Week 3 Challenge
- Question no. 1.
- Ans -
- Aim - to find out effective length factor.
- Introduction - effective length factor is for memebers in frames the effective length factor ke depends on the ratios of the compression member stiffness to the end restraint stiffnesses.
- Also figure attached below.
- Formula to find effective length factor = Le = Ke x L.
- Given - height of column is 10 m.
- Condition - effectively restrained in mid height in one direction (in Z direction) but free to move in other direction (Y direction). Bottom of column is Fixed and top of column is Free.
- a. braced frame & shear wall. As per given in condition from IS 800 table.
- Here given that frame with shear wall connection.
- Effective length factor ke = 1.2.
- So Le = 1.2 L.
- b. unbraced frame with pin based. As per given in condition from IS 800 table.
- Here is given that frame with pinbased one ends & footing type is spread footing.
- Effective length factor ke = 2.0.
- So Le = 2.0 L.
- c. unbraced frame with fixed base. As per given in condition from IS 800 table.
- Here is given that frame with fixbased one ends & footing type is spread footing.
- Effective length factor ke = 1.2.
- So Le = 1.2 L.
- Question no. 2.
- Ans -
- Aim - to find out design axial strength determination with its end condition.
- Introduction - axial strenght means that load carrying capacity of a member at a safe compressive stress corrosponding to the slenderness ratio of column, it depends on factors such as radius of gyration, effective length and area of cross section.
- Also figure attached below of the section member.
- Formula to find axial strength = P = σac x A.
- Given - Hot rolled ISMB 300.
- Flange cover plate of 200x12 mm.
- Length of beam as 3 m.
- Assume the bottom of the column is fixed and top as Roller (Rotation Fixed, translation free).
- Yield strength of material fy= 250 N / mm^2.
- From IS 800 table Le = 1.2 L.
- Le = 1.2 x 3 = 3.6 m.
- For ISMB 300 from steel table.
- Weight = W = 81.8 kg / m.
- Cross sectional area = A = 10426 mm^2.
- Least redius of gyration ryy = 44.4 mm.
- λ max = Le / r min = 3.6 x 1000 / 44.4 = 81.08.
- From table 5.1 of IS 800 for λ = 81.08 and fy = 250 N / mm^2 we get.
- σac = 101 - (101 - 90 / 10 x 1.08).
- σac = 99.81 N / mm^2.
- P = 99.81 x 10426 = 1041639.91 N = 1040.63 KN.
- Designed axial strength = 1040.63 KN.
- Question no. 3.
- Ans -
- Aim - to find out design circular tube strut member for certain load. Also find Equivalent Square hollow sections.
- Introduction - A strut is defined as a structural member subjected to compression in a direction parallel to its longitudinal axis. The term strut is commonly used for compression members in roof trusses. A strut may be used in a vertical position or in an inclined position in roof trusses. The compression members may be subjected to both axial compression and bending. here is circular tube strut member. design of long tabular strut member.
- Also figure attached below of the section member.
- Formula to find axial strength = P = σac x A.
- Given - Length of member = L = 2.0 m.
- Factored axial load = Pu = 400 KN.
- Axial load = P = 400 / 1.5 = 266.67 KN.
- 1.5 is partial safety factor.
- Assume the ends of the member are simply supported.
- Yield strength of material fy= 210 N / mm^2.
- From IS 800 table Le = 1.0 L.
- Le = 1.0 x 2.0 = 2.0 m.
- We can take built up circular I - section tabular long strut.
- λ for I section assumed between 60 to 90 mm.
- Assume λ = 75, hence from IS 800 table 5.1, fy= 210 N / mm^2.
- σac = 93.545 N / mm^2.
- A = 400 x 10^3 / 93.545 = 4276 mm^2.
- From steel table take I section is more economical since it has maximum value of minimum redius of gyration.
- Try ISMB 250.
- Weight = 50.5 Kg / m, having A = 4755 mm^2 and r min = 26.5 mm, t for flange thickness = 12.5 mm.
- λ = Le / r min = 2.0 x 1000 / 26.5 = 75.47.
- From table 5.1 of IS 800 for λ = 75.47 and fy = 210 N / mm^2 we get.
- σac = 92.66 N / mm^2.
- P = 92.66 x 4755 = 440.63 KN. this section is suitable to carry the load.
- P = 440.63 KN.
- Check for limiting width to thickness ratio.
- Width of outstanding leg = Nominal width / thickness = 250 / 12.5 = 20 mm.
- Since, 20 < 16 x t = 16 x 12.5 = 200 mm, hence ok.
- Buckling class is effective length = 1.0 x L.
- Slenderness ratio = 75.47.
- Compressive strength = 92.66 N / mm^2.
- Design strength = 440.63 KN.
- For equivalent square hollow section.
- We can take built up square hollow angle section.
- λ for single angle section assumed between 120 to 150 mm.
- Assume λ = 120, hence from IS 800 table 5.1, fy= 210 N / mm^2.
- σac = 58.22 N / mm^2.
- A = 400 x 10^3 / 58.22 = 6870.49 mm^2.
- From steel table take single angle section is more economical since it has maximum value of minimum redius of gyration.
- Try ISA 200 x 200 x 18.
- Weight = 54 Kg / m, having A = 6881 mm^2 and r min = 39 mm, t for thickness = 18 mm.
- λ = Le / r min = 2.0 x 1000 / 39 = 51.58.
- From table 5.1 of IS 800 for λ = 51.58 and fy = 210 N / mm^2 we get.
- σac = 116.89 N / mm^2.
- P = 116.89 x 6881 = 440.63 KN. this section is suitable to carry the load.
- P = 804.32 KN.
- Check for limiting width to thickness ratio.
- Width of outstanding leg = Nominal width / thickness = 200 / 18 = 11.11 mm.
- Since, 11.11 < 16 x t = 16 x 18 = 288 mm, hence ok.
- Buckling class is effective length = 1.0 x L.
- Slenderness ratio = 51.58.
- Compressive strength = 116.89 N / mm^2.
- Design strength = 804.32 KN.
- Question no. 4.
- Ans -
- Aim - design of laced column.
- Introduction - A laced column is most commonly used lateral system. lacing system used eccentric loading. most common sections used as lacing are flats, angles and channels. there are two types of lacing single lacing and double lacing. single lacing is one side lacing system. double lacing two side lacing system.
- Also figure attached below of the section member.
- Given - two channels back to back of 10 m height.
- Length of column = L = 10 m.
- Factored axial load = Pu = 1000 kN. The lacings are bolted.
- Assume fy = 250 N / mm^2.
- For built up column section λ may be assumed between 30 to 50.
- Let us assume λ = 40, corrosponding to which σac = 139 N / mm^2. from table 5.1 IS 800.
- Cross section area A = P / σac = 1000 x 10^3 / 139 = 7194.24 mm^2.
- Since, λ = Le / r min.
- 40 = 10 x 1000 / r min.
- So, r min = 250 mm.
- There was 2 channel section back to back.
- Let us we choose 2 ISLC 350 back to back.
- Weight = W = 38.8 kg / m.
- A = 4947 mm^2, rxx = 137.2 mm, ryy = 28.2 mm, cyy = 24.1 mm.
- Total area A = 2 x A = 2 x 4947 = 9894 mm^2.
- Specing S = 2 x [rxx - cyy] = 2 x [137.2 - 24.1] = 226.2 mm.
- Observe that for ISLC 350 from steel table @ 77.6 kg / m. When S = 220 mm, ryy = 13.70 < rxx = 13.72 cm.
- While S = 240 mm, ryy = 14.68 > rxx = 13.72 cm. keep S = 240 mm, for which ryy = 137 mm. for composite section.
- Hence r min = rxx = 137 mm.
- Since, λ = 10 x 1000 / 137 = 72.99 = 73.00.
- Hence from table 5.1 IS 800.
- For fy = 250 & λ = 73.00 we get.
- σac = 139 - (139 - 132) x 3.37 / 10 = 136.64 N / mm^2.
- P = A x σac = 9894 x 136.64 x 10^-3 = 1351.9 KN > 1000 KN. Hence satisfy.
- Single laced column ISLC 350, channels back to back placed.
- Compressive strength = 136.64 N / mm^2.
- Axial load capacity = 1351.9 KN.
- Slenderness ratio = 73.00 Within the limits.
- Hence ok.
- Question no. 5.
- Ans -
- Aim - design of battened column.
- Introduction - A battened column is most commonly used lateral system. more flexible than solid column used as axial loading. They carry transverse shear force when column deflects. most common sections used as battens are flats, plates, angles and channels also I section used. it is subjected to eccentric loading or applied moment in the plne of battens. end batten also known as tie plate
- Also figure attached below of the section member.
- Given - two channels back to back of 10 m height.
- Length of column = L = 5.0 m.
- Factored axial load = Pu = 2000 kN. The battenings are welded.
- Assume fy = 250 N / mm^2.
- For built up column section λ may be assumed between 30 to 50.
- For batened column effective length is incresed by 10 %.
- Le = 1.1 x L = 1.1 x 5.0 = 5.5 m = 5500 mm.
- Assumed λ = 40 to which σac = 139 N / mm^2.
- A = P / σac = 2000 x 10^3 / 139 = 14388.48 mm^2.
- So, r min = L / λ = 5500 / 40 = 137.5 mm.
- There is no availability of section greater than cross section area.
- Let us choose 2 ISLC 400 @ 45.7 kg / m each.
- A = 5825 mm^2.
- rxx = 15.50 cm = 155.0 mm.
- ryy = 2.81 cm = 28.1 mm.
- cyy = 2.36 cm = 23.6 mm.
- A = 2 x A = 2 x 5825 = 11650 mm^2.
- Total, The channels are placed back to back at a specing S given by
- S = 2 X [rxx - cyy] = 2 x [155 - 23.6] = 262.8 mm.
- Observe that for ISLC 400 When S = 260 mm, ryy = 15.62 > rxx = 15.5 cm.
- While S = 260 mm, ryy = 156.2 > rxx = 155 mm. keep S = 260 mm, for which ryy = 156.2 mm. for composite section.
- Hence r min = rxx = 155 mm.
- λ = 5500 / 155 = 35.48.
- Hence from table 5.1 IS 800.
- For fy = 250 & λ = 35.48 we get.
- σac = 145 - (145 - 139) x 5.48 / 10 = 141.71 N / mm^2.
- P = A x σac = 11650 x 141.71 x 10^-3 = 1650.92 KN. Hence ok.
- Design of battens
- 1. Specing of battens -
- The specing C of the batten is given by .
- C / rc min not greater than 50 or 0.7 times L / r of composite section.
- 0.7 x 35.48 = 24.83.
- Hence C / rc min < 24.83.
- C < 24.83 x rc min.
- rc min = ryy = 28.1.
- C < 24.83 x 28.1 < 697.72
- C = 700 mm.
- 2. Thickness of batten -
- t > ai / 50 from table II of handbook no. 1 [SP: 6(1)-1964] g= 60 mm for this section.
- Hence ai = 260 + 2 x 60 = 380 mm.
- t > 380 / 50 > 7.6 mm.
- Keep t = 8 mm.
- 3. Depth of batten -
- a = 260 + 2 x 23.6 = 307.2 mm, b = 100 mm.
- For intermediate battens d > 3 / 4 x 307.2 = 230.4 mm.
- For end batten d > a = 307.2 mm.
- For any batten d > 2b = 2 x 100 = 200 mm.
- Hence effective depth = 307.2 mm
- Total depth = 307.2 + edge distance 40 mm.
- Total depth = 387.2 mm.
- Hence try 400 x 8 mm battens at specing C = 800 mm.
- 4. Check for shearing and bending stresses.
- Transverse shear = V = 0.025 x 2000 = 50 KN.
- S = ai = 380.
- Longitudinal shear V1 = VC / NS = 50 x 800 / 2 x 380 = 25 KN.
- Moment M = VC / 2N = 50 x 1000 x 800 / 2 x 2 = 10 x 10^6 Nmm.
- Single battened column ISLC 400, channels back to back placed.
- Compressive strength = 141.71 N / mm^2
- Axial load capacity = 1650.92 KN.
- Slenderness ratio = 35.48 Within the limits.
- Transeverse shear force = 50 KN.
- Longitudinal shear force = 25 KN.
- Moment = 10 x 10^6 Nmm.
- Hence ok.