# Design and calculation example of the hottest fast

2022-09-30
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Design and calculation example of fastener type steel pipe scaffold

according to the provisions of the standard for safety inspection of building construction (JGJ) on external scaffolding; It is suggested that bamboo scaffold should be phased out and fastener type steel pipe scaffold should be popularized. The design principles and calculation methods of fastener type steel pipe scaffold are stipulated in JGJ safety technical specification for fastener type steel pipe scaffold in construction. Based on the code, the author systematically expounds the design and calculation of fastener steel pipe scaffold as follows

I. calculation of horizontal and longitudinal horizontal bars

1. The bending strength of horizontal and longitudinal horizontal bars is calculated according to the following formula:

σ= M/w ≤ f

where m - design value of bending moment, calculated according to m=1.2mgk+1.4mqk, MGK is the bending moment generated by the standard value of scaffold board self weight, and MqK is the bending moment generated by the standard value of construction load

w - section modulus, check the table φ forty-eight × 3.5mm steel pipe w=5.08cm3

f - calculated value of bending strength of steel, f=205n/mm2

(1) bending strength of transverse horizontal bar

when calculating the internal force of transverse horizontal bar, it is calculated as a simply supported beam, as shown in Figure 1. The calculation span is taken as the transverse distance of upright bar l0=80mm, and the calculated extension length of scaffold transverse horizontal bar is a1=300mm, a2=100mm

Figure 1 calculation diagram of horizontal bar

① the standard value of permanent load GK includes the standard value of structural self weight borne by each meter of vertical bar 0.136kn/m (longitudinal distance 1.5m, step distance 1.8m), the standard value of scaffold board self weight 0.35kn/m2 and the standard value of handrail and toe board self weight 0.14kn/m (as shown in Figure 2)

gk =0.136+0.35 × 1.2+0.14

=0.696kn/m

=696n/m

where, figure 2a) is equivalent to figure 2b)

Figure 2 Schematic diagram of structural self weight calculation

② standard value of uniform live load in construction

qk=3kn/m2 × 0.75m

=2.25kn/m=2250n/m

(the spacing between horizontal bars is 0.75m)

Figure 3 construction load calculation diagram

m=1.2mgk+1.4mqk

=1.2 × 11.31+1.4 × 180

=265.57nm

so the bending strength of the horizontal bar meets the safety requirements

(2) the bending strength of the longitudinal horizontal bar is calculated according to the three span continuous beam in Figure 4, and the longitudinal distance la=1500mm is taken as the calculation span. F is the maximum load at the midspan and support of the longitudinal horizontal bar, which is calculated according to the static load P and live load Q respectively. The f force acting on the support in Figure 4 can be ignored in the calculation of bending moment

Figure 4 calculation diagram of longitudinal horizontal bar

consider the maximum bending moment of midspan and bearing according to the static load arrangement in Figure 5

Figure 5 calculation diagram under static load

m1=0.175pla

mb=mc=-0.15pla

q=1/2qkl0=1/2 × two thousand two hundred and fifty × 0.8=900n

test the maximum bending moment at the midspan at the most unfavorable position of two kinds of live loads as shown in Figure 6 and 7

Figure 6 (1)

Figure 7 (2)

m1=0.213qla

according to figure 8 and Figure 9, the two live loads are the most unfavorable positions, and the maximum bending moment of the support is considered

figure 8 calculation of bearing bending moment under the most unfavorable condition of live load (1)

Figure 9 calculation of bearing bending moment under the most unfavorable condition of live load (2)

mb=mc=-0.175qla

according to the above analysis, it can be seen that M1 midspan bending moment is the largest when the most unfavorable combination of static load and live load is shown in Figure 5 and Figure 6 (or figure 7)

MGK=0.175Pla=0.175 × five hundred and twenty-two × 1.5=137.03Nm

MQK=0.213Qla=0.213 × nine hundred × 1.5=287.55Nm

M=1.2MGK+1.4MQK

=1.2 × 137.03+1.4 × 287.55

=567.01nm

2. The deflection of longitudinal and transverse horizontal bars is calculated according to the following formula:

υ ≤[ υ]

where n - deflection

[ υ]—— The allowable deflection is taken as l/150 according to the specification table

(1) deflection of horizontal bar

① considering the static load (Figure 2)

check the table of maximum deflection of beam under uniformly distributed load in the static calculation manual of building structure, and use K1 and K2 values to obtain the coefficient by inserting method

in the formula, e - elastic modulus of steel entering the experimental procedure,

e=2.06 × 105N/mm2

I—— φ forty-eight × The inertia moment of 3.5mm steel pipe, i=12.19cm4

② considering the live load (Figure 3)

the superposition of the two cases, the result is

so the deflection of the horizontal bar meets the safety requirements

(2) deflection of longitudinal horizontal bar

① considering the static load situation, it is composed of [strain gauge], elastic elements and some accessories (compensation elements, protective covers, wiring sockets, loading parts) (Fig. 5)

② considering the live load situation (Fig. 4)

the superposition of the two situations, the deflection of longitudinal horizontal bar meets the safety requirements

3. When the longitudinal horizontal bar is connected with the vertical bar, the anti sliding bearing capacity of the fastener that makes the pressure in the compressor system too high should comply with the following formula:

R ≤ RC

where R - the design value of the vertical bar force transmitted by the longitudinal horizontal bar to the vertical bar

RC - the design value of the anti sliding bearing capacity of the fastener, According to the specification table, rc=800kn

when the longitudinal horizontal bar is connected with the vertical bar, the vertical force on the fastener includes the sum of the transverse horizontal bar load F and M1 attached to the vertical bar, and the maximum shear force V in the same direction as f caused at the fastener with the increase of plastic deformation

F=1.2P+1.4

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