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Sanjivani Rural Education Society's
Sanjivani College of Engineering, Kopargaon 423603.
-Department of Civil Engineering-
By
Mr. Sudhir B. Gayake (Assistant Professor)
Mail Id- gayakesudhircivil@sanjivani.org.in
Prof. Sudhir B. Gayake, SCOE Kopargaon
Prof. Sudhir B. Gayake, SCOE Kopargaon
Question- Design a simply supported beam of effective span 4m carrying a factored
load of 350kN at mid span. The section is laterally supported throughout the span.
Design the cross section using I-section. Use steel grade Fe410.
1) Given data- 350kN
Span=l=4000mm,
Pu=350kN at mid span, 4000mm
fu=410MPa,
fy=250MPa, 175kN
ϒm0=1.1
2) Maximum SF and BM calculations- 175kN
Maximum SF=Vu=(Pu/2)
=(350/2)=175kN 350kN-m
Maximum BM=Mu=(Pu×l/4)
=(350×4/4)=350kN-m
3) Equating maximum BM with yield
strength of section= Md=Mu
𝑍𝑝×𝑓𝑦
ϒm0
= 350 × 106
+
-
+
Prof. Sudhir B. Gayake, SCOE Kopargaon
Zp required=
𝑀𝑢×𝛾𝑀0
𝑓𝑦
= (350×106×1.1/250)=1.54×106 mm3
select section from IS 800-annex H having plastic section modulus greater than above
calculated value.
Try ISLB 500 @ 75kg/m having properties-
h=500mm, bf=180mm, tf=14.1mm, tw=9.2mm, Izz=38549×104mm4 , h1=430.2mm,
Zp=1773.7×103 mm3 Ze= 1545.2×103 mm3
3) Classification of cross section- (Table No 2)
a) Flange= (bf/tf)=[(bf/2)/(tf)]=(90/14.1)= 6.38 <9.4ɛ ∴ flange is plastic
b) Web= (h1/tw)= (430.2/9.2) = 46.76< 84ɛ ∴ web is also plastic
4) Design shear strength of section-(Cl.No.8.4.1 Pg.No.59)
Vd=[(fyw×Aw)/(√3×ϒm0)]==[(250×500×9.2)/(√3×1.1)]= 603.6kN> Vu………..(OK)
Check for high or low shear condition= 0.6Vd= 0.6×603.6= 362.16kN > Vu……….(Low shear)
5) Design bending strength-(Cl.No. 8.2.1.2 Pg.No. 53)
Md=
βb 𝑍𝑝×𝑓𝑦
ϒm0
=(1×1773.7×103×250/1.1)= 402.95kN-m > Mu………………….(OK)
Section is
Plastic βb=1
Prof. Sudhir B. Gayake, SCOE Kopargaon
6) Design Checks-
a) Check for web buckling= Fwb= B×tw×fcd
Assume stiff bearing length= b=120mm
B=(120+(h1/2))=(120+(430.2/2))= 335.1mm…………………….for end reaction
tw=9.2mm
fcd- depends on slenderness ratio (λ)=2.45(h1/tw)=114.56≈ 115
From table 9 (c) Pg.no. 42 and fy=250MPa
fcd= 89.15MPa
Fwb= B×tw×fcd =(335.1 ×9.2 ×89.15)= 274.84kN>Vu…………………(OK)
b) Check for web crippling= Fw = Ae× (fyw/γm0)
fyw= 250MPa
Ae= b1×tw=(b+n1)×tw……………………………………………………for end reaction
=(120+ 2.5(tf+r1))× tw
=(120+2.5(14.1+17)) ×9.2
=1819.3mm2
Fw = Ae× (fyw/γm0)= (1819.3 ×250/1.1)=413.47kN>Vu…………………….(OK)
Prof. Sudhir B. Gayake, SCOE Kopargaon
c) Check for deflection-
Deflection due to external working load=δactual=(Pl3/48EI)
=(233.33×103×40003/48×2×105×38549×104 )
=4.03mm
Maximum deflection allowed as per IS800 δlimit =(Span/300 to Span/240 )
=(4000/300 to 4000/240)
=(13.33mm to 16.67mm)
As δactual < δlimit the deflection is within limit…………………(OK)
Therefore ISLB 500 @ 75kg/m can be used for given loading on
beam as it is safe for all checks.

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4a i numerical on design of laterally supported beams

  • 1. Sanjivani Rural Education Society's Sanjivani College of Engineering, Kopargaon 423603. -Department of Civil Engineering- By Mr. Sudhir B. Gayake (Assistant Professor) Mail Id- gayakesudhircivil@sanjivani.org.in Prof. Sudhir B. Gayake, SCOE Kopargaon
  • 2. Prof. Sudhir B. Gayake, SCOE Kopargaon Question- Design a simply supported beam of effective span 4m carrying a factored load of 350kN at mid span. The section is laterally supported throughout the span. Design the cross section using I-section. Use steel grade Fe410. 1) Given data- 350kN Span=l=4000mm, Pu=350kN at mid span, 4000mm fu=410MPa, fy=250MPa, 175kN ϒm0=1.1 2) Maximum SF and BM calculations- 175kN Maximum SF=Vu=(Pu/2) =(350/2)=175kN 350kN-m Maximum BM=Mu=(Pu×l/4) =(350×4/4)=350kN-m 3) Equating maximum BM with yield strength of section= Md=Mu 𝑍𝑝×𝑓𝑦 ϒm0 = 350 × 106 + - +
  • 3. Prof. Sudhir B. Gayake, SCOE Kopargaon Zp required= 𝑀𝑢×𝛾𝑀0 𝑓𝑦 = (350×106×1.1/250)=1.54×106 mm3 select section from IS 800-annex H having plastic section modulus greater than above calculated value. Try ISLB 500 @ 75kg/m having properties- h=500mm, bf=180mm, tf=14.1mm, tw=9.2mm, Izz=38549×104mm4 , h1=430.2mm, Zp=1773.7×103 mm3 Ze= 1545.2×103 mm3 3) Classification of cross section- (Table No 2) a) Flange= (bf/tf)=[(bf/2)/(tf)]=(90/14.1)= 6.38 <9.4ɛ ∴ flange is plastic b) Web= (h1/tw)= (430.2/9.2) = 46.76< 84ɛ ∴ web is also plastic 4) Design shear strength of section-(Cl.No.8.4.1 Pg.No.59) Vd=[(fyw×Aw)/(√3×ϒm0)]==[(250×500×9.2)/(√3×1.1)]= 603.6kN> Vu………..(OK) Check for high or low shear condition= 0.6Vd= 0.6×603.6= 362.16kN > Vu……….(Low shear) 5) Design bending strength-(Cl.No. 8.2.1.2 Pg.No. 53) Md= βb 𝑍𝑝×𝑓𝑦 ϒm0 =(1×1773.7×103×250/1.1)= 402.95kN-m > Mu………………….(OK) Section is Plastic βb=1
  • 4. Prof. Sudhir B. Gayake, SCOE Kopargaon 6) Design Checks- a) Check for web buckling= Fwb= B×tw×fcd Assume stiff bearing length= b=120mm B=(120+(h1/2))=(120+(430.2/2))= 335.1mm…………………….for end reaction tw=9.2mm fcd- depends on slenderness ratio (λ)=2.45(h1/tw)=114.56≈ 115 From table 9 (c) Pg.no. 42 and fy=250MPa fcd= 89.15MPa Fwb= B×tw×fcd =(335.1 ×9.2 ×89.15)= 274.84kN>Vu…………………(OK) b) Check for web crippling= Fw = Ae× (fyw/γm0) fyw= 250MPa Ae= b1×tw=(b+n1)×tw……………………………………………………for end reaction =(120+ 2.5(tf+r1))× tw =(120+2.5(14.1+17)) ×9.2 =1819.3mm2 Fw = Ae× (fyw/γm0)= (1819.3 ×250/1.1)=413.47kN>Vu…………………….(OK)
  • 5. Prof. Sudhir B. Gayake, SCOE Kopargaon c) Check for deflection- Deflection due to external working load=δactual=(Pl3/48EI) =(233.33×103×40003/48×2×105×38549×104 ) =4.03mm Maximum deflection allowed as per IS800 δlimit =(Span/300 to Span/240 ) =(4000/300 to 4000/240) =(13.33mm to 16.67mm) As δactual < δlimit the deflection is within limit…………………(OK) Therefore ISLB 500 @ 75kg/m can be used for given loading on beam as it is safe for all checks.