Universitatea Tehnica a Moldovei
Facultatea Urbanism si Arhitectura
Departament: Design și Design Urban
MEMORIU EXPLICATIV
La disciplina Reconstructia si Consolidarea Cladirilor
Tema: ”Cladire de locuit tip P+4E in
reconstructie cu o supraetajare”
A elaborat: st.gr. U-141 Botezatu Dorina
A verificat: Turcanu Iurie
Chisinău 2018
0
Date initiale:
H et = 2,7m
Rsol = 2,0kg / cm 2
Latimea fundatiei existente:
Sub peretii interior portanti-1,4m (pe axele 2;3)
Sub peretii exterior portanti-1,2m (pe axele 1;4)
Sub peretii autoportanti-0,8m (pe axele A;B;C)
Calculul sarcinii ce actioneaza pe fundatia existenta cu latimea 800
Pf = Pf .exist . + Pper .subsol + Pper .etaj + Pper .acop . + Pacop ;
1. Pf .exist . = b f �h f �g f �ks = 0.8 �0.3 �2.0 �1.1 = 0.528t / m s
2. Pper .subsol . = bper .subsol �hper .subsol �g �ks = 0.4 �1.95 �2 �1.2 = 1.872t / m
3.Pper .etaj . = bper .etaj. �hper .etaj �g �ks �nr. nivele = 0.4 �2.7 �2 �1.2 �6 = 15.552t / m
4.Pper .acop. = bper .acop �hper .acop �g �ks = 0.4 �0.25 �2 �1.2 = 0.24t / m
5. Pac = Pelem. port . + Pelem.inchid . + Pzap ;
Pelem.inchid = 0.07 �ks �C = 0.07 �1.2 �6.3 = 0.53t / m
�
Pelem. port . = 0.7 �Pelem.inchid . = 0.7 �0.53 = 0.37t / m
�
Pzap. = Pzap �ks �C = 0.05 �1.4 �6.3 = 0.44t / m
�
Pac = 0.53 + 0.37 + 0.37 = 1.34t / m
�
Pf = 0.66 + 1.872 + 15.552 + 0.24 + 1.34 = 19.664t / m
Calculul grinzii transversale
1
Pf �103 19664
b f .c. =
� = = 98.32cm = 120cm
Rsol �100 2.0 �100
Pf �103 19664
q=
� = = 1.639kg / cm 2 = 16.39t / m 2
b f .c. �100 12000
qg .tr . = q �a = 16.39 �1.7 = 27.86t / m
�
�gtr . = b �h = 400 �300
h0 = 300 - 30 = 270mm = 27cm
�
q �l 2 27.86 �0.42
M g .t . =
� = = 2.228t �m
2 2
M �105 222800
�A0 = = = 0.085 < 0.4
b �h0 �RB 40 �27 2 �90
2
h = 0.955
M �105 222800
Fs =
� = = 2.37cm2
h �h0 �RA 0.955 �27 �3650
Admitem 2�14 Fs = 3.08cm 2
Etriere �6 pas 50
Calculul grinzii longitudinale
qg .long . = q �bg .long . = 16.39 �0.2 = 3.278t / m
�
q �l 2 3.278 �1.7 2
M sup =
� = = 0.789t �m
12 12
M �105 78900
�A0 = = = 0.06
b �h0 �RB 20 �27 2 �90
2
h = 0.969
M �105 76100
Fs =
� = = 0.797cm 2
h �h0 �RA 0.969 �27 �3650
Admitem 2�10 Fs = 1.57cm 2
Etriere �6 pas 200
Calculul sarcinii ce actioneaza pe fundatia existenta cu latimea 1200
2
Pf = Pf .exist . + Pper .subsol + Pper .etaj + Pplanseu + Pper .acop. + Pacop ;
1. Pf .exist . = b f �h f �g f �ks = 1.2 �0.3 �2.5 �1.1 = 0.99t / m
2. Pper .subsol . = bper .subsol �hper .subsol �g �ks = 0.4 �1.95 �2 �1.2 = 1.872t / m
3. Pper .etaj. = bper .etaj . �hper .etaj �g �ks �nr. nivele = 0.4 �2.7 �2 �1.2 �6 = 15.552t / m
4. Pplanseu = Pppl + Piz + Psn + Ppard + Pprov ;
Pppl = q pl �ks �C �nr. niv. = 0.3 �1.1�3.0 �6 = 5.94t / m
�
Piz = hiz �g iz �k s �C �nr. niv. = 0.05 �0.8 �1.2 �3.0 �6 = 0.864t / m
�
Psn = hsn �g sn �k s �C �nr. niv. = 0.03 �1.8 �1.2 �3.0 �6 = 1,166t / m
�
Ppard = hpard �g pard �ks �C �nr. niv. = 0.01�0.8 �1.2 �3.0 �5 = 0.144t / m
�
Pprov = q prov �ks �C �nr. niv. = 0.15 �1.2 �3.0 �6 = 3, 24t / m
�
PPl = Pppl + Piz + Psn + Ppard + Pprov = 5.94 + 0.864 + 1,166 + 0.144 + 3, 24 = 11.354t / m
�
4. Pper .acop. = b per .acop �hper .acop �g �ks = 0.4 �0.25 �2 �1.2 = 0.24t / m
5. Pac = Pelem. port . + Pelem.inchid . + Pzap t / m
Pelem.inchid = 0.07 �ks �C = 0.07 �1.2 �3.15 = 0.265t / m
�
Pelem. port . = 0.7 �Pelem.inchid . = 0.7 �0.265 = 0.186t / m
�
Pzap. = Pzap �ks �c = 0.05 �1.4 �3.15 = 0.221t / m
�
Pac = 0.265 + 0.186 + 0.221 = 0.672t / m
�
Pf = Pf .exist . + Pper .subsol + Pper .etaj + Pplanseu + Pper .acop. + Pacop = 0.99 + 1.872 + 15.552 +
+11.354 + 0.24 + 0.672 = 30.68
Calculul grinzii transversale
3
Pf �103 30680
b f .c . =
� = = 153.4cm = 160cm
Rsol �100 2.0 �100
Pf �103 30680
q=
� = = 1.92kg / cm 2 = 19.2t / m 2
b f .c. �100 16000
qg .tr . = q �a = 19.2 �1.7 = 32.64t / m
�
�gtr . = b �h = 400 �300
h0 = 300 - 30 = 270mm = 27cm
�
q �l 2 32.64 �0.62
M=
� = = 5,88t �m
2 2
M �105 588000
�A0 = = = 0.224
b �h0 �RB 40 �27 2 �90
2
h = 0.8714
M �105 588000
Fs =
� = = 6.85cm 2
h �h0 �RA 0.8714 �27 �3650
Admitem 3�18 Fs = 7.63cm 2
Etriere �6 pas 50
Calculul grinzii longitudinale
qg .long . = q �bg .long . = 19.2 �0.2 = 3.84t / m
�
q �l 2 3.84 �1.7 2
M sup =
� = = 0.925t �m
12 12
M �105 92500
�A0 = = = 0.07 < 0.4
b �h0 �RB 20 �27 2 �90
2
h = 0.9639
M �105 92500
Fs =
� = = 0.974cm 2
h �h0 �RA 0.9639 �27 �3650
Admitem 2�10 Fs = 1.57cm 2
Etriere �6 pas 200
Calculul sarcinii ce actioneaza pe fundatia existenta cu latimea 1500
4
Pf = Pf .exist . + Pper .subsol + Pper .etaj + Pplanseu ;
1. Pf .exist . = b f �h f �g f �ks = 1.5 �0.3 �2.5 �1.1 = 1.238t / m
2. Pper .subsol . = bper .subsol �hper .subsol �g �ks = 0.4 �1.95 �2 �1.2 = 1.872t / m
3. Pper .etaj . = bper .etaj . �hper .etaj �g �ks �nr. nivele = 0.4 �2.7 �2 �1.2 �6 = 15.552t / m
4. Pplanseu = Pppl + Piz + Psn + Ppard + Pprov ;
Pppl = q pl �ks �C �nr. niv. = 0.3 �1.1�4.65 �6 = 9.207t / m
�
Piz = hiz �g iz �k s �C �nr. niv. = 0.05 �0.8 �1.2 �4.65 �6 = 1.339t / m
�
Psn = hsn �g sn �ks �C �nr. niv. = 0.03 �1.8 �1.2 �4.65 �6 = 1.807t / m
�
Ppard = hpard �g pard �k s �C �nr. niv. = 0.01�0.8 �1.2 �4.65 �5 = 0.223t / m
�
Pprov = q prov �ks �C �nr. niv = 0.15 �1.2 �4.65 �6 = 5.022t / m
�
PPl = Pppl + Piz + Psn + Ppard + Pprov = 9.207 + 1.339 + 1.807 + 0.223 + 5.022 = 17.598t / m
�
Pf = Pf .exist . + Pper .subsol + Pper .etaj + Pplanseu = 1.238 + 1.872 + 15.552 + 17.598 = 35.468
Calculul grinzii transversale
5
Pf �103 35468
b f .c . =
� = = 177.34cm = 180cm
Rsol �100 2.0 �100
Pf �103 35468
q=
� = = 1.97kg / cm = 19.7t / m
b f .c. �100 180 �100
qg .tr . = q �a = 19.7 �1.7 = 33.49t / m
�
�gtr . = b �h = 400 �300
h0 = 300 - 30 = 270mm = 27cm
�
q �l 2 33.49 �0.7 2
M=
� = = 8.205t �m
2 2
M �105 820500
�A0 = = = 0.313
b �h0 �RB 40 �27 2 �90
2
h = 0.8058
M �105 820500
Fs =
� = = 9.493cm 2
h �h0 �RA 0.877 �27 �3650
Admitem 2�25 Fs = 9.82cm 2
Etriere �8 pas 50
Calculul grinzii longitudinale
qg .long . = q �bg .long . = 19.7 �0.15 = 2.955t / m
�
q �l 2 2.955 �1.7 2
�M sup = = = 0.712t �m
12 12
M �105 71200
�A0 = = = 0.027
b �h0 �RB 40 �27 2 �90
2
h = 0.9865
M �105 71200
Fs =
� = = 0.732cm 2
h �h0 �RA 0.9865 �27 �3650
Admitem 2�12 Fs = 2.26cm 2
Etriere �6 pas 200
