Week 2 Challenge
- Question no 1.
- Plate drilled holed.
- Ans -
- Given - Plate size = 260 x 20 mm.
- Bolt size = 20 mm.
- Gross Diameter = 20 + 1.5 = 21.5 mm.
- Where gross diameter is 25 to 50 % more than net diameter.
- Plate - drilled hole.
- From figure pitch = p = 60 mm.
- Gauge distance = g = 60 mm.
- Thick of plate = t = 20 mm.
- Width of plate = b = 260 mm.
- No. of bolts = 12.
- n = 4.
- ∑ dh = sum of bolt diameter = 12 x 20 = 240 mm.
- Use formula to find out net section area (An).
- The net sectional area of the tension member is the gross sectional area of the member minus sectional area of the maximum number of holes.
- An = [b - n x d] x t.
- An = [260 - 4 x 21.5] x 20.
- An = 3480 mm^2.
- Net section area of plate section (An) = 3480 mm^2.
- Question no 1.
- Plate zigzag bolting.
- Ans -
- Given - Plate size = 260 x 20 mm.
- Bolt size = 20 mm.
- Plate - zigzag hole.
- From figure pitch = p = 60 mm.
- Gauge distance = g1 = 60 mm.
- Gauge distance = g2 = 60 mm.
- Staggered pitch from first to next bolt = s1 = 60 mm.
- Staggered pitch from first to last bolt = s2 = 50 = 50 mm.
- Thick of plate = t = 20 mm.
- Width of plate = b = 260 mm.
- No. of bolts = 12.
- n = 4.
- ∑ dh = sum of bolt diameter = 12 x 20 = 240 mm.
- Use formula to find out net section area (An).
- The net sectional area of the tension member is the gross sectional area of the member minus sectional area of the maximum number of holes.
- For AD & BF section.
- Gross Diameter = 20 + 1.5 = 21.5 mm.
- Where gross diameter is 25 to 50 % more than net diameter.
- An = [b - n x d] x t.
- An = [260 - 2 x 21.5] x 20.
- An = 4340 mm^2.
- For CEG section.
- An = [b - n x d] x t.
- An = [260 - 3 x 21.5] x 20.
- An = 3910 mm^2.
- For ACDG section.
- No. of bolts = n = 4.
- No. of zigzag lines = m = 3.
- Staggered pitch = s = 60.
- An = [(b - n x d) + m x s^2/ 4 x g] x t.
- An = [(260 - 4 x 21.5) + 3 x 60^2 / 4 x 60] x 20.
- An = 4380 mm^2.
- For BCFG section.
- No. of bolts = n = 4.
- No. of zigzag lines = m = 3.
- Staggered pitch = s = 50.
- An = [(b - n x d) + m x s^2/ 4 x g] x t.
- An = [(260 - 4 x 21.5) + 3 x 50^2 / 4 x 60 ] x 20.
- An = 4105 mm^2.
- Most critical net section area of plate zigzag section (An) = 3910 mm^2.
- Question no 1.
- Angle section 200 x 100 x 12 with M20 bolt.
- Ans -
- Angle section figure with abc section and abde section attached below.
- Given - Angle size = 200 x 100 x 12 mm.
- Bolt size = 20 mm.
- Angle - zigzag hole.
- From figure Staggered pitch = s = 50 mm.
- Gauge distance = g = 90 mm.
- Thick of plate = t = 12 mm.
- Width of equivalent plate = b = 200 + 100 - 12 = 288 mm.
- Distance between bolt holes = 90 + 70 + 60 - 12 = 208 mm.
- Edge distance for each line = 40 mm.
- Gross diameter of bolts = d = 20 + 1.5 = 21.5 mm.
- Where gross diameter is 25 to 50 % more than net diameter.
- No. of bolts = 4.
- For section abc.
- No. of bolts = n = 1.
- Width of angle section = b = 200 mm.
- Use formula to find out net section area (An).
- The net sectional area of the tension member is the gross sectional area of the member minus sectional area of the maximum number of holes.
- An = [b - n x d] x t.
- An = [200 - 1 x 21.5] x 12.
- An = 2142 mm^2.
- For section abde.
- No. of bolts = n = 2.
- No. of zigzag lines = m = 1.
- Width of angle section = b = 200 mm.
- An = [(b - n x d) + m x s^2/ 4 x g] x t.
- An = [(200 - 2 x 21.5) + 1 x 50^2/ 4 x 90] x 12.
- An = 1967.32 mm^2.
- Most critical net section area of ang;e section (An) = 1967.32 mm^2.
- Question no 3.
- Determine section size for given tensile load.
- Ans -
- Given - Design a single angle member of length L = 5 m.
- Figure attached below.
- Axial tensile force of 500 kN.
- Yield strength fy = 350 N / mm^2.
- Assume that steel section with thickness less than 20 mm will be used. Hence axial tension σat = 0.6fy as per IS 800 = 0.6 x 350 = 210 N / mm^2, so σat = 210 N / mm^2.
- As per tips steps given for calculation of member.
- Assume bolted connection used.
- 1. Required An = 500 x 1000/ 210 = 2380.95 mm^2.
- Required An = 2380.95 mm^2.
- 2. Choose a section having gross area equal to 40 % more than as per IS 800 net area. Choose ISA 150 x 150 x 12 mm @ 27.2 kg / m having gross area = 3459 mm^2. this is connected by longer leg using 20 mm diameter bolts. Gross diameter of bolts = 20 + 1.5 = 21.5 mm.
- 3. An = A1 + kA2.
- where A1 = (150 - 21.5 - 12/6) x 12 = 1470 mm^2.
- A2 = (150 - 12/2) x 12 = 1728 mm^2.
- k = as per IS 800 = 3A1 / 3A1 + A2.
- k = 3 x 1470 / 3 x 1470 + 1728.
- k = 0.7184.
- An = 1470 + 0.7184 x 1728.
- An = 2711.39 mm^2.
- Which is slightly more than net area required.
- 4. Load capacity of member = 2711.39 x 210 = 569392.992 N = 569.39 KN which is greater than 500 KN. Hence the section is ok.
- Block shear capacity = 569.39 KN.
- 5. check for slenderness ratio - since this is a tie number there is no chance of reversal stress.
- However as per code requirements the slenderness ratio should not exceed 350.
- For the angle section r min = 29.3 mm.
- λ = L / r min = 5 x 1000 / 29.3 = 170.64 not greater than 350 ok.
- 6. Design of end connections.
- Strength of 20 mm diameter hand bolts:
- Take permissible stresses at yielding & tearing = 80 & 250 N / mm^2.
- In yielding = Π / 4 x 21.5^2 x 80 x 10^-3 = 29.04 KN.
- In tearing = 21.5 x 12 x 250 x 10^-3 = 64.5 KN.
- Bolt value R = 29.04 KN.
- No. of bolts = 500 / 29.04 = 17.21 = 18 bolts.
- Provide 18 bolts at a pitch = 2.5 x 20 = 50 mm.
- Keep edge distance = 35 mm from edge distance hole table.
- figure angle section with bolts attached below.
- Single equal angle ISA 150 x 150 x 12 mm.
- Pitch distance = 50 mm.
- Gauge distance = 80 mm.
- Edge distance = 35 mm.
- Design strength of member in yielding = 29.04 KN.
- Design strength of member in tearing = 64.5 KN.
- Question no 3.
- Design of sag rod.
- Ans -
- Design sag rod figure attached below.
- Given - purlins supported at 1 / 3 rd points by sag rod.
- centre to centre distance of Rafter spaced at 8 m , spacing of purlin = 1.2 m.
- Self weight of roofing (including thermal insulation)= 60 kg / m^2.
- Wind pressure intensity = 150 kg / m^2, Slope of rafter = 15 degrees, Live Load =75 kg / m^2.
- Assume self weight of purlin as 100 kg / m.
- 1. Total gravity load.
- Computation of dead load (DL) on purlin.
- Load from roofing = 60 x 1.2 = 72 kg / m.
- Self weight of purlin = 100 kg / m.
- Total dead load (DL) per m = 72 + 100 = 172 kg / m.
- Total dead load = 172 x 8 = 1376 kg.
- Computation of live load (LL) on purlin.
- Live load = 75 x 1.2 = 90 kg / m.
- Total live load (LL) per m on purlin = 90 kg / m.
- Total live load (LL) = 90 x 8 = 720 kg.
- 2. Computation of wind load (WL).
- Load from roofing = 150 x 1.2 = 180 kg / m.
- Total wind load (WL) = 180 x 8 = 1440 kg.
- Total gravity load = DL + LL.
- Total gravity load = 1376 + 720 = 2096 kg.
- Total gravity load = 2096 kg.
- 3.Wind load perpendicular to member = 1440 cos 15 = 1390.93 kg.
- Wind load perpendicular to member = 1390.93 kg.
- 4. Wind load parallel to member = 1440 sin 15 = 372.69 kg.
- Wind load parallel to member = 372.69 kg.
- Total load = gravity load + wind load.
- Total load = 2096 + 1390.93 + 372.69.
- Total load = 3859.62 kg.
- 5. Force contribution to each sag rod at 1 / 3 points.
- Tensile force = 3859.62 / 3 = 1286.54 kg = 12620.95 N.
- Assume fy = 250 N / mm^2.
- σat = 0.6fy as per IS 800.
- σat = 150 N / mm^2.
- Required net area An = 12620.95 / 150.
- An = 84.13 mm^2.
- Required An = 84.13 mm^2.
- From steel table for equal angle section choose ISA 20 x 20 x 4 mm.
- Sag rod size = ISA 20 x 20 x 4.