Problem 2
Question 1. Bending

1. Determine the section moment capacity to check whether it is satisfactory against the design moment. (3)

Solution
Section moment capacity                  M*≤ φMs
Nominal section capacity is given by  Ms= fyZe
Where                                             fy= 320 MPa
From Table 5.2 As 4100 the values for Heavily Welded compact section are given as:
𝜆sp < 𝜆s≤ 𝜆sy
Where 𝜆s= (do/t)(fy/250), and 𝜆sp=𝜆ep
𝜆sy=𝜆ey from the table
From table 5.2 of As 4100 for Heavily Welded 𝜆ep= 𝜆sp= 30
𝜆ey= 𝜆sy=35
And 𝜆s= (do/t)(fy/250)=(450/25)(320/250)=23.04 hence it is a compact section
23.04<30
For compact section               = 𝜆s < 𝜆sp
And                                          Ze = effective section modulus is given by
Ze= (S, 1.5Z).
Z= I/Ymax
I= 2[bhf3 / 12]+ bhf(hw /2+hf/2)2
Ymax= hw/2+hf
where, b=450, hf=25, hw= 1610
I=2 [ 450*253 / 12]+ 450*25(1610/2+25/2)2
I= 1.524* 10^10
Y max= 1610/2+25=830
Z =1.524*10^10/830= 18.36*10^6
S=2[b*hf*(hf/2+hw/2)
S= 2[450*25(1610/2+25/2)= 18.39*10^6
hence we determine to take Zc as the minimum value between (S,1.5Z)
1.5Z= 1.5* 18.36*10^6=27.54*10^6
(18.39*10^6, 27.54*10^6), we take Ze as =18.39*10^6
Ms=Ze*fy
Ms=18.39*10^6*320*10^-6=5884.8
ΦMs=0.9*5884.8=5296.32= section capacity
Mx≤ φMsx
=3556≤5296.32 hence it is satisfactory against the design moment..

1. Determine the member capacity to check whether it is satisfactory against the design moment.

Solution

1. What are the segments and its corresponding lengths?

There are two segments on this beam which are FP, PF
Which have a length of each FP= 12m,
PF=10m

1. Are the segments same for top and bottom flanges? Yes they are similar
• Determine the member capacities for each segment and check whether the member moment capacity of each segment is satisfactory corresponding to its design moment?

Segment FP: Mx*≤φMbx
Mb =𝛼𝑚𝛼𝑠Ms
Mb =𝛼m𝛼𝑠fyZe
Where Ze=effective section modulus
Fy=minimum yield stress for the section
𝛼𝑚= the moment modification factor
𝛼𝑠=is the slenderness reduction factor
𝛼𝑚= the moment modification factor is obtained from table 5.6.1 as=
𝛼𝑚= 1.35+0.4(2a/l)with a=4 and l=12
=1.35+0.4(2*4/12)2 = 1.528
𝛼𝑠= 0.6[{(Ms/Moa)2+3]}1/2-(Ms/Moa)]
Where Ms=section moment capacity
Moa=Mo= reference buckling moment
𝑀oa=(𝜋2𝐸𝐼𝑦/ 𝑙𝑒2)1/2 * (𝐺J+𝜋2𝐸𝐼𝑤/ 𝑙𝑒 2)1/2
Where, The modulus of elasticity (for steel, usually E = 200 GPa)
G = The shear modulus (If E = 200 GPa, G = 80 GPa)
J = Torsion constant for a cross-section
Iy= Moment of inertia about y axis (the minor axis)
Iw = Warping constant
le = Effective length for flexural tensional buckling
le=ktklkrl=1*1.4*0.8*12=13.44
𝛼m𝛼𝑠 should not exceed 1.0 so take them as 1.0
Mb=1.0*Ms=1*5884=
Mx=3556<5884 hence it is satisfactory

1. ii) For segment PF

The same procedure is used as the above segment and we get that:
-1808<5884 hence it is also satisfactory

Question 2. Bearing

1. Calculate the maximum bearing capacity of the girder at support C. Determine whether the girder at this location requires stiffening (3)

Solution
R*≤ φRb
where Φ is the capacity factor (Table 3.4 in AS4100,
Φ=0.9)
Rb is the nominal bearing capacity of a web under concentrated or patch loading
R* is the design bearing or reaction force

The nominal bearing capacity a web shall be taken as the lesser of its

• nominal bearing yield capacity (Rby)
• nominal bearing buckling capacity (Rbb)

Rby=1.25𝑏𝑏𝑓𝑡𝑤𝑓𝑦
𝑏b𝑓= dispersion bearing length=bs+5tf
bbf= 2+5*25=127

Rby= 1.25*127*25*320*10^-6=1.27

Rbb=𝛼cbbtwfy

Rbb= 𝛼𝑐determination

1. Estimate bearing buckling width bb
2. Calculate section of web resisting buckling Awc= twx bb
3. Adopt the following values as per Cl 5.13.
4. αb= web compression constant = 0.5
5. Kf = web compression form factor = 1.0

iii. Le / r = 2.5 d1/ tw = web compression member slenderness ratio

1. Calculate λn = web compression member modified slenderness ratio = 𝜆𝜆𝑛𝑛=𝐿𝐿𝑒𝑒 𝑟𝑟 𝑘𝑘𝑓𝑓 𝑓𝑓𝑦𝑦 250=2.5𝑑𝑑1 𝑡𝑡𝑤𝑤 𝑘𝑘𝑓𝑓 𝑓𝑓𝑦𝑦 250
2. Read the web compression member slenderness reduction factor αc from AS4100 Tb 6.3.3(3).

Rbb=1.37 hence we take the lesser which is 1.27

0.9*1.27=1.161
R>1.161 hence the girder requires stiffeners.

1. Design load bearing stiffener: select a pair of intermediate transverse stiffener (symmetrically on both sides of the web)of dimension L* width * 12(thickness)mm of grade300 steel which can safely carry the design bearing force. Select the width of the stiffener, L, such a way that it is a multiple of 10mm eg (100, 110, 120,130, 140mm…etc).

Question 3. Shear

1. Shear capacity check of web: Determine the web shear capacity of the girderassuming that the intermediate transverse are fitted at 2m centre to centre distance to check whether it is satisfactory against design shear. (3)

Solution
V* ≤ ΦVv
First check If dp/tw ≤82/√(fy/250
dp=2m=200/10= 20
82/ (320/250)1/2 =72.47
20≤72.47 hence the web will not buckle and we use Vv = Vu=Vw=0.6 fyAw
Vv= 0.6*320*22*1.61=6800.64

1. Design of intermediate transverse stiffeners: Use the same as the load bearing stiffener.

solution

R*≤ φRsy
Rsy= nominal yield capacity
Rsy= Rby+Asfys
R by= nominal bearing yield capacity of the web= 125bbftwfy
As = the area of the stiffener in contact with flange = 2Ls.ts
Fys= yield stress of the stiffener
𝑅𝑠b=𝛼𝑐𝐴𝑐𝑓𝑦
𝛼𝑐determination: 1. Estimate effective web width (act on both side the stiffener) Lw 𝐿𝑤=lesser{17.5𝑡𝑤 ⁄(𝑓𝑦/250)1/2, S/2

1. Compression area, Ac= (2Lw.tw) + (2Ls.ts)
2. Adopt the following values as per Cl 5.13.4
3. αb= web compression constant = 0.5
4. kf = web compression form factor = 1.0

iii. Le = d1

1. Moment of Inertia, 𝐼𝑐=𝑡𝑠2𝐿𝑠+𝑡𝑤3 12 +2𝐿𝑤𝑡𝑤 312
2. Radius of gyration, 𝑙𝑐= 𝐼𝑐 𝐴𝐶
3. Calculate λn = slenderness ratio = •𝜆𝑛=𝐿𝑒 𝑟 𝑘𝑓 𝑓𝑦 250=𝑑𝑑1 𝑟𝑐 𝑘𝑓 𝑓𝑦 250
4. Read the web compression member slenderness reduction factor αc from AS4100 Tb 6.3.3(3).

Minimum Area Ac= 𝐴𝑠≥0.5𝛾𝐴𝑤(1−𝛼𝑣)( 𝑉∗ /𝜙𝑉𝑢)[(S/𝑑𝑝)−(S/𝑑)2/[1+(S/𝑑𝑝)2]1/2.
𝑣= determined from Cl 5.11.5.
𝐴 𝑤= gross-sectional web area = 𝑑1𝑡𝑤d

Question 4. Combined shear and bending:

1. Locate the critical combination of beam bending moment and shear from the SFD and BMD. Check the combined bending moment and shear capacity at this critical section using either the proportional method or the interaction method.

Solution
V* ≤ ΦVv
If dp/tw ≤82/√(fy/250
Vv = Vu=Vw=0.6 fyAw
Vv= 0.6*320*22*1.61=6800.64

The critical combination is at point B which is shear =1341, and moment= 3556.

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