Week 6 Challenge

Structural Systems of Industrial structures -

• Lateral force resisting system (LFRS) - LFRS resist lateral force in both orthogonal axes.
• Mixed LFRS (Moment frames and shear walls) resist force in taller buildings.
• In low or mid size building, LFRS resists force in one direction.
• LFRS is also function of plane rigidity of floor and roof diaphragms.
• The diaphrame action may be rigid or flexible.
• Vertical force resisting system - It is based on gravity load path.
• The load must be transferred from the source to the ground.
• Topdown approach is simple way to understand vertical loads are resisted by priliminary structure frame.
• Flooring system - Floor carries vertical dead load, live load and transmit load through beam to column.
• Floor acts as a horizontal diaphragm to resist lateral force in stabilizing the walls and column, transfers lateral load to rigid frame, shear walls or braced bays.
• Question no 1.
• Ans -
• Aim - Chequered Plate design.
• Introduction - Chequered plate are normally used as flooring system. It is used for application similar to grating.
• It is used in material storage, equipment monitoring and material handling floors. Roarks chart is used to design ans arrive at suitable thickness based on service loading.
• Given -
• uniformly distributed Factored live load = 15 kN/m^2.
• Figure attached below.
• The plate is spanned with dimension = 1200 mm x 5000 mm.
• Density of steel = 78.5 KN/m^3.
• All edges fixed.
• By reffering of Roarks chart table 11.4.
• Assume thickness of plate as 1 mm.
• Take E = 2 x 10^5 N/mm^2.
• Self weight of plate = 0.001 x 78.5 = 0.0785 KN/m^2.
• Total load = q = 0.0785 + 15 = 15.07 KN/m^2.
• At center -
• σ = β2 x q x b^2/t^2.
• From roarks chart table 11.4.
• a/b = 5000/1200 = 4.166.
• Take β2 = 0.2500, β1 = 0.5000, α = 0.0284.
• σ = 0.2500 x 15.07 x 10^3 x 1.2^2/0.001^2.
• σ = 5425.2 N/mm^2.
• At the center of long edge -
• σmax = -β1 x q x b^2/t^2.
• σmax = -0.5000 x 15.07 x 10^3 x 1.2^2/0.001^2.
• σmax = -10850.4 N/mm^2.
• ymax = α x q x b^4/E x t^3.
• ymax = 0.0284 x 15.07 x 10^3 x 1.2^2/2 x 10^5 x 0.001^3.
• ymax = 3.08 m. = 3080 mm.
• As per above we get.
• σ = 5425.2 N/mm^2.
• σmax = -10850.4 N/mm^2.
• ymax = 3.08 m. = 3080 mm.
• Hence ok.
• Size of chequered plate as 1200 x 5000 x 1 mm.
• Question no 2.
• Ans -
• Aim - Design of deck slab, Design of thickness of RC slab, optimum spacing of the secondary beam.
• Introduction - A slab deck is one where the deck is analyzed as a plate. A beam and slab deck is one where the beams may deflect somewhat independently, and any transverse forces are carried in the deck. A cellular deck is one where a number of thin slabs and webs will enclose cells within the deck.
• In normal decking ribbed sheet is consiedred only for shuttering purpose without taking composite behaviour.
• Traditional steel concrete composite floor consist of structural steel and cast in situ floors connected using shear connectors. They behave monolithically. Now profile sheeting with embossment to interlock with concrete slab is used.
• Given -
• Dead load of 5 kN/m^2 & Live Load of 8 kN/m^2 for a mezzanine floor in an Industrial structure with plan as shown below.
• Provide optimum spacing of the secondary beam and also choose the decking sheet profile to be used. Neglect the composite action of decking sheet and RC slab. Decking sheet is provided only for formwork purpose.
• Deck slab dimension - 32 x 20 m.
• For Lx = 32 m, Ly = 20 m.
• Ratio Ly / Lx = 0.625.
• From above 0.625 < 2.0 from IS 456 2000.
• It is two way framing slab.
• We take M20, fy = 415 N/mm^2, fs = 240 N/mm^2.
• Assume 10 m width of slab.
• From IS 456 modification curve.
• M.F. = 1.6.
• d = span / L/d x M.F.
• d = 32000 / 20 x 1.6.
• d = 1000 mm.
• Assume 8 mm diameter of bar and nominal 20 mm cover as per IS 456.
• Overall depth D = 1030 mm.
• Total load Wu = 1.5 (DL x 25 + LL).
• Wu = 1.5 (5 x 25 + 8).
• Wu = 200 KN.
• Effective span along longer span = Lx = 32000 + 1000 = 33000 mm.
• Effective span along shorter span = Ly = 20000 + 1000 = 21000 mm.
• β = Ly/Lx = 21000/33000 = 0.636.
• From IS 456 table 26, For simply supoort edge.
• αx = 0.062 ,αy = 0.062.
• Calculate moment.
• Mux = αx x Wu x Lx^2.
• Mux = 0.062 x 200 x 33^2.
• Mux = 13503.6 KNm.
• Muy = αy x Wu x Ly^2.
• Muy = 0.062 x 200 x 21^2.
• Muy = 5468.4 KNm.
• Take maximum moment
• Mux = 13503.6 KNm.
• Mumax = 0.138 x fck x b x d^2.
• 13503.6 x 10^6 = 0.138 x 20 x 10000 x d^2.
• d = 699.47 mm = 700 mm not greater than 1000 mm.
• Hence ok.
• Thickness of deck slab = 700 mm.
• Optimum specing (S) of secondary beam.
• Assume width of secondary beam as 250 mm
• S = 20 x 0.25.
• S = 5.0 m
• S = 5.0 m.
• Optimum specing (S) of secondary beam = 5.0 m.
• Question no 3.
• Ans -
• Aim - Draw the load path and identify the axial force type in different type of bracing system, also find type of bracing system.
• Introduction -
• Lateral load resisting system (LLRS) - LLRS resist lateral load in both orthogonal axes. Mixed LLRS (moment frames and shear walls) resist lateral load in taller buildings. In low or mid size building LLRS resist load in one direction. Depending on the architectural conditions LLRS may be located internally or external face of building.
• Gravity load resisting system (GLRS) - The load must be transferred from source to ground. Topdown approach is simple way to understand vertical loads are resisted by priliminary structure frame (rafter column frame). Connections are critical and vulnerable for force transfer.
• Bracing system - A braced frame is a really strong structural system commonly used in strctures subject to lateral load such as wind and seismic pressure. The members in a braced frame are generally made of structural steel, which can work effectively both in tension and compression. Lateral load resisted through axial tension or compression forces in diagonal bracing members. Beam to column connections are simple connections.
• Load path - Simply the direction in which each consecutive load will pass through connected members. The sequence commences at the highest point of the structure working all the way down to the footing system, ultimately transferring the total load of the structure to the foundation.
• Given -
• Figures attached below of load path.
• First system.
• Axial force type - lateral force or horizontal force with pinned support.
• Bracing type - Diagonal bracing in that diagonal brace back.
• Second system.
• Axial force type - lateral force or horizontal force with pinned support.
• Bracing type - K type bracing.
• Third system.
• Axial force type - lateral force or horizontal force with pinned support.
• Bracing type - X type bracing.
• Fourth system.
• Axial force type - lateral force or horizontal force with pinned support.
• Bracing type - Eccentric type bracing.
• Firth system.
• Axial force type - gravity force or vertical force with pinned support.
• Bracing type - K type bracing.
• Sixth system.
• Axial force type - gravity force or vertical force with pinned support.
• Bracing type - Diagonal bracing in that diagonal brace back.
• Seventh system.
• Axial force type - gravity force or vertical force with pinned support.
• Bracing type - X type bracing.
• Eighth system.
• Axial force type - gravity force or vertical force with pinned support.
• Bracing type - Eccentric type bracing.