# 四边固定矩形板计算书

LB-1矩形板计算

《建筑结构荷载规范》 GB50009-2001

《混凝土结构设计规范》 GB50010-2010

1.几何参数

2.材料信息

3.荷载信息(均布荷载)

4.计算方法:弹性板

5.边界条件(上端/下端/左端/右端):固定/固定/固定/固定

6.设计参数

1.计算板的跨度: Lo = 8500 mm

2.计算板的有效高度: ho = h-as=400-55=345 mm

1.X向底板钢筋

1) 确定X向板底弯矩

Mx = 表中系数(γG*ｑgk+γQ*ｑqk)*Lo2

= (0.0129+0.0298*0.200)*(1.200*39.500+1.400*0.000)*8.52 = 64.434 kN*m

2) 确定计算系数

αs = γo*Mx/(α1*fc*b*ho*ho)

= 1.00*64.434×106/(1.00*14.3*1000*345*345)

= 0.038

3) 计算相对受压区高度

ξ = 1-sqrt(1-2*αs) = 1-sqrt(1-2*0.038) = 0.039

4) 计算受拉钢筋面积

As = α1*fc*b*ho*ξ/fy = 1.000*14.3*1000*345*0.039/360

= 529mm2

5) 验算最小配筋率

ρ = As/(b*h) = 529/(1000*400) = 0.132%

ρ<ρmin = 0.200% 不满足最小配筋要求

2.Y向底板钢筋

1) 确定Y向板底弯矩

My = 表中系数(γG*ｑgk+γQ*ｑqk)*Lo2

= (0.0298+0.0129*0.200)*(1.200*39.500+1.400*0.000)*8.52 = 110.923 kN*m

2) 确定计算系数

αs = γo*My/(α1*fc*b*ho*ho)

= 1.00*110.923×106/(1.00*14.3*1000*345*345)

= 0.065

3) 计算相对受压区高度

ξ = 1-sqrt(1-2*αs) = 1-sqrt(1-2*0.065) = 0.067

4) 计算受拉钢筋面积

As = α1*fc*b*ho*ξ/fy = 1.000*14.3*1000*345*0.067/360

= 924mm2

5) 验算最小配筋率

ρ = As/(b*h) = 924/(1000*400) = 0.231%

ρ≥ρmin = 0.200% 满足最小配筋要求

3.X向支座左边钢筋

1) 确定左边支座弯矩

M o x = 表中系数(γG*ｑgk+γQ*ｑqk)*Lo2

= 0.0565*(1.200*39.500+1.400*0.000)*8.52

= 193.613 kN*m

2) 确定计算系数

αs = γo*M o x/(α1*fc*b*ho*ho)

= 1.00*193.613×106/(1.00*14.3*1000*345*345)

= 0.114

3) 计算相对受压区高度

ξ = 1-sqrt(1-2*αs) = 1-sqrt(1-2*0.114) = 0.121

4) 计算受拉钢筋面积

As = α1*fc*b*ho*ξ/fy = 1.000*14.3*1000*345*0.121/360 = 1659mm2

5) 验算最小配筋率

ρ = As/(b*h) = 1659/(1000*400) = 0.415%

ρ≥ρmin = 0.200% 满足最小配筋要求

4.X向支座右边钢筋

1) 确定右边支座弯矩

M o x = 表中系数(γG*ｑgk+γQ*ｑqk)*Lo2

= 0.0565*(1.200*39.500+1.400*0.000)*8.52

= 193.613 kN*m

2) 确定计算系数

αs = γo*M o x/(α1*fc*b*ho*ho)

= 1.00*193.613×106/(1.00*14.3*1000*345*345)

= 0.114

3) 计算相对受压区高度

ξ = 1-sqrt(1-2*αs) = 1-sqrt(1-2*0.114) = 0.121

4) 计算受拉钢筋面积

As = α1*fc*b*ho*ξ/fy = 1.000*14.3*1000*345*0.121/360 = 1659mm2

5) 验算最小配筋率

ρ = As/(b*h) = 1659/(1000*400) = 0.415%

ρ≥ρmin = 0.200% 满足最小配筋要求

5.Y向上边支座钢筋

1) 确定上边支座弯矩

M o y = 表中系数(γG*ｑgk+γQ*ｑqk)*Lo2

= 0.0704*(1.200*39.500+1.400*0.000)*8.52

= 241.089 kN*m

2) 确定计算系数

αs = γo*M o y/(α1*fc*b*ho*ho)

= 1.00*241.089×106/(1.00*14.3*1000*345*345)

= 0.142

3) 计算相对受压区高度

ξ = 1-sqrt(1-2*αs) = 1-sqrt(1-2*0.142) = 0.153

4) 计算受拉钢筋面积

As = α1*fc*b*ho*ξ/fy = 1.000*14.3*1000*345*0.153/360 = 2102mm2

5) 验算最小配筋率

ρ = As/(b*h) = 2102/(1000*400) = 0.526%

ρ≥ρmin = 0.200% 满足最小配筋要求

6.Y向下边支座钢筋

1) 确定下边支座弯矩

M o y = 表中系数(γG*ｑgk+γQ*ｑqk)*Lo2

= 0.0704*(1.200*39.500+1.400*0.000)*8.52

= 241.089 kN*m

2) 确定计算系数

αs = γo*M o y/(α1*fc*b*ho*ho)

= 1.00*241.089×106/(1.00*14.3*1000*345*345)

= 0.142

3) 计算相对受压区高度

ξ = 1-sqrt(1-2*αs) = 1-sqrt(1-2*0.142) = 0.153

4) 计算受拉钢筋面积

As = α1*fc*b*ho*ξ/fy = 1.000*14.3*1000*345*0.153/360

= 2102mm2

5) 验算最小配筋率

ρ = As/(b*h) = 2102/(1000*400) = 0.526%

ρ≥ρmin = 0.200% 满足最小配筋要求

Mk -------- 按荷载效应的标准组合计算的弯矩值

Mq -------- 按荷载效应的准永久组合计算的弯矩值

1.计算荷载效应

Mk = Mgk + Mqk

= (0.0298+0.0129*0.200)*(39.500+0.000)*8.52 = 92.436 kN*m

Mq = Mgk+ψq*Mqk

= (0.0298+0.0129*0.200)*(39.500+1.0*0.000)*8.52 = 92.436 kN*m 2.计算受弯构件的短期刚度 Bs

1) 计算按荷载荷载效应的两种组合作用下，构件纵向受拉钢筋应力

σsk = Mk/(0.87*ho*As) 混规(7.1.4-3)

= 92.436×106/(0.87*345*1231) = 250.174 N/mm

σsq = Mq/(0.87*ho*As) 混规(7.1.4-3)

= 92.436×106/(0.87*345*1231) = 250.174 N/mm

2) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率

ρte = As/Ate 混规(7.1.2-4)

= 1231/200000 = 0.615%

3) 计算裂缝间纵向受拉钢筋应变不均匀系数ψ

ψk = 1.1-0.65*ftk/(ρte*σsk) 混规(7.1.2-2)

= 1.1-0.65*2.01/(0.615%*250.174) = 0.252

ψq = 1.1-0.65*ftk/(ρte*σsq) 混规(7.1.2-2)

= 1.1-0.65*2.01/(0.615%*250.174) = 0.252

4) 计算钢筋弹性模量与混凝土模量的比值αE

αE = Es/Ec = 2.0×105/3.00×104 = 6.667

5) 计算受压翼缘面积与腹板有效面积的比值γf

6) 计算纵向受拉钢筋配筋率ρ

ρ = As/(b*ho)= 1231/(1000*345) = 0.357%

7) 计算受弯构件的短期刚度 Bs

Bsk = Es*As*ho2/[1.15ψk+0.2+6*αE*ρ/(1+ 3.5γf')](混规(7.2.3-1)) = 2.0×105*1231*3452/[1.15*0.252+0.2+6*6.667*0.357%/(1+3.5*0.0)]

= 4.637×104 kN*m2

Bsq = Es*As*ho2/[1.15ψq+0.2+6*αE*ρ/(1+ 3.5γf')](混规(7.2.3-1)) = 2.0×105*1231*3452/[1.15*0.252+0.2+6*6.667*0.357%/(1+3.5*0.0)]

= 4.637×104 kN*m2

3.计算受弯构件的长期刚度B

1) 确定考虑荷载长期效应组合对挠度影响增大影响系数θ

2) 计算受弯构件的长期刚度 B

Bk = Mk/(Mq*(θ-1)+Mk)*Bs (混规(7.2.2-1))

= 92.436/(92.436*(2.0-1)+92.436)*4.637×104

= 2.318×104 kN*m2

Bq = Bsq/θ (混规(7.2.2-2))

= 4.637×104/2.0

= 2.318×104 kN*m2

B = min(Bk,Bq)

= min(23184.291,23184.291)

= 23184.291

4.计算受弯构件挠度

f max = f*(q gk+q qk)*Lo4/B

= 0.00198*(39.500+0.000)*8.54/2.318×104

= 17.630mm

5.验算挠度

fmax=17.630mm≤fo=34.000mm，满足规范要求!

1.跨中X方向裂缝

1) 计算荷载效应

Mx = 表中系数(ｑgk+ψｑqk)*Lo2

= (0.0129+0.0298*0.200)*(39.500+1.00*0.000)*8.52

= 53.695 kN*m

2) 带肋钢筋,所以取值v i=1.0

3) 因为C > 65，所以取C = 65

4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力

σsq=Mq/(0.87*ho*As) 混规(7.1.4-3)

=53.695×106/(0.87*345*1026)

=174.360N/mm

5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率

ρte=As/Ate 混规(7.1.2-4)

=1026/200000 = 0.0051

6) 计算裂缝间纵向受拉钢筋应变不均匀系数ψ

ψ=1.1-0.65*ftk/(ρte*σsq) 混规(7.1.2-2)

=1.1-0.65*2.010/(0.0100*174.360)

=0.351

7) 计算单位面积钢筋根数n

n=1000/dist = 1000/150

=6

8) 计算受拉区纵向钢筋的等效直径d eq

d eq= (∑n i*d i2)/(∑n i*v i*d i)

=6*14*14/(6*1.0*14)=14

9) 计算最大裂缝宽度

ωmax=αcr*ψ*σsq/Es*(1.9*C+0.08*Deq/ρte) (混规(7.1.2-1) =1.9*0.351*174.360/2.0×105*(1.9*40+0.08*14/0.0100)

=0.1092mm ≤ 0.30, 满足规范要求

2.跨中Y方向裂缝

1) 计算荷载效应

My = 表中系数(ｑgk+ψｑqk)*Lo2

= (0.0298+0.0129*0.200)*(39.500+1.00*0.000)*8.52

= 92.436 kN*m

2) 带肋钢筋,所以取值v i=1.0

3) 因为C > 65，所以取C = 65

4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力

σsq=Mq/(0.87*ho*As) 混规(7.1.4-3)

=92.436×106/(0.87*345*1231)

=250.174N/mm

5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率

ρte=As/Ate 混规(7.1.2-4)

=1231/200000 = 0.0062

6) 计算裂缝间纵向受拉钢筋应变不均匀系数ψ

ψ=1.1-0.65*ftk/(ρte*σsq) 混规(7.1.2-2)

=1.1-0.65*2.010/(0.0100*250.174)

=0.578

7) 计算单位面积钢筋根数n

n=1000/dist = 1000/125

=8

8) 计算受拉区纵向钢筋的等效直径d eq

d eq= (∑n i*d i2)/(∑n i*v i*d i)

=8*14*14/(8*1.0*14)=14

9) 计算最大裂缝宽度

ωmax=αcr*ψ*σsq/Es*(1.9*C+0.08*Deq/ρte) (混规(7.1.2-1) =1.9*0.578*250.174/2.0×105*(1.9*40+0.08*14/0.0100)

=0.2582mm ≤ 0.30, 满足规范要求

3.支座上方向裂缝

1) 计算荷载效应

M o y = 表中系数((ｑgk+ψｑqk)*Lo2)

= 0.0704*(39.500+1.00*0.000)*8.52

= 200.908 kN*m

2) 带肋钢筋,所以取值v i=1.0

3) 因为C > 65，所以取C = 65

4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力

σsq=Mq/(0.87*ho*As) 混规(7.1.4-3)

=200.908×106/(0.87*345*3142)

=213.036N/mm

5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率

ρte=As/Ate 混规(7.1.2-4)

=3142/200000 = 0.0157

6) 计算裂缝间纵向受拉钢筋应变不均匀系数ψ

ψ=1.1-0.65*ftk/(ρte*σsq) 混规(7.1.2-2)

=1.1-0.65*2.010/(0.0157*213.036)

=0.710

7) 计算单位面积钢筋根数n

n=1000/dist = 1000/100

=10

8) 计算受拉区纵向钢筋的等效直径d eq

d eq= (∑n i*d i2)/(∑n i*v i*d i)

=10*20*20/(10*1.0*20)=20

9) 计算最大裂缝宽度

ωmax=αcr*ψ*σsq/Es*(1.9*C+0.08*Deq/ρte) (混规(7.1.2-1) =1.9*0.710*213.036/2.0×105*(1.9*40+0.08*20/0.0157)

=0.2554mm ≤ 0.30, 满足规范要求

4.支座下方向裂缝

1) 计算荷载效应

M o y = 表中系数(ｑgk+ψｑqk)*Lo2

= 0.0704*(39.500+1.00*0.000)*8.52

= 200.908 kN*m

2) 带肋钢筋,所以取值v i=1.0

3) 因为C > 65，所以取C = 65

4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力

σsq=Mq/(0.87*ho*As) 混规(7.1.4-3)

=200.908×106/(0.87*345*3142)

=213.036N/mm

5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率

ρte=As/Ate 混规(7.1.2-4)

=3142/200000 = 0.0157

6) 计算裂缝间纵向受拉钢筋应变不均匀系数ψ

ψ=1.1-0.65*ftk/(ρte*σsq) 混规(7.1.2-2)

=1.1-0.65*2.010/(0.0157*213.036)

=0.710

7) 计算单位面积钢筋根数n

n=1000/dist = 1000/100

=10

8) 计算受拉区纵向钢筋的等效直径d eq

d eq= (∑n i*d i2)/(∑n i*v i*d i)

=10*20*20/(10*1.0*20)=20

9) 计算最大裂缝宽度

ωmax=αcr*ψ*σsq/Es*(1.9*C+0.08*Deq/ρte) (混规(7.1.2-1) =1.9*0.710*213.036/2.0×105*(1.9*40+0.08*20/0.0157)

=0.2554mm ≤ 0.30, 满足规范要求

5.支座左方向裂缝

1) 计算荷载效应

M o x = 表中系数(ｑgk+ψｑqk)*Lo2

= 0.0565*(39.500+1.00*0.000)*8.52

= 161.344 kN*m

2) 带肋钢筋,所以取值v i=1.0

3) 因为C > 65，所以取C = 65

4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力

σsq=Mq/(0.87*ho*As) 混规(7.1.4-3)

=161.344×106/(0.87*345*3142)

=171.084N/mm

5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率

ρte=As/Ate 混规(7.1.2-4)

=3142/200000 = 0.0157

6) 计算裂缝间纵向受拉钢筋应变不均匀系数ψ

ψ=1.1-0.65*ftk/(ρte*σsq) 混规(7.1.2-2)

=1.1-0.65*2.010/(0.0157*171.084)

=0.614

7) 计算单位面积钢筋根数n

n=1000/dist = 1000/100

=10

8) 计算受拉区纵向钢筋的等效直径d eq

d eq= (∑n i*d i2)/(∑n i*v i*d i)

=10*20*20/(10*1.0*20)=20

9) 计算最大裂缝宽度

ωmax=αcr*ψ*σsq/Es*(1.9*C+0.08*Deq/ρte) (混规(7.1.2-1) =1.9*0.614*171.084/2.0×105*(1.9*40+0.08*20/0.0157)

=0.1774mm ≤ 0.30, 满足规范要求

6.支座右方向裂缝

1) 计算荷载效应

M o x = 表中系数(ｑgk+ψｑqk)*Lo2

= 0.0565*(39.500+1.00*0.000)*8.52

= 161.344 kN*m

2) 带肋钢筋,所以取值v i=1.0

3) 因为C > 65，所以取C = 65

4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力

σsq=Mq/(0.87*ho*As) 混规(7.1.4-3)

=161.344×106/(0.87*345*3142)

=171.084N/mm

5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率

ρte=As/Ate 混规(7.1.2-4)

=3142/200000 = 0.0157

6) 计算裂缝间纵向受拉钢筋应变不均匀系数ψ

ψ=1.1-0.65*ftk/(ρte*σsq) 混规(7.1.2-2)

=1.1-0.65*2.010/(0.0157*171.084)

=0.614

7) 计算单位面积钢筋根数n

n=1000/dist = 1000/100

=10

8) 计算受拉区纵向钢筋的等效直径d eq

d eq= (∑n i*d i2)/(∑n i*v i*d i)

=10*20*20/(10*1.0*20)=20

9) 计算最大裂缝宽度

ωmax=αcr*ψ*σsq/Es*(1.9*C+0.08*Deq/ρte) (混规(7.1.2-1) =1.9*0.614*171.084/2.0×105*(1.9*40+0.08*20/0.0157)

=0.1774mm ≤ 0.30, 满足规范要求