exercício 1
considereig, ✗ o topo da liga
150mm
14mm
14mm
I-Σ (I. + Adg)
1
Retângulos:
144,4mm
¼a
<
In;½ (14011157 + 2290/154,4-7,517 48,59106mm²
Retângulo 2:
Soma
Faz ½(14) (140 Pt 2250 (154,4-90
P = 13,45×106mm²
Círculo:
Ins = ¼# (8) "+ (* só)/154,4-204,412=19,64 ✗ 106mm"
Itot = In, + In, + In,
81,78✗ 106mm"
exercício 2
l
150 mm
150mm
soam/
25mm
%-275mm
Ar-250×50-12500mm²
Ar-300×50=15000 Km²
c
240km
¾ = 125mm
7=125×12900 +275×14000 = 206,82mm
27500
125mm
Ix- ½ (g) (2507+ 12500×1206,82-1257 + ½ (300) (50) ¾ 15000 (20682-275)'
Ir = 221, 63 ✗ 106mm²
Iy = ½ (2501/50)' + ½ 60163007
In = 115,10×10%44