Page 12 - Mechanics of Structures – Chapter 1
P. 12
Mechanics of Civil Engineering Structures – Chapter 5
4. To calculate τ 2 and τ 6
Areas considered is above section 2 @ below section 6
A = 200 x 25 = 5000 mm
2
y = distance of C.G of area considered from N.A
= (100 – 25/2) = 87.5 mm
b = width of section (consider flange) = 200 mm
. .
Substitute, τ 2 =
.
= ( 12500)(5000)(87.5) = 0.33 N/mm
2
6
((84.11 10 )(200))
5. To calculate τ 3 and τ 5
Areas considered is above section 3 @ below section 5
2
A = 200 x 25 = 5000 mm
y = distance of C.G of area considered from N.A
= (100 – 25/2) = 87.5 mm
b = width of section (consider web) = 25 mm
. .
Substitute, τ 3 =
.
( 12500)(5000)(87.5)
2
= = 2.6 N/mm
((84.11 10 )(25))
6
6. To calculate τ 4 @ τ max at Neutral Axis
τ4 = τ3 + additional stress due to web portion
25
where τ3 = stress due to flange 75
2
A = 75 x 25 = 1875 mm
75
y = = 37.5 mm
2
b = 25 mm
NHO JKA PSAS Page 12
