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Led display base date

Led display size:1792mm(5.97ft)*1280mm(4.26ft)

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Led display with cabinet size:L 1792mm(5.97ft)*W 1280mm(4.26ft)* D 400mm(1.34ft)

Pillar size:(D)120mm*(T)5mm*(L)1780mm  to Inch: (D)4.73 inch*(T)0.2inch*(L)1495mm70inch

Basic plate size:L 400mm*W 400mm*T 20mm

Ground cage:L400mm* W 400mm * D 1300mm,Ground size:1200mm*1200mm*1600mm

Permanent load

Led display weight:mg=250*9.8=2450(N)

Pillar weight:7.8*3.14*(6*6-5.5*5.5).178*9.8/1000=245(N)

Total weight:(Take permanent load calculation coefficient 1.20)

G=1.2*(2450+245)=3234(N)

Wind load

Typhoon category 15,V=50.9meter/s,

Gust factor  βz=1.0  (Reference:Total heigth≤20meter)

Wind load shape coefficient  Us=1.6  Reference『建筑结构荷载规范 GB50009-2012』

Height variation factor of wind pressure   Uz=0.88 led display height(m) ≤10m

Windward area:Af  unite㎡

Wo= Wp(Basic wind pressure)

由 Wp=V/1600 ==> Wo=50.9*50.9/1600=1.62 KN/㎡

Led display Standard value of wind load: Wk1=βz*Us*Uz*Wo*Af=1*1.6*0.88*1.62*1.792*1.28=5.23 KN

Pillar Standard value of wind load:       Wk2=βz*Us*Uz*Wo*Af=1*1.6*0.88*1.62*0.12*1.78=0.49 KN

Strength checking

Pillar external diameter 120mm,thickness 5mm,A=1.8*10-3(㎡),

area moment of inertia:I=7.22*10-6(m4

Section bending modulus:W=9.967*10-5 (m3)

Bending moment at column root caused by wind load

M=Wk1*Hb+Wk2*Hp=5.23*2.24+0.49*0.8=12.1KN*m)

Shear force at column root caused by wind load

F=Wk1+Wk2=5.23+0.49=5.72(KN)

Checking calculation of maximum direct stress:

Pillar material:Q235, maximum tension:235MPa, Maximum shear force :117.5Mpa

Maximum stress at column root caused by wind load:

σmax=M/W=12.1*10^3/9.967*10^-5=121(MPa)<( σd,235 MPa)

Checking calculation of maximum shear force:

Maximum shear force at column root caused by wind load:

Tmax=2*F/A=5.57/1.8*10^3=6.2Mpa<(Td,117.5Mpa)

Stress checking calculation of dangerous points:

For the section of cylindrical column, the normal stress and shear stress are large at the intersection of the straight line whose center is 45 degrees from the x-x axis and the section centerline, and the complex stress state. The stress state analysis should be carried out at this point

Location of dangerous points:X=y=(D-0.05)/2*sin(兀/4)=0.0408(M)

Sx(Static moment of dangerous point section of column)=1.948*10^ -5

direct stress at dangerous point:σ=My/I=12.1*0.0408/7.22*10^-6=68.4(Mpa)

Shear force at dangerous point: T=F*Sx/(I*2*t)=57200*1.948*10^5/(7.22*10^-6*2*5)=1.54(Mpa)

According to the fourth strength theory:

σ4=(σ2+3*T21/2(68.42+3*1.5421/2 =68.45Mpa(σd,235 MPa),fulfill a request

Checking calculation of pillar deformation:

This sign includes a display screen. The wind load on it is regarded as the concentrated load acting on the geometric centroid, and the wind load on the column clamped between the display screen and the foundation is regarded as the uniformly distributed load.

Deflection due to concentrated load:

Fb=[P*H2/(6*E*I)]*(3*L-H)=3735.8*2.242 /(6*206*109*7.22*10-6)*(3*1.6-2.24)=0.0054(m)

Deflection due to uniformly distributed load:

F1=q*h4/(8*E*I)=(1/2*ρ* c*V2)*W*1.64/(8*206*109*7.22*10-6)=0.0001(m)

Corner caused by uniformly distributed load:

Θ=q*h3/(6*E*I)= (1/2*ρ* c*V2)*W*1.53/(6*206*109*7.22*10-6)= 0.0000729°

To sum up, the total deformation deflection at the top of the column is calculated:

F=Fb+F1+tanΘ(Lp-h)=0.0054+0.0001+tan(0.0000721)*(2.88-1.6) ≈0.0055m

F/L=0.0055/2.88=0.0019<0.01, fulfill a request

Checking calculation of column base strength

External load on column base

Vertical force: G=Y0*YG*G=1.0*1.0*3234=3234(N)

Horizontal force F=5.72(KN)

Bending moment caused by wind load:M=12.1(KN*m)

eccentric distance:   e=M/G=12100/3234=3.74(m)

Base plate geometry:L 400mm*W 400mm*T 20mm

The anchor bolt is proposed to adopt 8M22 specification, the number of anchor bolts on the tension side is n = 3, and the total effective area is:

Ae=3*3.03=9.09(Cm2)=9.09*10-4(m2)

length of pressure Xn:It is solved by trial calculation according to the following formula:

(Xn)3+3*(e-L/2)*(Xn)2-[6*n*Ae*(e+L/2-Lt)*(L-Lt-Xn)]/W=0

(Xn)3+10.6*(Xn)2-658.9*10-4* (0.038-Xn)=0

(Xn)3+10.6*(Xn)2+0.16(Xn)-0.029=0

Xn=0.045(m)

Maximum compressive stress of concrete

σC=2*G(e+L/2-Lt)/[B*Xn*(L-Lt-Xn/3)]

  =2*3234*(3.74+0.4/2-0.02)/[0.4*0.045*(0.4-0.02-0.045/3)]

 ≈3.86(Mpa)< Fcc*βc=[(1.2*1.2)/(0.4*0.4)] ^0.5*9.6=28.8(Mpa),fulfill a request

Checking calculation of anchor bolt strength

Ta=G*(e-L/2+Xn/3)/ (L-tt-Xn/3)

  =3234*(3.74-0.4/2+0.045/3)/(0.4-0.02-0.045/3)

   ≈31.5(KN)<n.T0 =3*55.2=165.6(KN), fulfill a request

Verification of horizontal shear force

fb=K*(G+ Ta) =0.4*(3.2+31.5)=13.88(KN)> 5.72KN(Shear force at pillar root caused by wind load)

Checking pill of pillar Basic plate thickness

The influence coefficient of concrete strength is that the number of flange ribs is 4, and the support conditions of flange on the compression side are considered as two adjacent support plates.

M=α*σC*(a2) 2=0.089*3860000*0.12*0.12=4947N*m/m)

Basic plate thickness

t=(6*Mamx/fb1) ^0.5=(6*4947/210*106) ^0.5=0.014(m)=14mm<20mm,fulfill a request

Tension side Basic plate thickness

T1={6*Ma*D(basic)/[ (D(bolt)+sin45*D(basic)+ D(basic)*210*106]) ^0.5=18.4mm<20mm,fulfill a request

Foundation checking calculation

The thickness of thebasic plate at the tension side shall be set as the foundation:

Wf=1.2m,Hf=1.6米,Lf=1.2m,

Weight of foundation concrete:r=24KN/m3

Base allowable stress:fa=100KPa.

Load on base

The total vertical loadN=G+rV=3.234+24*(1.2*1.2*1.6)=58.53 (KN)

horizontal load: H=5.72(KN)

Bending moment caused by wind load

M=Wk1*(h1+Hf)+Wk2*(h2+Hf)=5.23*(2.28+1.6)+0.49*(1.6+1.6)=21.9(KN.m)

Stress checking calculation of base

Average base stress calculated under axial compression:

Pk=N/A=58.53/[(1.2)*(1.2)]=40.7 (Kpa)<fa=100(KPa), fulfill a request

Maximum base stress:

σmax=N/A+M/W=58.53/1.2*1.2+21.9/(1/6*1.2*(1.2)2) =117(KPa)<1.2fa=120(KPa), fulfill a request

Minimum base stress:

σmin=N/A-M/W=58.53/(1.2*1.2)-21.9/ [1/6*1.2*(1.2)2])=-35.4(KPa)

The base appears in negative stress, and the distribution width of negative stress:

Lx=IσminI*Lf/(IσminI+:σamx)=37*1.2/(37+106)=0.279 Standard value Lf/40.3

Checking calculation of anti overturning stability of basement

K0=Lf/(2*e)=1.2/(2*e)=1.6 >Standard value 1.10

Base sliding stability coefficient

Kc=n*N/F=0.3*58.53/5.72=3.06>Standard value 1.20

Hunter

Hunter

Hello, I am Hunter Pang, who obtained a master's degree with excellent achievements in 2006 and completed the study of mechanical design, theoretical mechanics, material mechanics, structural mechanics, metal structure, and other professional disciplines. I joined DHX 15 years ago, and now I am the company's business director and technical consultant. I have unique views on the selection of materials, production technology, and installation method of LED display screens. Welcome friends to comment on my microblog and articles. Thank you!

For more new product installed way, please fill in the following form,or add my WhatsAp,Thanks.

WhatsApp:+86 1802540589

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Hunter

Hunter

Hello, I am Hunter Pang, who obtained a master's degree with excellent achievements in 2006 and completed the study of mechanical design, theoretical mechanics, material mechanics, structural mechanics, metal structure, and other professional disciplines. I joined DHX 15 years ago, and now I am the company's business director and technical consultant. I have unique views on the selection of materials, production technology, and installation method of LED display screens. Welcome friends to comment on my microblog and articles. Thank you!

For more new product installed way, please fill in the following form,or add my WhatsAp,Thanks.

WhatsApp:+86 1802540589

Your email information is absolutely safty and we must not disclose it to third parties for any reason!
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