Week 4 Challenge
- Question no. 1.
- Ans -
- Aim - To calculate section classification according to beam shapes.
- Introduction - bending moment reqiured to make whole section reach yield stress is called plastic moment (Mp). if the section is thin walled it may fail before reaching yield stress by local buckling. section can be classified as plastic, compact, semicompact, slender.
- Plstic section - Cross section which develops plastic hinges and rotational capacity requred for failure of structure.
- Compact section - Cross section which develops plastic moment resistance but have inadequate plastic hinge rotation capacity because of local buckling.
- Semicompact section - Local buckling can prevent the development of plastic moment resistance with extreme compression fibre reaching yield strength.
- Slender section - Local buckling will occur in cross section even before attaining yield stress.
- Given - section classification of following beam shapes of ISMB 500, ISMC 200, PG 1300 x 450 x 50 x12.
- Also figures attached below.
- For ISMB 500 - it is I section as per steel table. as per welded connection consideration.
- Width of flange - b - 180 mm.
- Depth of section - D - 500 mm.
- Depth of web - d - 465.6
- Thickness of flange - tf - 17.2 mm.
- Thickness of web - tw - 10.2 mm.
- according to ratio - b / tf = 180 / 17.2 = 10.46.
- Also d / tw = 465.6 / 10.2 = 45.64.
- From above according to flange section classification - semicompact section.
- From web of I section classification - Plastic section.
- For ISMC 200 - it is C section as per steel table. as per welded connection consideration.
- Width of flange - b - 75 mm.
- Depth of section - D - 200 mm.
- Depth of web - d - 177.2 mm.
- Thickness of flange - tf - 11.4 mm.
- Thickness of web - tw - 6.1 mm.
- according to ratio - b / tf = 75 / 11.4 = 6.57.
- Also d / tw = 177.2 / 6.1 = 29.049.
- From above according to flange section classification - plastic section.
- From web of C section classification - Plastic section.
- For PG 1300 x 450 x 50 x 12 - it is I section as per steel table. as per welded connection given.
- Width of flange - b - 450 mm.
- Depth of section - D - 1300 mm.
- Depth of web - d - 1200 mm.
- Thickness of flange - tf - 50 mm.
- Thickness of web - tw - 12 mm.
- according to ratio - b / tf = 450 / 50 = 9.0.
- Also d / tw = 1200 / 12 = 100.
- From above according to flange section classification - compact section.
- From web of PG section classification - compact section.
- For Double angle (Back to back) - 2 x ISA 75 x 75 x 6. it is angle section as per steel table.
- Assume welded connection.
- Width of flange - b - 75 mm.
- Depth of section - D - 75 mm.
- Depth of web - d - 63 mm.
- Thickness of flange - t - 6 mm.
- Thickness of web - t - 6 mm.
- according to ratio - b / t = 75 / 6 = 12.5.
- Also d / tw = 69 / 6 = 11.5.
- From above according to flange section classification - semicompact section.
- From web of angle section classification - semicompact section.
- For RHS 150 x 100 x 10. it is angle section as per steel table.
- Assume welded connection.
- Width of flange - b - 100 mm.
- Depth of section - D - 150 mm.
- Depth of web - d - 140 mm.
- Thickness of flange - t - 10 mm.
- Thickness of web - t - 10 mm.
- according to ratio - b / t = 100 / 10 = 10.
- Also d / t = 140 / 10 = 14.
- From above according to flange section classification - compact section.
- From web of angle section classification - semicompact section.
- For Builtup box – 1000 x 450 x 20 x 12. it is two angle section box of same size.
- Assume welded connection.
- Width of flange - b - 426 mm.
- Depth of section - D - 1000 mm.
- Depth of web - d - 960 mm.
- Thickness of flange - tf - 20 mm.
- Thickness of web - tw - 12 mm.
- according to ratio - b / tf = 426 / 20 = 21.3.
- Also d / tw = 960 / 12 = 80.
- From above according to flange section classification - semicompact section.
- From web of angle section classification - semicompact section.
- Question no. 3.
- Ans -
- Aim - To calculate deflection at center of beam
- Introduction - Deflection is defined as the vertical displacement of a point on a loaded beam. There are many methods to find out the slope and deflection at a section in a loaded beam. Double Integration Method: This is most suitable when concentrated or udl over entire length is acting on the beam.
- Given -
- First case beam.
- Figure attached below.
- Udl - w - 6 N/m.
- Point load - P1 - P2 - 100 N.
- b1 - b2 - 2.0 m.
- Total length - L - 7.0 m.
- EI - 20 x 10^6 Nm^2.
- δ - P1 x b1 / 48 EI (3L^2 - 4b1^2) + P2 x b2 / 48 EI (3L^2 - 4b2^2) + 5 wL^4 / 384 EI.
- δ = 100 x 2 / 48 x 20 x 10^6 (3 x 7^2 - 4 x 2^2) + 100 x 2 / 48 x 20 x 10^6 (3 x 7^2 - 4 x 2^2)
- + 5 x 6 x 7^4 / 384 x 20 x 10^6.
- δ = 9.487 m.
- Deflection at center of beam = 9.487 m.
- Second case beam.
- Figure attached below.
- Udl - w - 1.5 KN/m.
- Point load - P1 - 2 KN.
- P2 - 5 KN.
- b1 - 1.5 m.
- b2 - 2.5 m.
- Total length - L - 5.0 m.
- EI - 25 x 10^6 Nm^2.
- δ - P1 x b1 / 48 EI (3L^2 - 4b1^2) + P2 x b2 / 48 EI (3L^2 - 4b2^2) + 5 wL^4 / 384 EI.
- δ = 2 x 1.5 x 10^3 / 48 x 25 x 10^6 (3 x 5^2 - 4 x 1.5^2) + 5 x 2.5 x 10^3 / 48 x 25 x 10^6 (3 x 5^2 - 4 x 2.5^2)
- + 5 x 1.5 x 5^4 / 384 x 25 x 10^6.
- δ = 0.165 + 0.520 + 0.000488 m.
- Deflection at center of beam = 0.685 m.
- Third case beam.
- Figure attached below.
- Aim - To find out EI flexure stiffness of beam.
- Introduction - Flexural stiffness is a standard for measuring deformability. It is based on two crucial properties: the elastic modulus of the material and the moment of inertia (function of geometry).
- Udl - w - 10 KN/m.
- Deflection - δ - 5.0 mm - 5 x 10^-3 m.
- Total length - L - 10.0 m.
- δ - 5 wL^4 / 384 EI.
- 5 x 10^-3 = 5 x 10 x 10^3 x 10^4 / 384 x EI.
- EI = 26.04 x 10^4 Nm^2.
- Flexure stiffness EI = 260.41 x10^6 Nm^2.
- Question no. 4.
- Ans -
- Aim - To calculate moment capacity of beam.
- Introduction - The moment capacity is maximum bending moment that can be resisted by an element before it fails in bending.
- Given -
- A standard hot rolled section for the beam.
- Simply supported of length L = 6 m.
- Spacing at 3 m c/c.
- Floor Dead load FDL = 5 KN / m^2 (including Self weight).
- Live load LL = 6 KN / m^2.
- Concentrated load W = 100 kN at mid span of beam.
- Assume fy = 250 N / mm^2, E = 2 x 10^5 N / mm^2.
- The beam is laterally unrestrained and assumes bearing length of 50 mm.
- Figure attached below.
- 1. find maximum bending moment -
- Total udl = FDL + LL = 5 x 6 + 6 x 6 = 96 KN/m.
- Max. M1 with udl = WL/8.
- M1 = 96 x 6 / 8 = 72 KNm.
- Max. M2 with point load M2 = WL/4.
- M2 = 100 x 6 / 4 = 150 KNm.
- Total Max. M = M1 + M2.
- M = 72 + 150 = 222 KNm.
- Max bending moment M = 222 KNm. = 222 x 10^6 Nmm.
- 2. find plastic modulus -
- Take σbc = σbt = 0.66fy = 0.66 x 250 = 165 N / mm^2.
- Z = 222 x 10^6 / 165 = 1345.45 x 10^3 mm^3.
- Reuired plastic modulus Z = 1345.45 x 10^3 mm^3.
- 3. select suitable section -
- From steel table try ISWB 500 @ 95.2 Kg / m having the following parameters.
- Zxx = 2091.6 cm^3 = 2091.6 x 10^3 mm^3.
- Ixx = 52209.9 x 10^4 mm^4, A = 12122 mm^2.
- h = 500 mm, tf = 14.7 mm, tw = 9.9 mm.
- 3. design shear force and shear strength -
- Shear force with udl = V1 = WL / 2 = 96 x 6 / 2 = 288 KN.
- Shear force with point load = V2 = W/2 = 100 / 2 = 50 KN.
- Total Shear force = 288 + 50 = 338 KN.
- Design shear force = 338 KN.
- Shear strength for avarage = τva = V / h x tw.
- τva = 338 x 10^3 / 500 x 9.9 = 68.28 N / mm^2.
- Shear strength for permissible = 0.4 x fy = 0.4 x 250 = 100 N / mm^2.
- Hence ok.
- 4. design capacity of section -
- Capacity = area x fy = 12122 x 250 = 3030.5 KN.
- Design capacity = 3030.5 KN.
- 5. check for deflection -
- Generally limiting deflection should not exceed 1/325 of the span.
- For udl δ - 5 wL^4 / 384 EI.
- δ = 5 x 96 x 6000^4 / 384 x 2 x 10^5 x 52209.9 x 10^4.
- δ = 15.51 mm.
- For point load δ - wL^3 / 48 EI.
- δ - 100 x 10^3 x 6000^3 / 48 x 2 x 10^5 x 52209.9 x 10^4.
- δ - 4.30 mm.
- Total deflection δ = 15.51 + 4.30 = 19.81 mm.
- Allowable deflection = δ = span / 325 = 18.46 mm.
- Hence safe.
- 6. check for web buckling -
- slenderness ratio λw = h1 / tw x √3 where distance between roots of fillet h1 = 431 mm.
- λw = 431 / 9.9 x √3 = 75.4.
- Hence from table 7.5 for fy = 250 and λ = 75.4 we have
- σac = 106.06.
- Hence allowable reaction Ra = σac x tw x B.
- Ra = 106.06 x 9.9 x B.
- B = b + 1/2 x h = 50 + 1/2 x 500 = 300 mm.
- Ra = 106.06 x 9.9 x 300 = 315 KN.
- Actual reaction R = w/2 udl + w/2 point load.
- R = 98 KN.
- Hence safe.
- Question no. 5.
- Ans -
- Aim - Design of purlin section.
- Introduction - The purlins which support roof covering, batten or common rafters, are supported on principal rafters of trusses. They may therefore be considered as continuous beams. The purlin may of rectangular, I section, channel section, angle section. Purlins are subjected to dead load including self weight, roof covering, battens etc, acting in vertical direction and wind load normal to slope of roof.
- Design channel purlin, considering sag rod at mid span for the roof truss having following data.
- Figure attached below.
- Span of truss - 20 m.
- Spacing of Truss - 4 m c/c.
- Roof slope - θ - 20 degrees.
- Spacing of purlin - 2 m.
- Wind suction - 1 KN/m^2.
- Live Load - 0.75 KN/m^2, Dead Load - 1.4 KN/m^2.
- sinθ = sin20 = 0.342, cosθ = cos20 = 0.939.
- 1. Calculation of dead load on purlin -
- Dead load DL = 1.4 x 2 = 2800 N/m.
- Live load LL = 0.75 x 2 = 1500 N/m.
- Total load Wd = 2800 + 1500 = 4300 N/m.
- 2. Calculation of wind pressure/m run -
- Wind pressure Ww = 1 KN/m^2.
- Ww = 1 x 10^3 x 2 = 2000 N/m.
- 3. Design of bending moment for channel section purlin.
- It is assumed that dead load + wind load combination would produce more effect on purlins than the combination dead load + live load combination.
- For channel (C) section purlin -
- Muu = Mxx = (Ww + Wdcosθ)L^2/10.
- Muu = Mxx = (2000 + 4300 x 0.939) x 4^2/10.
- Muu = Mxx = 9660.32 x 10^3 Nmm.
- Mvv = Myy = (Wdsinθ)L^2/10.
- Mvv = Myy = 4300 x0.342 x 4^2/10.
- Mvv = Myy = 2352.96 x 10^3 Nmm.
- 4. Selection of C - section.
- Taking Zxx/Zyy = 8.
- Taking σbt = 165 N/mm^2.
- Zxx = Mxx / σbt (1 + Zxx/Zyy x Myy/Mxx).
- Zxx = 9660.32 x 10^3 / 165 (1 + 8 x 2352.96 x 10^3/9660.32 x 10^3).
- Zxx = 172.63 x 10^3 mm^3.
- Hence provide ISLC 225 @ 24 Kg/m, having Zxx = 226.5 cm^3 and Zyy = 32.0 cm^3.
- Check for stresses fb = Mxx/Zxx + Myy/Zyy.
- fb = (9660.32 x 10^3/226.5 x 10^3) + (2352.96 x 10^3/32 x 10^3).
- fb = 116.18 N/mm^2 < 165 N/mm^2.
- Hence ok.
- Designed Channel purlin section -
- Use ISLC 225 @ 24 Kg/m.
- Question no. 2.
- Ans -
- Aim - To find out effective length.
- Introduction - Effective length is a critical concept in structural design which relates to the length of a component which is effectively restrained. When designing structural members they can be subject to a number of forces such as Shear, Tension, Compression, Torsion and combinations of these to name but a few
- First case.
- Calculate effective length Top chord beams (double angle) having length L1,L2.
- Vertical member (single angle) having length L3.
- Inclined member (single angle) having length L4.
- Figure attached below.
- Effective length for top chord = L1,L2 is top chord member with built up section continuous construction with restrained against torsion at end.
- Effective length = L1,L2 = 0.75 x L.
- For vertical member = L3 = 0.7 x L.
- For inclined member = L4 = 0.7 x L.
- Second case.
- Aim - To find effective length of rafter made up of built up I section about major x-x and minor y-y axis.
- Given -
- Span of truss - 40 m.
- Spacing of purlin - 1.2 m.
- Pinned base.
- Total height = H = 5 + 5.36 = 10.36 m.
- Roof slope = 15 degree.
- Length of rafter = L.
- sin15 = 5.36/L.
- L = 20.70 m.
- Length of rafter L = 20.70 m.
- Effective length Le = 0.7 x 20.70 = 14.49 m.
- Effective length about major axis x-x = Le = 14.49 x cos15 = 14.0 m.
- Effective length about minor axis y-y = Le = 14.49 x sin15 = 3.75 m.
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