Design of girder formwork and support for the tran

Design of girder formwork and support of structural transfer floor

1. project overview

Xiamen Jinqiu Haoyuan project has a total construction area of 26405m2, with 29 floors of ground structure on the first floor of the basement, of which the second floor of the podium is used as a shopping mall, and the total height of the building is 98.8m. The third floor of the project is used as the structural transfer floor, with a floor height of 4.8m. The maximum section size of the transfer floor girder is B × h=600mm × 1500mm, with a span of 9.6m, and 250mm thick transfer floor. It is proposed to use 18mm thick plywood for girder formwork, and 70mm horizontal joist for supporting formwork × 100mm square timber, spacing 300mm; Longitudinal joist adopts 100mm × 150mm square timber, spacing 800mm. Mf1219 portal frame is used for the support at the bottom of the beam, which is set parallel to the beam axis. The spacing of the portal frame is 600mm and the span is 800mm. See Figure 1 for details

Figure 1 Schematic diagram of longitudinal section of girder support

2. Design and calculation of girder formwork

2.1 load determination

① self weight of formwork and support: 1.0 × one point two × 0.6=0.72kn/m

(the standard value of self weight of formwork and support is 1.0kn/m2)

② self weight of newly poured concrete: 24 × one point two × zero point six × 1.5=29.92kn/m

③ self weight of reinforcement: 2.6 × one point two × zero point six × 1.5=2.81kn/m

(the standard value of reinforcement self weight is 2.6kn/m3 according to the actual reinforcement)

④ load generated during concrete vibration: 2.0 × one point four × 0.6=1.68kn/m (horizontal)

2.2 calculation of beam bottom formwork

is calculated as four span equidistant continuous beam, and its calculation diagram is shown in Figure 2

Figure 2 calculation diagram of beam bottom formwork (side formwork)

① calculation of bending capacity of beam bottom formwork:

mmax=-0.107q1l2=-0.107 × thirty-one point one three × 0.32=-0.30kN·m

б= M/W=(0.30 × 106)/(1/6 × six hundred × 182)

=9.25n/m the entry threshold is also low M2 ＜ fm=13n/mm2

the strength meets the requirements

② deflection calculation of beam bottom formwork

i= (1/12) B · h3=1/12 × six hundred × 183=291600mm4

ν max=0.632·(q2l4)/100EI=(0.632 × twenty-nine point four five × 3004)/(100 × nine × one hundred and three × 291600)=0.57mm

[ ν]= l/400 =306/400=0.75mm

ν max＜[ ν]， The rigidity meets the requirements

2.3 calculation of beam side formwork

① load determination

a. concrete side pressure: F1 '=0.22rc · to· β 1· β 2·V1/2

=0.22 × twenty-four × five × one point two × one point one five × 21/2

=51.52kN/m2

F1＇＇= γ c·H=24 × 1.5=36kn/m2

the smaller of the two is f1=36kn/m2 × 1.2

=42.2kn/m2

b. load generated during concrete vibration, that is, water vapor dispersion resistance will be very high: 4.0 × 1.4=5.6kn/m2 (vertical)

f2=43.2+5.6=48.8kn/m2

c. convert it into line uniformly distributed load (take the clear height of beam =1250mm as the calculation unit)

q1=48.8 × 1.25=61kn/m (checking bearing capacity)

q2=43.2 × 1.25=54kn/m (checking stiffness)

② calculation of bending strength of beam side formwork (the calculation diagram is the same as that in Figure 3)

mmax=-0.107q1l2=-0.107 × sixty-one × 0.32

=-0.587kN·m

б= M/W=(0.587 × 106)/(1/6 × one thousand two hundred and fifty × 182)

=8.7n/mm2

the strength meets the requirements

③ deflection calculation of beam bottom formwork

i= (1/12) B · h3=1/12 × one thousand two hundred and fifty × 183=607500mm4

υ max=0.632·(q2t4)/100EI·0.632 × (54 × 3004)/(100 × nine × one hundred and three × 607500)

=0.51mm＜[ υ]= L/400 =0.75mm

the stiffness meets the requirements

2.4 calculation of vertical stops

① load determination (vertical stop spacing is 300mm, section size 6 × h=100 × 130mm2）

q1=48.8 × 0.3=14.64kn/m (checking bearing capacity)

until the specimen deformation reaches the specified amount q2=43.2 × 0.33 star, apple, Google and Huawei began to accumulate various patents related to graphene =12.96kn/m (checking stiffness)

calculated as a simply supported beam, and its calculation diagram is shown in Figure 3

Figure 3 Schematic diagram of vertical block calculation

② vertical block strength calculation

mmax=1/8q1l2=-0.125 × fourteen point six four × 1.252

=2.86kN·m

б= M/W=(2.86 × 106)/(1/6 × one hundred × 1302) =10.15n/mm2 ＜ fm

=12n/mm2

the strength meets the requirements

③ calculation of vertical deflection

υ max=5q2·l4/384EI=(5 × twelve point nine six × 12504)/[384 × 9000(1/12) × one hundred × 1303]

=2.50mm

[ υ]= l/400=1250/400=3.13mm

υ max＜[ υ]， The rigidity meets the requirements

2.5 calculation of joists across the beam:

① load determination

q1=31.13 × 0.3/6=15.57kn/m (checking bearing capacity)

q2=29.45 × 0.3/6=14.73kn/m (checking stiffness)

calculated as a simply supported beam, and the calculation diagram is shown in Figure 4

Figure 4 calculation diagram of beam transverse joist

② bending strength calculation of transverse joist

mmax=15.57 × 0.6/2 × 0 fifty-seven × zero point three × 0.3/2=1.17kN·m

б= M/W=(1.17 × 106)/(1/6 × seventy × 1002)

=10.03n/mm2

the strength meets the requirements

③ calculation of deflection of transverse joist

it is calculated according to the full paving of load Q2. At this time, the stress state is unfavorable compared with the original stress state

υ max=S·q2·l4/384·E·I=(5 × fourteen point seven three × 8004)/(384 × nine thousand × 1/12 × seventy × 1003)

=1.50mm

[ υ]= l/400=800/400=2mm

υ max＜[ υ]， The rigidity meets the requirements

2.6 calculation of longitudinal joist of beam

① load determination

is calculated according to three span non equidistant continuous beam, and its calculation diagram is shown in Figure 5

Figure 5 calculation diagram of beam longitudinal joist

f1=31.13 × 0.3/2=4.67kn/m (checking bearing capacity)

f2=29.45 × 0.3/2=4.42kn/m (checking stiffness)

② calculation of longitudinal joist strength of beam (according to the most unfavorable state)

mmax=4.67 × 3/2 × 0 sixty-seven × 0.3=2.8kN·m

б= M/W=(2.8 × 106)/(1/6 × one hundred × 1502) =7.47n/mm2 ＜ fm=13n/mm2

the strength meets the requirements