Challenge

Week 5 Challenge

dtw�� = (1200 -2x50)/10 =111 > 67 ε� (ε=1)(�=1)

check for shear buckling required

2) Minimum thickness requirment of web-

dtw��< 200ε�

11100/10 = 111 < 200ε� ( 0K)

3) Shear capacity Check -

Vd=Av×fyw/√3×γmo��=��×��/3×��

= 1100x10x250/1.732x1.1 = 1443418 N = 1443 KN

0.6Vd�� = 865 KN < V = 1000 KN

Not safe in shear

4) Check for shear buckling as per IS 800: 2007 , Page-59 -

Vcr=��=A_v xxtau_B`

shear stress correspondind to web buckling

τB=fyw√3×λ2w��=��3×��2

λ2w=√fyw√3×τcr��2=��3×��

τcr=Kv×π2×E/12(1−μ2)(dtw)2��=��×�2×�/12(1-�2)(��)2

Kv�� = 5.35 when transverse stiiffners are provided only at support

E = 2x105N/mm2105�/��2 , μispoisson'sratio=0.3

`nbsp; τcr�� = 5.35x π2�2 x 2x105105/ 12 ( 1- 0.320.32) (110010)2(110010)2

= 79.84

λw = √fyw√3×τcr/3×tcr

=√2501.732×79.842501.732×79.84 = 1.3445

τB = 250/1.732x 1.344521.34452 = 79.849

Vcr = 1100x10 x 79.849 = 878339 N = 878.339 < 1000 KN

The web shear force corresponding to the web buckling is less than maximum Factored shear force , hence not safe

Result- Thus it is concluded that web is not adequate to carry a factored shear force of 1000 KN.

dtw�� = (1500-2x30)/12 = 120 > 67ε�

shear buckling check requred

2) Minimum web thickness requirment-

dtw�� 120 < 200 ε� (ε=1)(�=1) ( OK)

3) Shear capacity Check -

Vd=Av×fyw/√3×γmo��=��×��/3×��

= 1440x12x250/1.732x1.1 = 2267478.47 N = 2267. 47 KN

0.6Vd�� = 1360.48 KN < V = 2000 KN ( Not safe in shear)

4) Check for shear buckling as per IS 800: 2007 , Page-59

Vcr=��=A_v xxtau_B`

shear stress correspondind to web buckling

Ship work like this
in your own course.