Page 9 - Mechanics of Structures – Chapter 1
P. 9
Mechanics of Civil Engineering Structures – Chapter 5
# Thus τ 1 = τ 5 = 0 (because no area)
90 mm
4. To calculate τ 2 – term S and I are same
10
2
A = 90 x 10 = 900 mm
y = distance of C.G of area considered from N.A y 26.18
Neutral Axis
= (90 - 63.82) - 10/5 = 21.18 mm
b = width of section = 90 mm
Substitute, τ 2 = . .
.
(40 10 )(900)(21.18)
3
2
= = 6.62 N/mm
((1.28 10 )(90)
6
5. To calculate τ 3 – term S and I are same
A = same as that for τ 2 = 900
y = distance of C.G of area considered from N.A 3
3
= (90 - 63.82) - 10/5 = 21.18 mm
b = width of section = 10 mm
. .
Substitute, τ 3 = .
3
2
= (40 10 )(900)(21.18) = 59.56 N/mm
6
((1.28 10 )(10)
6. To calculate τ 4 – term S and I are same and consider
area under N.A
2
A = 63.82 x 10 = 638.2 mm 31.91
63.82 63.82
y = = 31.91 mm
2
b = width of section = 10 mm
Substitute, τ 3 = . .
. 10
(40 10 )(638.2)(31.91)
3
2
= ((1.28 10 )(10) = 63.64 N/mm
6
NHO JKA PSAS Page 9
