cm⁴+8451.43+1214.44

= 18117.3 cm⁴---- Ans

(iv) MI of ISMB500 with ISMC 300 cap channel

Properties of standard rolled section ISMB 500:

Depth=500 mm, b_f=180 mm, t_w= 10.2mm, t_f =17.2 mm,

A=111 cm² I_x= 45200 cm⁴, I_y = 1370 cm²,

Properties of standard rolled section ISMC 300

Depth=300 mm, b_f =90 mm, t_w= 7.8mm, t_f =13.6mm,

A=46.3 cm² I_x=6420 cm⁴, I_y = 313 cm⁴ C_y= 2.35 mm

Total MI: I_XX = I_XXI + I_XX2 + I_XX3

CG of the composite section-

Y = (A₁Y₁+A₂Y₂)/( A₁+A₂)

Y₁= CG of ISMB 500 = 25 cm with respect to the reference point

(0, 0) which is the left most and bottom most point of

Composite section

Y₂ = 50 cm- C_{yy =} 50 cm - CG of ISMC 300 which is 2.35 mm

Y₂ = 49.765 cm

= (111x 25+ 46.3x 49.765)/ (111+46.3)

= 2775 + 2304.1195/ 157.3

Y = 5079.1195/157.3 = 32.289 cm

MI of Standard sections(ISMB 500)-

I_xx(1)= I_G (or I_XX) + Ah² = 45200 +111 (32.289-25)² ( Parallel axis theorem)

= 45200 +5897.38 = 51097.38 cm⁴

I_xx(2) = I_G or (I_YY) + Ah²

= 313 + 46.3 (49.765-32.289)²

= 313 + 14140.51 = 14453.51 cm⁴

Total MI of Composite section (I_xx) = I_XX1 + I_xx2

=51097.38 +14453.51 cm⁴

65550.89 cm⁴---Ans