It's necessary to go to the supervision center
Publish: 2021-03-27 18:34:40
1. Design of integral one-way slab rib floor
I. design data
1. The floor plan is as shown in Figure 1, with wall thickness of 240mm, structure transverse length L1 = 26.4m, structure longitudinal length L2 = 27m, and the staircase is located outside the floor plan, so it is not considered. The floor adopts integral one-way slab rib structure
2. The building is located in non seismic area
3. The safety level of buildings is grade II
4. Structural environment category
5. Building material grade: C25 concrete for beam and slab
steel bar: steel bar in slab, steel bar in beam, secondary structural steel bar, secondary stressed steel bar in beam and primary stirrup
6. Load: the gravity density of reinforced concrete is 25kn / m3,
the floor surface is 20 mm terrazzo, the self weight is 0.65 kn / m2,
the beam slab ceiling is 15 mm plastered with mixed mortar, the gravity density is 20 kn / m3,
the standard value of floor live load is 2.5 kn / m2
7. Structural plane layout and preliminary estimated size:
the supporting length of the plate is 120mm, the supporting length of the beam is 240mm, and the supporting length of the main beam is 240mm
plate thickness: H = 100 mm
secondary beam: height: H = 450 mm, width: B = 220 mm, spacing: L1 = 3.0 M
main girder: H = 750mm, B = 250mm, spacing L2 = 6.6m
column: B * H = 400mm * 400mm, column grid size is 6.6m * 9.0m
8. Application requirements: the deformation and crack width of beam and slab are not considered in this design
9. Adopted specifications: Code for design of hydraulic concrete structures (SL / t191-96)
code for load of building structures (GBJ9-87)
II. Floor structure layout
the main beam span is 9m, the section size b * H = 250mm * 750mm
the secondary beam span is 6m, the section size b * H = 220mm * 450mm
the slab span is 2m, Thickness h = 80m
III. design of slab (internal force is calculated according to plastic internal force redistribution)
1. Load
design value of dead load:
terrazzo surface 1.05 * 0.65 = 0.683 kn / m2
dead weight of slab 1.05 * 25 * 0.1 = 2.625 kn / m2
plastering at slab bottom 1.05 * 20 * 0.015 = 0.315 kn / m2
total g = 3.623 kn / m2
design value of live load: q = 1.2 * 2.5 = 3 0 kn / m2
P = G + q = 3.623 + 3.0 = 6.623 kn / m2
2. Calculation diagram
3. Internal force calculation
plate thickness h = 100 mm, Secondary beam B * H = 220 mm * 450 mm
calculation span: side span L01 = ln + H / 2 = 3000-120-220 / 2 + 100 / 2 = 2820 mm
L01 = ln + A / 2 = 3000-220 / 2-120 + 120 / 2 = 2830 mm
choose the smaller L01 = 2820 mm
middle span L02 = ln = 3000-220 = 2780 mm
span difference (l01-l02) / L02 = (2820-2780) / 2780 = 1.4% & lt; The bending moment calculation table of the slab:
the first inner support in the middle of the side span and the middle support in the middle of the middle span
calculation span / M 2820 2820 2780 2780
bending moment coefficient 1 / 11 - 1 / 11 1 / 16 - 1 / 14
bending moment kn · m 4.788 - 4.788 3.199 - 3.656
4. Reinforcement calculation
take a = 25, B = 1000mm, H = 100mm, H0 = H-A = 100-25 = 75mm
FC = 12.5n/mm2, FY = 210n / mm2
calculation table of slab reinforcement:
location ① ~ ② and ④ ~ ⑤ slab ② ~ ③ and ③ ~ ④ slab
section side span middle first support middle middle support middle support side span middle first support middle support middle support middle support middle span
m / kn · m 4.788 - 4.788 3.199 - 3.565 4.788 - 4.788 2.559 - 2.925
m / kn · m 4.788 - 4.788 3.199 - 3.565 4.788 2.559 - 2.925 α s=rdM/(fcbh02) 0.082 0.082 0.055 0.062 0.082 0.082 0.044 0.050
ζ= 1-√(1-2* α s) 0.086 0.086 0.057 0.064 0.086 0.086 0.045 0.051
As=fcb ζ H0 / FY (mm2) 383.93 383.93 254.46 285.71 383.93 383.93 200.89 227.68
reinforcement φ 6@70 φ 6@70 φ 6@100 φ 6@100 φ 6@70 φ 6@70 φ 6@120 φ 6@120
actual matching area (mm2) 404 404 283 283 404 404 236 ζ 086 & lt; ζ b=0.614;
the minimum reinforcement area selected in the table is as = 217 mm2
check the minimum reinforcement ratio ρ= As/(bh0)=217/1000*75=0.29%> ρ Load calculation
dead load plate transfer load 3.623 * 3 = 10.869 kn / M
side plastering 1.05 * (450-100) * 3 * 20 * 10-3 * 0.015 = 0.331 kn / M
dead weight 1.05 * (450-100) * 10-3 * 25 * 0.22 = 2.201 kn / m
secondary beam/ >Total g = 13.221kn/m
live load q = 3 * 3 = 9 kn / M
total load p = G + q = 13.221 + 9 = 22.221 kn / M
2. Calculation diagram
calculation span: side span ln = 6600-240 / 2-250 / 2 = 6355 mm
L01 = (6600-240 / 2-250 / 2) * 1.025 = 6475 mm
L01 = ln + A / 2 = 6355 + 240 / 2 =6514 mm
so the smaller one is L01 = 6475 mm
middle span L02 = 6600-250 = 6350 mm
3. Internal force calculation
span difference (l01-l02) / L02 * 100% = (6475-6350) / 6350 * 100% = 1.97% & lt; 10%
calculated as equal span continuous beam
secondary beam bending moment calculation table (M= α Mbpl02)
the first support in the side span of the section, the middle support in the middle span and the middle inner support in the middle span
α MB 1 / 11 - 1 / 11 1 / 16 - 1 / 14
lo (mm) 6475 6475 6350 6350
m (KN / M) 84.69 - 84.69 56.00 - 64.00
secondary beam simple force calculation table (V= α Vbpln)
ar BL br CL CR
α VB 0.450.600.5500.550.55
ln (mm) 6355 6355 6350 6350 6350
V (KN) 63.55 84.73 77.61 77.61 77.61
4. Reinforcement calculation
(1) bending calculation of normal section of secondary beam
midspan section is calculated as T-section, flange width is
side span BF '= L01 / 3 = 6475 / 3 = 2158 mm & lt; B + Sn = 220 + (3000-120-125) = 2975 mm
mid span BF '= L02 / 3 = 6350 / 3 = 2117 mm & lt; b+Sn=220+2780=3000 mm
hf’=100 mm,h=450 mm,a=35 mm, h0=450-35=415 mm
fc bf’hf’(h0- hf’/2)=12.5*2117*100*(415-100/2)*10-6=965.88 kN·m
γ dM=1.2*84.69=101.63 kN·m
fc bf’hf’(h0- hf’/2)> γ Therefore, the midspan section of the secondary beam is calculated according to the first type of T-section, and the support is calculated according to the rectangular section
b = 220 mm, FC = 12.5 n / mm2, FY = 310 n / mm2
calculation table of bending resistance of normal section:
the first inner support in the middle of side span, the middle support in the middle of middle span
m (kn · m) 84.69-84.69 56.00-64.00
b (mm) 2158 220 2117 220
m (KN · m) 84.69-84.69 56.00-64.00 α s=rdM/(fcbh02) 0.022 0.215 0.015 0.162
ζ 0.022 0.245 0.015 0.178
As=fcbf’ ζ H0 / FY 794 902 509 763
reinforcement 3 Φ 18 4 Φ 18 2 Φ 18 3 Φ 18
actual matching area (mm2) 763 1018 509 763
of which ζ= X / H0 is less than 0.35, which is in line with the condition of plastic internal force redistribution
(2) calculation of shear reinforcement of inclined section of secondary beam
HW / b = (H0 - HF ') / b = (415-100) / 220 = 1.432 & lt; 4
1/ γ d0.25fcbh0=1/1.2*0.25*12.5*220*415=237.76kN> Vmax = 84.73 kn
meet the requirements of section size
shear calculation table of inclined section:
section ar BL br CL CR
V (KN) 63.55 84.73 77.61 77.61 77.61
RDV (KN) 76.26 101.68 93.13 93.13 93.13
VC = 0.07fcbh0 79.89 79.89 79.89 79.89 φ 6
Asv(mm2) 56.6
S(mm) 214 283 466 466 466
Smax 300 200 200 200 200
ρ Svmin 0.12%
actual reinforcement φ 6@200
v. design of main beam (calculated according to elastic theory)
1. Load
dead load secondary beam transfer load: 13.154 * 6.6 = 86.816 kn
dead weight of main beam: 1.05 * (0.75-0.1) * 0.25 * 25 * 3 = 12.797 kn
plastering of main beam: 1.05 * (0.75-0.1) * 0.015 * 2 * 20 * 3 = 1.229 kn
total g = 100.842 kn
live load q = 9 * 6.6 = 59.4 kn
Total load p = G + q = 100.842 + 59.4 = 160.242 kn
2. Calculation diagram
calculation span: side span ln = 9000-240 / 2-400 / 2 = 8680mm
l0 = ln + A / 2 + B / 2 = 8680 + 240 / 2 + 400 / 2 = 9000mm
l0 = 1.025 * ln + B / 2 = 1.025 * 8680 + 400 / 2 = 9097mm
the smaller l0 = 9000mm
middle span B = 400 & lt; 0.05lc = 0.05 * 9000 = 450mm
l0 = LC = 9000mm
3. Internal force calculation
span difference is zero, so it is calculated according to equal span continuous beam
bending moment diagram 1 (KN & # 8226; m)
bending moment diagram 2 (KN & # 8226; m)
bending moment diagram 3 (KN & # 8226; m)
bending moment diagram 4 (KN & # 8226; m)
bending moment diagram 5 (KN & # 8226; m)
shear diagram 1 (KN)
shear diagram 2 (KN)
Shear Diagram 3 (KN)
shear diagram 4 (KN)
shear diagram 5 (KN)
bending moment envelope diagram:
① + ②
① + ③
① + ④
& 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; ① + 5
moment envelope diagram
shear envelope diagram:
shear envelope diagram
4. Reinforcement calculation
(1) bending capacity calculation of normal section
midspan section is calculated as T-beam, flange width is:
midspan: BF '= l0 / 3 = 9000 / 3 = 3000mm & lt; B + Sn = 250 + (9000-250) = 9000mm
side span: BF '= l0 / 3 = 8880 / 3 = 2960mm & lt; b+Sn=250+(9000-120-250/2)=9005mm
hf’=100mm,h0=750-60=690mm
fcbf’hf’ (h0-hf’/2)=12.5*3000*100*(690-100/2)*10-6=2400kN
> RDM = 1.2 * 438.07 = 525.684kn · m
so it is calculated according to the first type T-section of mid span section, and the support is calculated according to rectangular section
b = 250mm, FC = 12.5n/mm2, FY = 310n / mm2
because the beam is connected with the support as a whole, The calculated bending moment of the bearing B is as follows: < br /
MB = 9475; MB 9475; MB ┃ MB ┃ MB ┃ MB ┃ MB ┃ MB ┃-0.02lnand VBR ┃┃ - 441.01-441.01 9475, - 0.025 ┃┃┃ MB ┃ MB ┃ MB 9475; MB 9475;┃┃┃┃\\\\/ FY 2479 2911 1344
reinforcement 6 Φ 22+1 Φ 18 6 Φ 22+3 Φ 18 4 Φ 22
x values were less than ζ bh0=0.544*670=364mm
ρ= As/(bh0)=1344/(250*670)=0.802%> ρ Min = 0.12%
meet the requirements< (2) shear capacity calculation of inclined section
b = 250mm, H0 = 750-80 = 670, FC = 12.5n/mm2, FY = 310n / mm2, HW = h0-h, f '= 670-100 = 570mm
HW / b = 570 / 250 = 2.28 & lt; 4.0
0.25fcbh0/rd=0.25*12.5*250*670=436.20kN> 209.24kn
the section size meets the requirements
shear reinforcement calculation table
section side support a B support left B support right
V (KN) 122.65 - 209.24 174.5
RDV / (fcbh0) 0.070 0.120 0.100
stirrup method construction reinforcement calculation reinforcement calculation reinforcement calculation reinforcement
primary stirrup double limb φ 8
ASV (mm2) 100.6
s 330 160 280
Smax 250
actual stirrup φ@ one hundred and sixty
I. design data
1. The floor plan is as shown in Figure 1, with wall thickness of 240mm, structure transverse length L1 = 26.4m, structure longitudinal length L2 = 27m, and the staircase is located outside the floor plan, so it is not considered. The floor adopts integral one-way slab rib structure
2. The building is located in non seismic area
3. The safety level of buildings is grade II
4. Structural environment category
5. Building material grade: C25 concrete for beam and slab
steel bar: steel bar in slab, steel bar in beam, secondary structural steel bar, secondary stressed steel bar in beam and primary stirrup
6. Load: the gravity density of reinforced concrete is 25kn / m3,
the floor surface is 20 mm terrazzo, the self weight is 0.65 kn / m2,
the beam slab ceiling is 15 mm plastered with mixed mortar, the gravity density is 20 kn / m3,
the standard value of floor live load is 2.5 kn / m2
7. Structural plane layout and preliminary estimated size:
the supporting length of the plate is 120mm, the supporting length of the beam is 240mm, and the supporting length of the main beam is 240mm
plate thickness: H = 100 mm
secondary beam: height: H = 450 mm, width: B = 220 mm, spacing: L1 = 3.0 M
main girder: H = 750mm, B = 250mm, spacing L2 = 6.6m
column: B * H = 400mm * 400mm, column grid size is 6.6m * 9.0m
8. Application requirements: the deformation and crack width of beam and slab are not considered in this design
9. Adopted specifications: Code for design of hydraulic concrete structures (SL / t191-96)
code for load of building structures (GBJ9-87)
II. Floor structure layout
the main beam span is 9m, the section size b * H = 250mm * 750mm
the secondary beam span is 6m, the section size b * H = 220mm * 450mm
the slab span is 2m, Thickness h = 80m
III. design of slab (internal force is calculated according to plastic internal force redistribution)
1. Load
design value of dead load:
terrazzo surface 1.05 * 0.65 = 0.683 kn / m2
dead weight of slab 1.05 * 25 * 0.1 = 2.625 kn / m2
plastering at slab bottom 1.05 * 20 * 0.015 = 0.315 kn / m2
total g = 3.623 kn / m2
design value of live load: q = 1.2 * 2.5 = 3 0 kn / m2
P = G + q = 3.623 + 3.0 = 6.623 kn / m2
2. Calculation diagram
3. Internal force calculation
plate thickness h = 100 mm, Secondary beam B * H = 220 mm * 450 mm
calculation span: side span L01 = ln + H / 2 = 3000-120-220 / 2 + 100 / 2 = 2820 mm
L01 = ln + A / 2 = 3000-220 / 2-120 + 120 / 2 = 2830 mm
choose the smaller L01 = 2820 mm
middle span L02 = ln = 3000-220 = 2780 mm
span difference (l01-l02) / L02 = (2820-2780) / 2780 = 1.4% & lt; The bending moment calculation table of the slab:
the first inner support in the middle of the side span and the middle support in the middle of the middle span
calculation span / M 2820 2820 2780 2780
bending moment coefficient 1 / 11 - 1 / 11 1 / 16 - 1 / 14
bending moment kn · m 4.788 - 4.788 3.199 - 3.656
4. Reinforcement calculation
take a = 25, B = 1000mm, H = 100mm, H0 = H-A = 100-25 = 75mm
FC = 12.5n/mm2, FY = 210n / mm2
calculation table of slab reinforcement:
location ① ~ ② and ④ ~ ⑤ slab ② ~ ③ and ③ ~ ④ slab
section side span middle first support middle middle support middle support side span middle first support middle support middle support middle support middle span
m / kn · m 4.788 - 4.788 3.199 - 3.565 4.788 - 4.788 2.559 - 2.925
m / kn · m 4.788 - 4.788 3.199 - 3.565 4.788 2.559 - 2.925 α s=rdM/(fcbh02) 0.082 0.082 0.055 0.062 0.082 0.082 0.044 0.050
ζ= 1-√(1-2* α s) 0.086 0.086 0.057 0.064 0.086 0.086 0.045 0.051
As=fcb ζ H0 / FY (mm2) 383.93 383.93 254.46 285.71 383.93 383.93 200.89 227.68
reinforcement φ 6@70 φ 6@70 φ 6@100 φ 6@100 φ 6@70 φ 6@70 φ 6@120 φ 6@120
actual matching area (mm2) 404 404 283 283 404 404 236 ζ 086 & lt; ζ b=0.614;
the minimum reinforcement area selected in the table is as = 217 mm2
check the minimum reinforcement ratio ρ= As/(bh0)=217/1000*75=0.29%> ρ Load calculation
dead load plate transfer load 3.623 * 3 = 10.869 kn / M
side plastering 1.05 * (450-100) * 3 * 20 * 10-3 * 0.015 = 0.331 kn / M
dead weight 1.05 * (450-100) * 10-3 * 25 * 0.22 = 2.201 kn / m
secondary beam/ >Total g = 13.221kn/m
live load q = 3 * 3 = 9 kn / M
total load p = G + q = 13.221 + 9 = 22.221 kn / M
2. Calculation diagram
calculation span: side span ln = 6600-240 / 2-250 / 2 = 6355 mm
L01 = (6600-240 / 2-250 / 2) * 1.025 = 6475 mm
L01 = ln + A / 2 = 6355 + 240 / 2 =6514 mm
so the smaller one is L01 = 6475 mm
middle span L02 = 6600-250 = 6350 mm
3. Internal force calculation
span difference (l01-l02) / L02 * 100% = (6475-6350) / 6350 * 100% = 1.97% & lt; 10%
calculated as equal span continuous beam
secondary beam bending moment calculation table (M= α Mbpl02)
the first support in the side span of the section, the middle support in the middle span and the middle inner support in the middle span
α MB 1 / 11 - 1 / 11 1 / 16 - 1 / 14
lo (mm) 6475 6475 6350 6350
m (KN / M) 84.69 - 84.69 56.00 - 64.00
secondary beam simple force calculation table (V= α Vbpln)
ar BL br CL CR
α VB 0.450.600.5500.550.55
ln (mm) 6355 6355 6350 6350 6350
V (KN) 63.55 84.73 77.61 77.61 77.61
4. Reinforcement calculation
(1) bending calculation of normal section of secondary beam
midspan section is calculated as T-section, flange width is
side span BF '= L01 / 3 = 6475 / 3 = 2158 mm & lt; B + Sn = 220 + (3000-120-125) = 2975 mm
mid span BF '= L02 / 3 = 6350 / 3 = 2117 mm & lt; b+Sn=220+2780=3000 mm
hf’=100 mm,h=450 mm,a=35 mm, h0=450-35=415 mm
fc bf’hf’(h0- hf’/2)=12.5*2117*100*(415-100/2)*10-6=965.88 kN·m
γ dM=1.2*84.69=101.63 kN·m
fc bf’hf’(h0- hf’/2)> γ Therefore, the midspan section of the secondary beam is calculated according to the first type of T-section, and the support is calculated according to the rectangular section
b = 220 mm, FC = 12.5 n / mm2, FY = 310 n / mm2
calculation table of bending resistance of normal section:
the first inner support in the middle of side span, the middle support in the middle of middle span
m (kn · m) 84.69-84.69 56.00-64.00
b (mm) 2158 220 2117 220
m (KN · m) 84.69-84.69 56.00-64.00 α s=rdM/(fcbh02) 0.022 0.215 0.015 0.162
ζ 0.022 0.245 0.015 0.178
As=fcbf’ ζ H0 / FY 794 902 509 763
reinforcement 3 Φ 18 4 Φ 18 2 Φ 18 3 Φ 18
actual matching area (mm2) 763 1018 509 763
of which ζ= X / H0 is less than 0.35, which is in line with the condition of plastic internal force redistribution
(2) calculation of shear reinforcement of inclined section of secondary beam
HW / b = (H0 - HF ') / b = (415-100) / 220 = 1.432 & lt; 4
1/ γ d0.25fcbh0=1/1.2*0.25*12.5*220*415=237.76kN> Vmax = 84.73 kn
meet the requirements of section size
shear calculation table of inclined section:
section ar BL br CL CR
V (KN) 63.55 84.73 77.61 77.61 77.61
RDV (KN) 76.26 101.68 93.13 93.13 93.13
VC = 0.07fcbh0 79.89 79.89 79.89 79.89 φ 6
Asv(mm2) 56.6
S(mm) 214 283 466 466 466
Smax 300 200 200 200 200
ρ Svmin 0.12%
actual reinforcement φ 6@200
v. design of main beam (calculated according to elastic theory)
1. Load
dead load secondary beam transfer load: 13.154 * 6.6 = 86.816 kn
dead weight of main beam: 1.05 * (0.75-0.1) * 0.25 * 25 * 3 = 12.797 kn
plastering of main beam: 1.05 * (0.75-0.1) * 0.015 * 2 * 20 * 3 = 1.229 kn
total g = 100.842 kn
live load q = 9 * 6.6 = 59.4 kn
Total load p = G + q = 100.842 + 59.4 = 160.242 kn
2. Calculation diagram
calculation span: side span ln = 9000-240 / 2-400 / 2 = 8680mm
l0 = ln + A / 2 + B / 2 = 8680 + 240 / 2 + 400 / 2 = 9000mm
l0 = 1.025 * ln + B / 2 = 1.025 * 8680 + 400 / 2 = 9097mm
the smaller l0 = 9000mm
middle span B = 400 & lt; 0.05lc = 0.05 * 9000 = 450mm
l0 = LC = 9000mm
3. Internal force calculation
span difference is zero, so it is calculated according to equal span continuous beam
bending moment diagram 1 (KN & # 8226; m)
bending moment diagram 2 (KN & # 8226; m)
bending moment diagram 3 (KN & # 8226; m)
bending moment diagram 4 (KN & # 8226; m)
bending moment diagram 5 (KN & # 8226; m)
shear diagram 1 (KN)
shear diagram 2 (KN)
Shear Diagram 3 (KN)
shear diagram 4 (KN)
shear diagram 5 (KN)
bending moment envelope diagram:
① + ②
① + ③
① + ④
& 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; 8226; ① + 5
moment envelope diagram
shear envelope diagram:
shear envelope diagram
4. Reinforcement calculation
(1) bending capacity calculation of normal section
midspan section is calculated as T-beam, flange width is:
midspan: BF '= l0 / 3 = 9000 / 3 = 3000mm & lt; B + Sn = 250 + (9000-250) = 9000mm
side span: BF '= l0 / 3 = 8880 / 3 = 2960mm & lt; b+Sn=250+(9000-120-250/2)=9005mm
hf’=100mm,h0=750-60=690mm
fcbf’hf’ (h0-hf’/2)=12.5*3000*100*(690-100/2)*10-6=2400kN
> RDM = 1.2 * 438.07 = 525.684kn · m
so it is calculated according to the first type T-section of mid span section, and the support is calculated according to rectangular section
b = 250mm, FC = 12.5n/mm2, FY = 310n / mm2
because the beam is connected with the support as a whole, The calculated bending moment of the bearing B is as follows: < br /
MB = 9475; MB 9475; MB ┃ MB ┃ MB ┃ MB ┃ MB ┃ MB ┃-0.02lnand VBR ┃┃ - 441.01-441.01 9475, - 0.025 ┃┃┃ MB ┃ MB ┃ MB 9475; MB 9475;┃┃┃┃\\\\/ FY 2479 2911 1344
reinforcement 6 Φ 22+1 Φ 18 6 Φ 22+3 Φ 18 4 Φ 22
x values were less than ζ bh0=0.544*670=364mm
ρ= As/(bh0)=1344/(250*670)=0.802%> ρ Min = 0.12%
meet the requirements< (2) shear capacity calculation of inclined section
b = 250mm, H0 = 750-80 = 670, FC = 12.5n/mm2, FY = 310n / mm2, HW = h0-h, f '= 670-100 = 570mm
HW / b = 570 / 250 = 2.28 & lt; 4.0
0.25fcbh0/rd=0.25*12.5*250*670=436.20kN> 209.24kn
the section size meets the requirements
shear reinforcement calculation table
section side support a B support left B support right
V (KN) 122.65 - 209.24 174.5
RDV / (fcbh0) 0.070 0.120 0.100
stirrup method construction reinforcement calculation reinforcement calculation reinforcement calculation reinforcement
primary stirrup double limb φ 8
ASV (mm2) 100.6
s 330 160 280
Smax 250
actual stirrup φ@ one hundred and sixty
2. 1. Supervision work content in construction preparation stage: 1. Prepare supervision plan and supervision rules, organize and hold the first supervision meeting, and carry out supervision disclosure. ② Review the qualification of general contractor and subcontractor. ③ Preside over or participate in drawing review and design disclosure. ④ Check the quality assurance system, quality management system, technical management system and safety management system of the construction unit, and improve the quality management proceres. ⑤ Examine and approve the construction organization design and construction scheme, focus on the review of the organization of personnel, data and machinery, progress plan and various guarantee measures, and review its economy at the same time. ⑥ Review the positioning and setting out data and measurement results, and control the retention and protection of piles. ⑦ Check and implement the commencement conditions, examine and approve the commencement report, and the chief supervision engineer shall sign the commencement report and report it to the construction unit for future reference. 2. The main contents of quality control in the construction stage are as follows: (1) check and confirm the quality of raw materials, semi-finished procts, components and equipment, and do a good job of witness sampling and delivery. ② The key parts and key links of the construction process should be well inspected and on-site work, including energy-saving parts and nodes. ③ Check and sign all concealed works, check and confirm the quality of completed inspection lot, sub item, sub part and sub part works, and sign the project quality acceptance record. ④ Inspect the construction survey, setting out, setting out, etc., timely inform the rectification if any quality problems are found, and make supervision records. ⑤ Supervise the construction unit to carry out the construction in strict accordance with the specifications, standards and design drawings, and strictly implement the construction contract. ⑥ The inspection and testing equipment, instruments and quality scales of the construction unit shall be inspected regularly and sampled irregularly to ensure the accuracy of quality data. ⑦ Check the quality "three inspection" work of the construction unit, and supervise the implementation of the quality "three inspection" work. ⑧ Check and spot check all kinds of engineering tests and test pieces according to the mandatory provisions and standards, and register the account. ⑨ Handle the quality accident according to the procere. ⑩ According to the acceptance procere, do a good job in the quality acceptance of the inspection lot, sub item, sub division and sub division projects, organize the pre acceptance inspection of the project with the participation of the company departments, and participate in the completion acceptance of the project organized by the owner. 3. The main contents of the progress control work in the construction stage are as follows: (1) review the general schele and the construction schele of each stage, and review the specific measures to ensure the progress. ② The supervision progress account often checks, compares and analyzes the actual progress. Supervise and urge the construction unit to adjust the schele in time; Report the implementation of construction schele and existing problems to the owner monthly. ③ Coordinate the progress arrangement of each construction unit, make adjustment measures in time, and strive for harmony; Issue order to start work, order to stop work and order to resume work, and approve the delay of construction period. 4. The main contents of project investment control in the construction stage are as follows: (1) to check whether the construction methods and technical measures adopted in the construction organization design and construction scheme are necessary and reasonable. ② Review the capital use plan of the construction unit; Sign on site visa to handle claims and counterclaims. ③ Carefully measure the completed projects with qualified quality; Audit the monthly settlement statement and issue the project payment certificate; Audit the completion settlement and input the settlement payment voucher. ④ Review the engineering design change according to the owner's authorization and the construction contract. ⑤ Establish the measurement and payment visa account, check and settle with the construction unit regularly. 5. The main contents of supervision work in the stage of completion acceptance are as follows: (1) supervise and urge the construction unit to timely sort out and report the completion documents and acceptance data, accept the completion acceptance report of the unit project, and put forward the opinions of the supervision unit. ② According to the actual situation of the project, prepare the quality (inspection) evaluation report of divisional and unit projects in time. ③ Organize pre acceptance inspection of project supervision and participate in project completion acceptance organized by the owner. 6. The main contents of contract management are as follows: (1) supervision contract management system, including the working system and process of contract formulation, countersignature, negotiation, modification, approval, signing and storage. ② Assist the owner to draw up contract terms and participate in contract negotiation. ③ Analysis and tracking management of contract execution. ④ Assist the owner in handling claims and contract disputes related to the project. What are the main tasks of supervision?
3. The necessity of supervision engineer's work license
4. ID
5. 1. Supervisors shall have primary professional title and supervisor certificate, supervision engineers shall have intermediate professional title and provincial supervision engineer certificate or national registered supervision engineer certificate, and directors shall have intermediate or senior professional title and national registered supervision engineer certificate
2. The title is to go to the local personnel department, and the post certificate is to go to the construction administrative department.
2. The title is to go to the local personnel department, and the post certificate is to go to the construction administrative department.
6. The plug-in punctuation 31.57 on the map is right. On the first floor of the Jasper mine, I killed 5-6 miners in the pit, and the quests came out directly
7. Hello, it's OK
in the skill book (shortcut key K), select a major. There is a small red fork beside the medicine collection. Just click it< br />
8. The project needs a large number of Mines
mining first, it is recommended to leave 8 groups for each type, and at least 15 groups for magic iron
when the mining is full, the cloth should also be left. First, remember which one can brush more FB cloth, and then use it all
our bank doesn't have a bag of less than 22 grids. Now, we don't sell anything we have before, and the materials are still not enough. You'd better save thousands of money in your bag to study. You're three generations poor in engineering, and archaeology will ruin your life
mining first, it is recommended to leave 8 groups for each type, and at least 15 groups for magic iron
when the mining is full, the cloth should also be left. First, remember which one can brush more FB cloth, and then use it all
our bank doesn't have a bag of less than 22 grids. Now, we don't sell anything we have before, and the materials are still not enough. You'd better save thousands of money in your bag to study. You're three generations poor in engineering, and archaeology will ruin your life
9. Hello, I am an AFK player, below level 20, according to your
skill proficiency
should be about your level * 5 (about below 100), you can mine copper (skill requirement 1),
tin mine
(skill requirement 65), and
Silver Mine
(skill requirement 75) below level 20. In the 1980s-85s, when each group (i.e. 20 pieces) was 4G, 6G, 8g, 13g, tin ore was the most expensive, which was about 40g, silver ore was rare,
can be met or not,
, but it was 3G. A vein can proce 6 pieces of ore when it is good. At present, a 4000 minute point card costs about 1 W6. According to the average copper price of 6 g (actually less than 6 g), it takes about 2666.67 groups, that is, 53333.3 pieces. Each vein costs about 3-4 pieces on average. If you count it as four pieces, you need to mine 13333 times. Every time you mine a piece, it takes about 5 minutes, 66666 minutes, which is about 1111 hours. It is calculated day and night, A point card takes about 46 days
therefore, I advise you to buy
point cards
......
in addition, although tin mines are expensive, the places where grade 20 tin mines come out are more dangerous for you, so I don't give it to you. According to the preliminary estimation, there is no copper mining value for you
P.S.: I am more interested in Mathematics in University, so I like to quantify everything. If I don't want to see it, I can ignore the calculation part and look at the conclusion directly
the model ignores the time when you need to go back to the city to put ore on the shelves, the market price fluctuation caused by oversupply e to your continuous supply, and the competition of mining. In addition, the brother on the first floor said that you may not be able to put the ore on the shelves.
skill proficiency
should be about your level * 5 (about below 100), you can mine copper (skill requirement 1),
tin mine
(skill requirement 65), and
Silver Mine
(skill requirement 75) below level 20. In the 1980s-85s, when each group (i.e. 20 pieces) was 4G, 6G, 8g, 13g, tin ore was the most expensive, which was about 40g, silver ore was rare,
can be met or not,
, but it was 3G. A vein can proce 6 pieces of ore when it is good. At present, a 4000 minute point card costs about 1 W6. According to the average copper price of 6 g (actually less than 6 g), it takes about 2666.67 groups, that is, 53333.3 pieces. Each vein costs about 3-4 pieces on average. If you count it as four pieces, you need to mine 13333 times. Every time you mine a piece, it takes about 5 minutes, 66666 minutes, which is about 1111 hours. It is calculated day and night, A point card takes about 46 days
therefore, I advise you to buy
point cards
......
in addition, although tin mines are expensive, the places where grade 20 tin mines come out are more dangerous for you, so I don't give it to you. According to the preliminary estimation, there is no copper mining value for you
P.S.: I am more interested in Mathematics in University, so I like to quantify everything. If I don't want to see it, I can ignore the calculation part and look at the conclusion directly
the model ignores the time when you need to go back to the city to put ore on the shelves, the market price fluctuation caused by oversupply e to your continuous supply, and the competition of mining. In addition, the brother on the first floor said that you may not be able to put the ore on the shelves.
10. As far as I know, looking for you & RLM; Play Bee & RLM; There are more nests. The assistance here is much easier to use. It doesn't need root and prison break. It has many functions and can be used well. Civilization restart is a sandbox survival game with an open super large map. It is a grand world view plot, unrestrained imagination, ubiquitous sense of crisis, friends living together, real shooting experience and so on. What kind of sparks will these elements bring together? The story background of the game is the future doomsday theme, doomsday comes, the terrorist virus sweeps the world, as a survivor you come to the island
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