Week 4 Challenge

^{1. Section Classification Problem}

^{(a) ISMB 500}

• ^{From IS 808:1989, dimensions of ISMB 500 are:}
• ^{Flange width bf=180 mmbf​=180mm, flange thickness tf=17.2 mmtf​=17.2mm,}
• ^{Web height hw=465.6 mmhw​=465.6mm, web thickness tw=9.4 mmtw​=9.4mm.}
• ^{Flange Classification (b/t):}

^{bf2tf=1802×17.2=5.232tf​bf​​=2×17.2180​=5.23}

^{Compare with limit for plastic classification: bt≤9.4tb​≤9.4 (from Table 2, IS 800:2007).
→ Flange is Plastic.}

• ^{Web Classification (h/tw):}

^{hwtw=465.69.4=49.52tw​hw​​=9.4465.6​=49.52}

^{Compare with limit for plastic classification: hwtw≤72tw​hw​​≤72.
→ Web is Plastic.}

^{Conclusion: ISMB 500 is Plastic.}

^{(b) ISMC 200}

• ^{Dimensions: bf=70 mm,tf=8.6 mm,hw=200 mm,tw=5.4 mmbf​=70mm,tf​=8.6mm,hw​=200mm,tw​=5.4mm.}
• ^{Flange Classification (b/t):}

^{bf2tf=702×8.6=4.072tf​bf​​=2×8.670​=4.07}

^{Compare with limit for plastic classification: bt≤9.4tb​≤9.4.
→ Flange is Plastic.}

• ^{Web Classification (h/tw):}

^{hwtw=2005.4=37.04tw​hw​​=5.4200​=37.04}

^{Compare with limit for plastic classification: hwtw≤72tw​hw​​≤72.
→ Web is Plastic.}

^{Conclusion: ISMC 200 is Plastic.}

^{(c) PG 1300 × 450 × 20 × 12}

• ^{Flange Classification (b/t):}

^{bf2tf=4502×20=11.252tf​bf​​=2×20450​=11.25}

^{Compare with limit for compact classification: bt≤10.5tb​≤10.5.
→ Flange is Compact.}

• ^{Web Classification (h/tw):}

^{hwtw=130012=108.33tw​hw​​=121300​=108.33}

^{Compare with limit for slender classification: hwtw≤72tw​hw​​≤72.
→ Web is Slender.}

^{Conclusion: PG section is Compact/Slender.}

^{(d) Double Angle (2 × ISA 75 × 75 × 6)}

• ^{Flange Classification (b/t):bftf=756=12.5tf​bf​​=675​=12.5 Compare with limit for slender classification: bt≤9.4tb​≤9.4.
  → Flange is Slender.}

^{Conclusion: Double Angle is Slender.}

^{(e) RHS 150 × 100 × 10}

• ^{Width-to-Thickness Ratio (B/t):Bt=10010=10tB​=10100​=10 Compare with limit for plastic classification: Bt≤15.7tB​≤15.7.
  → RHS is Plastic.}

^{Conclusion: RHS section is Plastic.}

^{(f) Built-up Box 1000 × 450 × 20 × 12}

• ^{Flange Classification (b/t):}

^{bf2tf=4502×20=11.252tf​bf​​=2×20450​=11.25}

^{Compare with limit for compact classification: bt≤10.5tb​≤10.5.
→ Flange is Compact.}

• ^{Web Classification (h/tw):}

^{hwtw=100012=83.33tw​hw​​=121000​=83.33}

^{Compare with limit for slender classification: hwtw≤72tw​hw​​≤72.
→ Web is Slender.}

^{Conclusion: Built-up Box is Compact/Slender.}

^{2. Effective Length Problems}

^{For the rafter with purlins spaced at 1.2 m:}

• ^{Effective length about major axis (x-x):}
• ^{Purlins provide lateral torsional restraint, so K=1.0K=1.0.}

^{Leff, x-x=1.0×1.2=1.2 mLeff, x-x​=1.0×1.2=1.2m}

• ^{Effective length about minor axis (y-y):}
• ^{No intermediate lateral restraint, so K=1.2K=1.2.}

^{Leff, y-y=1.2×1.2=1.44 mLeff, y-y​=1.2×1.2=1.44m}

^{Effective Lengths:}

• ^{About major axis: Leff, x-x=1.2 mLeff, x-x​=1.2m}
• ^{About minor axis: Leff, y-y=1.44 mLeff, y-y​=1.44m}

^{3. Deflection of Beam Problem}

^{Case 1: Simply Supported Beam with UDL}

• ^{Given: w=10 kN/m,L=10 m,EI=25×106 Nm2w=10kN/m,L=10m,EI=25×106Nm2.}

^{δmax=5wL4384EI=5×10×104384×25×106}

^{δmax​=384EI5wL4​=384×25×1065×10×104​}

^{δmax=5.208 mm}

^{δmax​=5.208mm}

^{4. Moment Capacity Determination Problem}

^{Problem Statement:}

^{Design a simply supported beam of span 6 m spaced at 3 m c/c, carrying:}

• ^{Floor Dead Load = 5 kN/m25kN/m2 (including self-weight),}
• ^{Live Load = 6 kN/m26kN/m2,}
• ^{Concentrated Load = 100 kN100kN at midspan.
  Assume the beam is laterally restrained and has a bearing length of 50 mm50mm.}

^{Step 1: Calculate Maximum Loads}

1. ^{Floor Area Supported by Each Beam:
   Spacing of beams = 3 m3m.}

^{Area supported = 3×6=18 m23×6=18m2.}

2. ^{Distributed Load on Beam (UDL):
   Dead Load (DL) = 5×3=15 kN/m5×3=15kN/m,
   Live Load (LL) = 6×3=18 kN/m6×3=18kN/m.}

^{Total UDL = 15+18=33 kN/m15+18=33kN/m.}

3. ^{Total Load on Beam (including concentrated load):
   Concentrated load = 100 kN100kN at midspan.}

^{Step 2: Calculate Maximum Bending Moment}

1. ^{Moment Due to UDL:}

^{MUDL=wL28=33×628=148.5 kNm.MUDL​=8wL2​=833×62​=148.5kNm.}

2. ^{Moment Due to Concentrated Load:}

^{MPoint Load=P⋅L4=100⋅64=150 kNm.MPoint Load​=4P⋅L​=4100⋅6​=150kNm.}

3. ^{Total Maximum Bending Moment:}

^{Mmax=MUDL+MPoint Load=148.5+150=298.5 kNm.Mmax​=MUDL​+MPoint Load​=148.5+150=298.5kNm.}

^{Step 3: Required Plastic Section Modulus}

^{Using: M=fbd⋅ZpM=fbd​⋅Zp​,
where fbd=250 MPafbd​=250MPa (design bending strength),}

^{Zp=Mmaxfbd=298.5×106250=1,194×103 mm3.Zp​=fbd​Mmax​​=250298.5×106​=1,194×103mm3.}

^{Step 4: Select Suitable Trial Section}

^{From IS 808:1989, try ISMB 400:}

• ^{Zp=1,257×103 mm3Zp​=1,257×103mm3,}
• ^{hw=400 mm,bf=140 mm,tf=21 mm,tw=8.6 mm.hw​=400mm,bf​=140mm,tf​=21mm,tw​=8.6mm.}

^Check:

^{Zp(ISMB 400)>Zp(required).Zp​(ISMB 400)>Zp​(required).}

^{So, ISMB 400 is adequate.}

^{Step 5: Check Shear Strength}

1. ^{Design Shear Force:
   Total Shear = Total Load2=33×6+1002=199 kN.2Total Load​=233×6+100​=199kN.}
2. ^{Shear Capacity of Section:
   For ISMB 400, tw=8.6 mmtw​=8.6mm, depth hw=400−2⋅21=358 mm hw​=400−2⋅21=358mm.}

^{Shear strength:}

^{Vd=fvd⋅Aw3=250⋅8.6⋅3583⋅1,000=444 kN.Vd​=3​fvd​⋅Aw​​=3​⋅1,000250⋅8.6⋅358​=444kN.}

^{Vd(capacity)>V(applied).Vd​(capacity)>V(applied).}

^{Step 6: Check for Deflection}

^{Deflection due to UDL:}

^{δUDL=5wL4384EI.δUDL​=384EI5wL4​.}

^{Deflection due to Concentrated Load:}

^{δPoint Load=PL348EI.δPoint Load​=48EIPL3​.}

^{Use E=2×105 MPaE=2×105MPa and I=103.9×106 mm4}

^{I=103.9×106mm4 (for ISMB 400).
Check that total deflection δ<L/325δ<L/325.}

^{5. Purlin Design Problem}

^{Problem Statement:}

^{Design a channel purlin for:}

• ^{Span of truss = 20 m20m,}
• ^{Spacing of truss = 4 m4m,}
• ^{Roof slope = 20∘20∘,}
• ^{Spacing of purlins = 2 m2m,}
• ^{Wind suction = 1 kN/m21kN/m2,}
• ^{Live Load = 0.75 kN/m20.75kN/m2, Dead Load = 1.4 kN/m21.4kN/m2.}

^{Step 1: Calculate Loads}

1. ^{Dead Load (DL):}

^{DL=1.4×2=2.8 kN/m.DL=1.4×2=2.8kN/m.}

2. ^{Live Load (LL):}

^{LL=0.75×2=1.5 kN/m.LL=0.75×2=1.5kN/m.}

3. ^{Wind Load (WL):
   Wind suction acts perpendicular to the roof surface. Horizontal wind suction:}

^{WL=1.0×2=2.0 kN/m.WL=1.0×2=2.0kN/m.}

^{Step 2: Bending Moment}

^{For a simply supported beam of span L=4 mL=4m,}

^{M=wL28,M=8wL2​,}

^{where w=Resultant Load=DL+LL+WLcos⁡20∘w=Resultant Load=DL+LL+WLcos20∘.}

^{Substitute values:}

^{w=2.8+1.5+2.0cos⁡20∘=5.678 kN/m.}

^{w=2.8+1.5+2.0cos20∘=5.678kN/m.}

^{M=5.678×428=11.36 kNm.}

^{M=85.678×42​=11.36kNm.}

^{Step 3: Section Modulus}

^{Using M=fbd⋅ZM=fbd​⋅Z:}

^{Z=Mfbd=11.36×106250=45.44×103 mm3.Z=fbd}

^{​M​=25011.36×106​=45.44×103mm3.}

^{From IS 808, select a channel section with Z≥45.44 cm3Z≥45.44cm3.
Use ISMC 200, Z=63.1 cm3Z=63.1cm3.}

^{Step 4: Check Deflection}

^{Check deflection for purlin under combined loading using:}

^{δ=5wL4384EI.δ=384EI5wL4​.}

^{Ensure δ<L/325δ<L/325.}

^{hence safe.}