Exercicio 2 Laje Alveolar

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EXERCÍCIO 2 – RESOLUÇÃO 
 
 
i) Características geométricas 
 2,679.1
4
20526125A
2
c =
×pi
×−×= cm2 
13
2
26yy supinf === cm 
4,813.143
64
205
12
26125I
43
=
×pi
×−
×
= cm4 
 
 
ii) Carga permanente 
 
p.p. : 25168,0 × = 4,20 kN/m 
capa : 2525,105,0 ×× = 1,56 kN/m 
rev. : 25,125,1 × = 1,56 kN/m 
 –––––––––––––– 
 g = 7,32 kN/m 
 
 
iii) Tensões devidas à protensão 
 
09,10009,110Ap =×= cm2 
7,199.19,11809,10AP pp =×=σ×= kN 
5,95,313e =−= cm 
316,0
4,813.143
135,97,199.1
2,679.1
7,199.1
psup, +=
××
+−=σ kN/cm2 
745,1
4,813.143
135,97,199.1
2,679.1
7,199.1
pinf, −=
××
−−=σ kN/cm2 
Capítulo 1 – Exercício 2 
 
 
 
a) q = ? para l = 9,50 m e 0,1inf +=σ MPa: 
 
800
950)q25,132,7(
8
10)q25,1g(M
222
qg
××+
=
×××+
=
−
+
l
 
)q25,132,7(125,128.1M qg ×+×=+ 
4,813.143
13M qg
qginf,
×
=σ
+
+ 
1,0pinf,qginf,inf +=σ+σ=σ + kN/cm2 
1,0745,1
4,813.143
13)q25,132,7(125,128.1
+=−
××+×
 
 
62,8q = kN/m2 
 
b) l = ? para q = 12 kN/m2 e 0,1inf +=σ MPa: 
 
32,221225,132,7qg =×+=+ kN/m 
8423,358
1032,22M
222
qg
ll
=
××
=
−
+ ( l em cm) 
4,813.143
13M qg
qginf,
×
=σ
+
+ 
1,0pinf,qginf,inf +=σ+σ=σ + kN/cm2 
1,0745,1
4,813.1438423,35
132
+=−
×
×l
 
 
3,855=l cm 553,8= m 
 
c) No apoio: 
 
psup,psup,qgsup,supmax,t σ=σ+σ=σ=σ + 
316,0sup +=σ kN/cm2 16,3= MPa (tração) 
 
 
 
	Capitulo 1 Exercicio 2 Laje Alveolar
	Capitulo 1 Exercicio 2