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  Force analysis of aluminum alloy derrick for household ladders

In the whole elevator structure, the aluminum alloy derrick of the household ladder is only fixed by the guide rail support, and the force of the guide rail support is transmitted to the derrick, resulting in bending and deformation. Therefore, before analyzing the force of the aluminum alloy derrick, it is necessary to analyze the force of the guide rail support.
One end of the guide rail bracket is fixed with a pressure plate, and the other end is fixed on the derrick crosspiece by a standard part such as bolts. The guide rail bracket can be divided into three categories according to the use: car guide rail bracket, counterweight guide rail bracket and car counterweight combination bracket. Among them, the stress situation of the car counterweight combination bracket is the most complicated, and it bears the eccentric load capacity of both the car and the counterweight at the same time. Assuming that 1000 mm ∗ 1000 mm 1000 mm ∗ 1000 mm 1000mm**1000mm1000 \mathrm{~mm} * 1000 \mathrm{~mm} the net size of the car is , the force analysis of the car counterweight combination is now carried out directly.
  1. Mechanical analysis of car counterweight combined support
  Figure 1: Bracket structure
1-bracket rung, 2-bracket hard gear, 3-bottom yard, 4-derrick rung, 5-pair of weight guide rails, 6-pair of weight, 7-car guide rail
Set the elevator parameters: center suspension design, progressive safety gear, the rest of the parameters are as follows: car weight P=600kg, rated load Q = 400 kg Q = 400 kg Q=400kg\mathrm{Q}=400 \mathrm{~kg} , car width D y = 1.05 m D y = 1.05 m D_(y)=1.05m\mathrm{D}_{\mathrm{y}}=1.05 \mathrm{~m} , car depth D x = 1.05 m D x = 1.05 m D_(x)=1.05m\mathrm{D}_{\mathrm{x}}=1.05 \mathrm{~m} , car guide shoe distance h = 2.71 m h = 2.71 m h=2.71m\mathrm{h}=2.71 \mathrm{~m} , counterweight guide rail distance h 1 = 2.29 m h 1 = 2.29 m h1=2.29m\mathrm{h} 1=2.29 \mathrm{~m} , car guide rail number n = 2 n = 2 n=2\mathrm{n}=2 , counterweight guide shoe number n 1 = 2 n 1 = 2 n1=2\mathrm{n} 1=2 .
  The size and force of the car counterweight combination bracket are shown in Figure 2:
  Fig.2. Bracket size and force
The guiding force generated by the elevator is the largest when the safety gear is acted, so this situation is directly taken for analysis. According to Appendix G of GB7588-2003 "Safety Code for Elevator Manufacturing and Installation" [3], the force of the guiding force of the safety gear action condition on the Y axis: the center suspension considers the distance of the center of gravity P of the car relative to 10 % 10 % 10%10 \% the suspension center deviation of the car from the car's size, and the center X p = D x / 10 , Y p = D y / 10 X p = D x / 10 , Y p = D y / 10 Xp=Dx//10,Yp=Dy//10X p=D x / 10, Y p=D y / 10 of gravity P of the car relative to the Cartesian coordinate system of the guide rail
According to Table G4.4 in Appendix G7588-2003 GB2, the impact coefficient k1 is selected as 2.0
According to the provisions of G7.2.1.1 of the Appendix of GB7588-2003, the distance of the rated load Q relative to the Cartesian coordinate system of the guide rail shall be taken when the car with central suspension and guidance is operated by the XQ = Dx / 8 , YQ = Dy / 8 XQ = Dx / 8 , YQ = Dy / 8 XQ=Dx//8,YQ=Dy//8\mathrm{XQ}=\mathrm{Dx} / 8, \mathrm{YQ}=\mathrm{Dy} / 8 safety gear
  The force of the guiding force on the Y-axis of the guide rail under the operating condition of the safety gear:
Fx = k 1 g n ∗ ( Q xQ + P xP ) / nh = 2.0 ∗ 9.8 ∗ ( 400 ∗ 1.05 ÷ 8 + 600 ∗ 1.05 ÷ 10 ) / 2 × 2.71 ≈ 418 N Fx = k 1 g n ∗ Q xQ + P xP / nh = 2.0 ∗ 9.8 ∗ ( 400 ∗ 1.05 ÷ 8 + 600 ∗ 1.05 ÷ 10 ) / 2 × 2.71 ≈ 418 N Fx=k1g_(n)**(Q_(xQ)+P_(xP))//nh=2.0^(**)9.8^(**)(400**1.05-:8+600**1.05-:10)//2xx2.71~~418N\mathrm{Fx}=\mathrm{k} 1 \mathrm{~g}_{\mathrm{n}} *\left(\mathrm{Q}_{\mathrm{xQ}}+\mathrm{P}_{\mathrm{xP}}\right) / \mathrm{nh}=2.0^{*} 9.8^{*}(400 * 1.05 \div 8+600 * 1.05 \div 10) / 2 \times 2.71 \approx 418 \mathrm{~N}
   The force of the guiding force on the x-axis of the guide rail under the operating condition of the safety gear:
Fy = k 1 g n ∗ ( Q YQ + P Yp p ) / nh = 2.0 ∗ 9.8 ∗ ( 400 ∗ 1.05 ÷ 8 + 600 ∗ 1.05 ÷ 10 ) / 2 ∗ 2.71 ≈ 418 N Fy = k 1 g n ∗ Q YQ + P Yp p / nh = 2.0 ∗ 9.8 ∗ 400 ∗ 1.05 ÷ 8 + 600 ∗ 1.05 ÷ 10 / 2 ∗ 2.71 ≈ 418 N Fy=k1g_(n)^(**)(Q_(YQ)+P_(Yp_(p)))//nh=2.0^(**)9.8^(**)(400^(**)1.05-:8+600^(**)1.05-:10)//2**2.71~~418N\mathrm{Fy}=\mathrm{k} 1 \mathrm{~g}_{\mathrm{n}}{ }^{*}\left(\mathrm{Q}_{\mathrm{YQ}}+\mathrm{P}_{\mathrm{Yp}_{\mathrm{p}}}\right) / \mathrm{nh}=2.0^{*} 9.8^{*}\left(400^{*} 1.05 \div 8+600^{*} 1.05 \div 10\right) / 2 * 2.71 \approx 418 \mathrm{~N}
According to Article G2.6 of the Appendix to GB7588-2003, for counterweights or counterweights of central suspension and guidance, the deviation of the point of action of gravity with respect to its center of gravity should be taken into account, and the eccentricity on the horizontal section should be at least 10\% in the width direction 5 % 5 % 5%5 \% and 10\% in the depth direction.
Take the depth of X cwt = 0.52 m X cwt = 0.52 m X_(cwt)=0.52m\mathrm{X}_{\mathrm{cwt}}=0.52 \mathrm{~m} the weight, the width of Y cwt = 0.1 m Y cwt = 0.1 m Y_(cwt)=0.1m\mathrm{Y}_{\mathrm{cwt}}=0.1 \mathrm{~m} the weight, the distance x cwt = X cwt / 10 , y cwt = Y cwt / 20 x cwt = X cwt / 10 , y cwt = Y cwt / 20 x_(cwt)=X_(cwt)//10,y_(cwt)=Y_(cwt)//20\mathrm{x}_{\mathrm{cwt}}=\mathrm{X}_{\mathrm{cwt}} / 10, \mathrm{y}_{\mathrm{cwt}}=\mathrm{Y}_{\mathrm{cwt}} / 20 of the heavy force application point relative to its center of gravity
Take K = 0.45 K = 0.45 K=0.45K=0.45 the balance factor, then the counterweight weight is:
M cwt = P + K Q = 600 + 0.45 ∗ 400 = 780 kg M cwt  = P + K Q = 600 + 0.45 ∗ 400 = 780 kg M_("cwt ")=P+KQ=600+0.45^(**)400=780kgM_{\text {cwt }}=P+K Q=600+0.45^{*} 400=780 \mathrm{~kg}
According to Table G2 in Clause G4.4 of Appendix GB7588-2003, the impact coefficient K2 selects 1.2,
Counterweight guiding force on the guide Y Y YY shaft:
F X 1 = K 2 g n M cwt x cwt / n 1 h 1 = 1.2 ∗ 9.8 ∗ 780 ∗ 0.052 / 2 ∗ 2.29 ≈ 104 N F X 1 = K 2 g n M cwt x cwt / n 1 h 1 = 1.2 ∗ 9.8 ∗ 780 ∗ 0.052 / 2 ∗ 2.29 ≈ 104 N F_(X1)=K_(2)g_(n)M_(cwt)x_(cwt)//n1h1=1.2^(**)9.8^(**)780^(**)0.052//2^(**)2.29~~104N\mathrm{F}_{\mathrm{X} 1}=\mathrm{K}_{2} \mathrm{~g}_{\mathrm{n}} \mathrm{M}_{\mathrm{cwt}} \mathrm{x}_{\mathrm{cwt}} / \mathrm{n} 1 \mathrm{~h} 1=1.2^{*} 9.8^{*} 780^{*} 0.052 / 2^{*} 2.29 \approx 104 \mathrm{~N}
Counterweight guiding force on the guide x x xx shaft:
F Y 1 = K 2 g n M cwt Y cwt / 0.5 n 1 h 1 = 1.2 ∗ 9.8 ∗ 780 ∗ 0.005 / 0.5 ∗ 2 ∗ 2.29 ≈ 20 N F Y 1 = K 2 g n M cwt Y cwt / 0.5 n 1 h 1 = 1.2 ∗ 9.8 ∗ 780 ∗ 0.005 / 0.5 ∗ 2 ∗ 2.29 ≈ 20 N F_(Y1)=K2g_(n)M_(cwt)Y_(cwt)//0.5n1h1=1.2^(**)9.8^(**)780^(**)0.005//0.5^(**2)**2.29~~20N\mathrm{F}_{\mathrm{Y} 1}=\mathrm{K} 2 \mathrm{~g}_{\mathrm{n}} \mathrm{M}_{\mathrm{cwt}} \mathrm{Y}_{\mathrm{cwt}} / 0.5 \mathrm{n} 1 \mathrm{~h} 1=1.2^{*} 9.8^{*} 780^{*} 0.005 / 0.5^{* 2} * 2.29 \approx 20 \mathrm{~N}
The force on the rail bracket is simplified as follows. Since the counterweight is usually suspended and guided in the center, the force generated by it has little effect on the rail support, so its force is ignored.
For the rung, the force produces only a small bending moment F X F X F_(X)\mathrm{F}_{\mathrm{X}} for it, which is negligible in the calculation. Therefore for the rung, only calculate the force F y F y F_(y)\mathrm{F}_{y} acting on it, so it is simplified to the mechanical model of simply supported beam, and its force diagram is shown in Figure 3. The maximum displacement and stress are at the F y F y F_(y)F_{y} force.
  Fig. 3 rungs force diagram
   2. Force analysis of aluminum alloy rungs
  Fig.4. Simplified diagram of the force of the aluminum alloy rung
   Mechanical properties of aluminum alloy rungs:
   Tensile strength: 175 MPa
   Yield strength: 130 MPa
  Modulus of Elasticity: 68600 M / mm 2 68600 M / mm 2 68600M//mm^(2)68600 \mathrm{M} / \mathrm{mm}^{2}
   Hardness: 65 HB
   Aluminum alloy rung parameters:
  m Aluminum alloy rung line density: 2.77 kg / cm 3 2.77 kg / cm 3 2.77kg//cm^(3)2.77 \mathrm{~kg} / \mathrm{cm}^{3}
F Load: 418 N
l Unsupported length of aluminum alloy profile: 1230 mm
  E Modulus of Elasticity: 70000 M / mm 2 70000 M / mm 2 70000M//mm^(2)70000 \mathrm{M} / \mathrm{mm}^{2}
  Set inertia: 46.1cm 4 4 ^(4){ }^{4}
  Z-section inertia: 15.2 cm 3 15.2 cm 3 15.2cm^(3)15.2 \mathrm{~cm}^{3}
g 9.8 N / kgf 9.8 N / kgf 9.8N//kgf9.8 \mathrm{~N} / \mathrm{kgf}