Week 1 Challenge

QUESTION 1:

AIM: To find out in the given truss types, which is the most efficient in carrying gravity load.

INTRODUCTION: Trusses are designed to withstand gravity and uplift load, by transfering loads in the form of tension and compression to the top and bottom chords with vertical and diagonal bracings.

ANSWER: Pratt and Howe truss are the types given above. When comparing, Truss A is more efficient to carry gravity loads than Truss B.

REASON:

- In truss A, compression loads will be taken by top chord while tension taken by both diagonal and bottom chords, which makes it efficient to carry gravity loads.

- In truss B, tension will be occur on vertical and bottom chord, while compression will be more on top and diagonals. This is more suitable for uplift loads.

QUESTION 3:

AIM: To find MOI, C/S Area and Weight for given sections.

SOLUTION:

d) Channel Section:

A1 = 800 mm2	x1 = 40 mm	y1 = 5 mm
A2 =1440 mm2	x2 = 4 mm	y2 = 100 mm
A3 = 800 mm2	x3 = 40 mm	y3 = 195 mm

X̅ = = 22.95mm

Y̅ = = 100mm

I_xx = I_xx1 + I_xx2 + I_xx3;

I_xx1 = +A₁(y₁-y̅)² = + [800*(5-100)²] = 72.27x10⁹mm⁴

I_xx2 = 38.88x10⁵ mm⁴ I_xx3 = 72.27x10⁹ mm⁴

Ixx = 14.45x1010 mm4

I_yy = I_yy1 + I_yy2 + I_yy3;

I_yy1 = +A₁(x₁-x̅)² = 4.27x10¹⁰ mm⁴

I_yy2 = 52.48x10⁵ mm⁴; I_yy3 = 4.27x10¹⁰ mm⁴

Iyy = 8.54x1010 mm4

RESULT: I_xx = 14.45x10¹⁰ mm⁴; I_yy = 8.54x10¹⁰ mm⁴

C/S Area = 3040 mm²; Weight = 0.216kg/m

a) Plate Girder:

A1 = 9000 mm2	x1 = 150 mm	y1 = 15 mm
A2 = 8000 mm2	x2 = 150 mm	y2 = 530 mm
A3 = 9000 mm2	x3 = 150 mm	y3 = 1045 mm

X̅ = = 150mm

Y̅ = = 530mm

I_xx = I_xx1 + I_xx2 + I_xx3;

I_xx1 = +A₁(y₁-y̅)² = + [9000*(15-530)²] = 2.49x10⁹mm⁴

I_xx2 = 66.67x10¹² mm⁴ I_xx3 = 2.49x10⁹ mm⁴

Ixx = 6.67x1013 mm4

I_yy = I_yy1 + I_yy2 + I_yy3;

I_yy1 = +A₁(0) = 6.75x10⁷ mm⁴

I_yy2 = 4266.67x10⁷ mm⁴ I_yy3 = 6.75x10⁷ mm⁴

Iyy = 4280.17x107 mm4

RESULT: I_xx = 6.67x10¹³ mm⁴; I_yy = 4280.17x10⁷ mm⁴

C/S Area = 26000 mm²; Weight = 4.75kg/m

QUESTION 4:

AIM: To find load on, a) Interior beam b) Exterior beam c) Interior girder d) Exterior girder

GIVEN: Floor load = 10kN/m²

SOLUTION:

a) Interior beam = 2x8x10 = 160kN

= 160/8 = 20kN/m

b) Exterior beam = 8x1x10 = 80kN

= 80/8 = 10kN/m

c) Interior girder = 8x6x10 = 480kN

= 480/6 = 80kN/m

d) Exterior girder = 4x6x10 = 240kN

= 240/6 = 40kN/m

RESULT: Interior beam = 20kN/m; Exterior beam = 10kN/m

Interior girder = 80kN/m; Exterior girder = 40kN/m

QUESTION 5:

AIM: To find the change in length due to temperature differences.

GIVEN: Initial temperature = -20°C

Initial length = 2000m

Final temperature = 40°C

SOLUTION:

Change in temperature, ΔT = 40 – (-20) = 60°C

Change in length, ΔL = 60 x 12 x 10^-6 x 2000 = 1.44m

Total length, L = 2000 + 1.44 = 2001.44m

RESULT: ΔL = 1.44m; L = 2001.44m

QUESTION 2: