Structural Analysis and Design Solutions for Gantry and Beams

This article provides solutions for structural analysis and design of a gantry and beams. It covers topics such as truss bracing, beam size selection, and reaction calculations. The solutions are obtained using Staad Pro software. The article also mentions the properties of steel used in the gantry preparation. Desklib offers solved assignments, essays, and dissertations for various courses and universities.

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ASSIGNMENT SOLUTION
Question 1
a) As per the requirement , a structure is formed to support billboard hoading.
The length of truss is 24 mtr and height is 2.5 mtr .As the structure is too long
and heavy load need to carry by this structure so the truss is provided with
bracing at 3 mtr to support the load of 495 kN . This Zig zag bracing is
considered to be the best for transferring the load.

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b)
As in above figure , the fixed support at the end and the 3 loads of 165 kN each
is placed at equidistant from each other at 6 meter apart. Apart from the load
the bracing also shown to transfer the heavy loads at the end .

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c)
As all the forces are added in previous figure and the reactions at the support
end is shown above.
From above figure the result is obtain .
The reaction at left side 508.499 kN in x (Horizontal) direction .
The reaction at left side 264.147 kN in y (Vertical) direction .
The reaction at Right side -508.499 kN in x (Horizontal) direction .
The reaction at Right side 264.147 kN in y (Vertical) direction .

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d)
The above figure shows the deflection in the overall structure , it is in the safe
zone . the material used of manufacturing this structure is Steel
Major properties of steel is shown below
Elastic Modulus 2.1e+011 N/m^2
Poisson's Ratio 0.28 NA
Shear Modulus 7.9e+010 N/m^2
Mass Density 7800 kg/m^3
Tensile Strength 399826000 N/m^2
Yield Strength 220594000 N/m^2
Thermal Conductivity 43 W/(m·K)
Specific Heat 440 J/(kg·K)

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The total amount of steel used for the gantry preparation is shown above.
Question 2
a)
As per the provided information , the floor plan of 42 mtr x 30 mtr is created ,
in which the beams of 7 mtr centre to centre distance is maintained
horizontally and 6 mtr centre to centre distance is maintained vertically. There
are six beams added horizontally and five beams added vertically Total
numbers of beams are 30 .

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b)
As per the provided details of concrete the size select for all beams are 1 meter
by 1 meter .As the density of concrete as 25 kN/m3 , fck as 30 N/mm2 , fyd as
500 N/mm2 and the cover as 30 mm and 40 mm respectively.
From the left side of 30 mtr ,
Dead load = Density x length x thickness
= 25 x 42 x 40/1000
= 42 kN / m
Live load = 30 x30
= 9 kN / m
Total load = Dead + live = 42 + 9 =51 kN/ m
For the UDL CASE Maximum bending moment at centre
BM = Wl^2 / 8

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= 51 x 30 x 30 /8
= 5737.5 kN / m
AS we are using around 6 member on one side to support the load,
So size of concrete = 5737.5 /6 = 956.25
So for safe size member size will be 1000 x1000. This selection also shown on
the figure provided above.
Question 3
As the data given in question provided below.
Span = 7+5 =12 mtr
Height left side =7 mtr
Height right side = 5 mtr
Supports :-Hinge on both sides
Loading
At a = 13 kN in vertical downward direction
a b

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At b = 15 kN at angle 50 Deg
Load at b is converted to 2 components 15x cos 50 in vertical downward
direction and 15 x sin 50 in horizontal direction.
Applied all the condition in Staad pro and get the reactions at two supports.
As shown in above figure , all the conditions are applied like loading condition ,
dimension and support. The load at an angle 50 is divided into 2 components
like vertical and horizontal component.

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From above figure the result is obtain .It is statically indetermined so added
one more support to carry out the reaction.
The reaction at roller support is 0 kN in x (Horizontal) direction .
The reaction at roller support is 14.102 kN in y (Vertical) direction .
The reaction at hinged support 1 is 9.699 kN in x (Horizontal) direction .
The reaction at hinged support 1 is 12.411 kN in y (Vertical) direction .
The reaction at hinged support 2 -0.59 kN in x (Horizontal) direction .
The reaction at hinged support 2 1.048 kN in y (Vertical) direction .

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Question 4
As the data given in question provided below.
Span = 3+1+1.25+1.25+7.5 =14 mtr
Supports :-Hinge on B and C side
Loading
Between A & B = 11 kN in vertical downward direction
At B = 27 kN /mtr uniformly distributed load on 2.5 mtr length.
Applied all the condition in Staad pro and get the reactions at two supports.

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As shown in above figure , all the conditions are applied like loading condition ,
dimension and support. The first load is 11 kN point load and Second load UDL
of 27 kN /mtr for 2.5 meter length in like vertical direction.

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From above figure the result is obtain .
The reaction at hinged support is 0 N in x (Horizontal) direction .
The reaction at hinged support is 83.428 kN in y (Vertical) direction .
The reaction at roller support is 0 N in x (Horizontal) direction .
The reaction at roller support is -2.304 kN in y (Vertical) direction .