column subjected to bending numerical problem

1. Sanjivani Rural Education Society's
Sanjivani College of Engineering, Kopargaon 423603.
-Department of Civil Engineering-
Course Title: (Design of Steel Structures- Third Year B.Tech)(CE-302)
Unit 3(a) i. Eccentrically loaded column
(Check for section strength)
By
Mr. Santosh R. Nawale(Assistant Professor)

2. 1)Given Data-
Pu=N= 750kN
ey= 270mm
Leff=KL=3000mm
ISHB 450 @ 85.4kg/m
2) To find-
Check for section strength=? [(N/Nd)+(Mz/Mdz)≤ 1]
3) Solution-
ISHB 450 @ 85.4kg/m having properties;
h=450, A=11114, bf=250, tf=13.7, tw=9.8, rz=187.8, ry=51.8, h1=386.2, zpz=1955.03×103
i) Classification of cross section-
[(bf/2)/tf]=[125/13.7]=9.12 < 9.4 Flange is plastic
[h1/tw]=[386.2/9.8]=39.40< 84 web is plastic ………………….. βb= 1.0
270

3. ii) Pu= 750kN
Mz= 750 X 0.270= 202.5kNm
iii) Design strength of section-
Nd= (AgXfy/ϒm0)= (11114X250/1.10)= 2525.90kN
Mdz= βbX ZpzXfbd
For fbd
(KL/rz)=( 3000/187.8)=15.97=16, (h/tf)=(386.2/13.7)=28.18 ,Table no 14 pg.no. 57 & Table 13a for fbd
Mdz= 1.0 X 1955.03×106
X 227.3= 444.378kNm
[(N/Nd)+(Mz/Mdz)≤ 1]
[(750/2525.90)+(202.5/444.378)] ≤ 1
[0.2969+0.45569]= 0.75 <1.0 ok safe to transfer given eccentric load.
(h/tf)
KL/rz
16
20 5637.8
28.18 3166.54
30 2616.7
fy
Fcr,b
250
4000 227.3
3166.54 227.3
2000 227.3

4. Plastic Hinge-
 A plastic hinge is a zone of yielding due to fracture in a structural member. A strain
hardening action usually occurs at the location of maximum BM and hinges are formed
causing large deformations and rotation by slightly increase in load.
 A structure can support the computed ultimate load due to formation of plastic hinges. The
member remains elastic until the moment reaches a value of Mp, any additional
increase in moment will cause the beam to rotate with little increase in stress.

5. Plastic moment of a section-
 The maximum moment of resistance of a section which is fully yielded.
 MR=[(I/Y) × σ]
 MR=[Z × σ]
 Z= Section modulus [Geometrical property of cross section used for design ]
 Ze= Elastic section modulus= This is applied upto the yield point of the material.
 Zp=Plastic section modulus= Elastic yielding is assumed and plastic behavior is assumed to be
an acceptable limit for design.
 Consider a fully yielded cross section of a
beam having bending stress distribution
rectangular.
 For SS condition top fibers will be in
compression while below NA the bottom
fibers will be in tension.
 C= fyXA1
 T= fyXA2

6.  For equilibrium the force in compression must be equal to tension.
C=T
fy×A1= fy×A2
But we have from fig. A1=A2=A/2 (As NA= Equal area Axis)
Mp= fy×(A/2)× Ӯ1 +fy×(A/2)×Ӯ2
Mp= fy×(A/2)×(Ӯ1 + Ӯ2 )
Mp= fy × Zp
Zpz= (A/2)×(Ӯ1 + Ӯ2 )
Zpy= (A/2)×(x1 + x2 )

7. Zpz= (A/2)×(Ӯ1 + Ӯ2 )
Ӯ1=[(A1y1+A2y2)/(A1+A2)]
Ӯ2=[(A3y3+A4y4)/(A3+A4)]
Zpy= (A/2)×(z1 + z2 )
z1=[(A1z1+A2z2)/(A1+A2)]
z2=[(A3z3+A4z4)/(A3+A4)]
A1
A2
y1
y2 A1
A1
A
2
z1
z2

8. Question- A corner column , located in the bottom storey of a braced frame is
subjected to factored loads: Pu=1100kN, Mz=135kNm, My=70kNm. The effective
length of column if 3.75m. Design the beam-column assuming steel grade
fe410.