QUESTION 2 Figure 2 shows the cross section of a jetty toget

QUESTION 2 Figure 2 shows the cross section of a jetty together with the unfactored loads applied to the columns. Spanning between the columns is a transom (standard hot-rolled section 180UB 22.2) which has been fully welded to the columns providing partial restraint. The transom has services attached to it and the connection of these services can be assumed to be fully restrained. Furthermore, the unfactored dead load due to the services and the factored horizontal thermal load of 100 kN are shown in Figure 2. The bearing plate situated under the services is strong enough so as to not deform in bending. Assuming G300 steel, undertake the following design checks a) Verify the adequacy of transom for the combined thermal and services loading. Assume that there are no moment amplifications. Basic design capacities such as 0N.Nc.M, and Mb can be taken directly from ASI tables. (26 marks) M* for a simply supported bean under a uniform load = WL18, M* (hogging & sagging) for a fixed ended beam with a central point load is - PL/8 Note: b) The column is made of an imported non-standard hot-rolled compact I-section and its section and material properties are given as follows (A 4010 mm2, f,- 300 MPa, r 105 mm ry 33.4 mm). Calculate the member capacity of the column for the compression loads only, ignoring any shearing loads transferred from the transom. Is the column structurally adequate? Assume fixed ends at the location of the transom and base of columns, while the (14 marks) top of the columns can be considered to be pinned (Total 40 Marks) DL 30 kN LL 50 kN Transom 18OUB 22.2 30 kN 2500 100 kN 200 1750 6000 SECTIONA-A Column TYPICAL CROSS SECTION

Solution

Ans:

b)

Key Given points in que are as follows:

Section Properties of sections: As=4010mm^2, fy=300Mpa, r_x=105mm and r_y=33.4mm

Ignoring any shearing loads transferred from the transom

Assume fixed ends at the location of the transom and base of column

Top of column can be consider to be pinned

Free Body Diagram for Given frame:

         DL 30kN and LL 50kN

Solution:

Compression capacity of column: Pd=As*fcd

Where Pd, Design compression capacity of member

              As, Effective sectional Area of section

              Fcd, Design compressive stress

In above formula As=4010mm² ………………………………………………………(Key Given Point 1)

Calculation of Fcd,

fcd=(fy/rmo)/x

where, fy=300Mpa=300N/mm²………………………………………………………..(Key Given Point 1)

              rmo, Partial safety factor for material, Assume 1.1

              x= +SQT[ ²-²]

where =0.5[1-(-0.2)+ ²

             =SQT[fy/fcc]

             fcc=²*E/(Max. Effective slenderness ratio²)

Effective Slenderness ration of column is depends on Unsupported length of member on both axis, Radius of gyration on both axis and support condition of column

Calculation of Effective slenderness ration of column on both axis:

About x-x:

Unsupported length of column=6000mm………………..(Key Given point 3, Assume fixed ends at location of trasom)

Rx=105mm…………………………………………………...(Key Given Point 1)

Since the boundary condition as both end fixed,

Effective length=0.65*Unsupported length

                             =0.65*6000

                             =3900mm

Effective Slenderness ratio=3900/105=37.14

Assume buckling class, b

About y-y:

Unsupported length of column=6000+2500=8500mm

ry=33.4mm…………………………………………………………....................(Key Given Point 1)

Since the boundary condition as one end fixed and one end pinned……….( Key Given Point 4)

Effective length=0.8*Unsupported length

                             =0.8*8500

                             =6800mm

Effective Slenderness ratio=6800/33.4=203.59

Assume buckling class, c

Therefore, fcc=²*(2*10⁵)/(203.59²)=47.62 N/mm²

=SQT(250/47.62)=2.29

=0.5[1-0.49*(2.29-0.2)+(2.29²)=2.61…………........................………..( =0.49 for buckling class c)

x= 2.61+SQT[ 2.61²-2.29²]=3.86

fcd=(300/1.1)/3.86=70.615 N/mm²

Compression capacity of column: Pd=As*fcd=4010*70.615=283.16 kN

Calculation of Actual Load on Column:

Since the shear transfer from transom asked to ignore, load from transom of 30 kN will not come on column.

Fixed ends at the location of transom, Hence 100 kN load will not affect column

Hence the total loads on column are DL and LL

Assume Partial safety factor for loads= 1.5

Factored load on column= 1.5* (30+50)=120 kN

Conclusion:

Applied load on column is 120 kN which is less than capacity of column which is equal to 283.16 kN

Hence the column is structurally adequate to take loads