Week 1 Challenge

Question 1:-

Truss type reasoning

Out of the two through type truss systems shown below, which is the most efficient in carrying gravity load? Reason it out.

Answer:-

Truss A:

• Truss A is also said as Pratt Truss.
• Pratt trusses are normally utilized for span 10 m to 30 m.
• In Pratt truss, diagonal members are in tension for gravity loads.
• This truss is utilized where gravity loads are prevalent.

Truss B:

• Truss B is also said as Howe Truss.
• Howe truss is commonly used in light roofing so that the longer diagonal members are in Compression.
• It experiences tension under reversal of stresses due to wind load.
• It is just a converse of Pratt truss. They can be used for 10 m to 30 m span.

Conclusion:

• Hence, Truss A is the most efficient in carrying gravity loads.

Question 2:-

Deflection shape & Bending Moment Drawing (Qualitative approach)

For the beam & Frame below draw deflected shape and bending moment drawing. (Qualitative approach). E & I are constants.

Answer:-

Question 3:-

Determine MOI, Weight, C/S area of sections below

Answer:-

Moment of Inertia along X-Axis = Sum of I_xx of all members in a section.

Moment of Inertia along Y-Axis = Sum of I_yy of all members in a section.

I_xx of all member = bd³ /12 x A(d')²

where,

b is width of member

d is Depth of member

A is Area of member

d' is │Y - y│

y is centroidal axis in y direction of each member

Y is centroidal axis in y direction of Whole member

I_yy of all member = db³ /12 x A(d')²

where,

b is width of member

d is Depth of member

A is Area of member

d' is │X - x│

x is centroidal axis in x direction of each member

X is centroidal axis in x direction of Whole member

Weight of Section = Cross Section x Unit Weight of Steel

Where,

Unit Weight of Steel = 7850 kg/m³

For A. Channel Section,

MOI Along X-Axis = 18.34 x 10⁶ mm⁴

MOI Along Y-Axis = 1.843 x 10⁶ mm⁴

Area of Cross Section = 3040 mm²

Weight of Section per meter = 23.864 kg/m

For a. Plate Girder Section,

MOI Along X-Axis = 830.46 x 10⁶ mm⁴

MOI Along Y-Axis = 135.04 x 10⁶ mm⁴

Area of Cross Section = 26000 mm²

Weight of Section per meter = 204.100 kg/m

For b. Reinforced Section,

MOI Along X-Axis = 210.57 x 10⁶ mm⁴

MOI Along Y-Axis = 13.87 x 10⁶ mm⁴

Area of Cross Section = 9562 mm²

Weight of Section per meter = 75.062 kg/m

For c. ISMB 500 with ISMC 300 Section,

MOI Along X-Axis = 335.44 x 10⁶ mm⁴

MOI Along Y-Axis = 6.22 x 10⁶ mm⁴

Area of Cross Section = 15638 mm²

Weight of Section per meter = 122.758 kg/m

Question 4:-

Load calculation problem

For the floor framing below, Calculate the load on interior beam, Exterior beam, Interior Girder and exterior girder having Floor Load of 10kN/m²

^{Given Data:}

Floor Load - 10kN/m²

Tributary Width of Interior Beam = 1/2 x (2+2) = 2 m.

Tributary Width of Exterior Beam = 1/2 x (0+2) = 1 m.

Tributary Width of Interior Girder = 1/2 x (8+8) = 8 m.

Tributary Width of Exterior Girder = 1/2 x (0+8) = 4 m.

Answer:-

Load on beam = Tributary width x floor load

Load on Interior Beam = 2 x 10 = 20 kN/m.

Load on Exterior Eeam = 1 x 10 = 10 kN/m.

Load on Interior Girder = 8 x 10 = 80 kN/m.

Load on Exterior Girder = 4 x 10 = 40 kN/m

Question 5:-

Expansion joint size determination for temperature stated problem

Assuming a bridge, made of steel, with an initial total length of 2000 m at -20 ºC, calculate the length difference at 40 ºC and its total length.

Coefficient of Thermal expansion of steel = 12x10^-6/°C

Step 1.

Write down the known parameters of the problem:

L₀ = 2000 m

T₀ = -20 ºC

T_α = 12x10^-6/°C

Step 2.

Calculate the temperature difference

ΔT = T–T₀ = 40–(−20)=40+20=60 ºC

Step 3. .

Calculate the length difference

ΔL = α x L₀ x ΔT

=12x10^-6 x 2000 x 60 = 1.44 m

Step 4.

Calculate the total final length

L = L₀ + ΔL

= 2000 + 1.44

L = 2001.44 m