正截面受弯承载力计算例题
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例
题
3－1 解： fc = 14.3 N mm2 ； ft = 1.43 N mm2 ； f y = 360 N mm2 ；
h0
=
h
−c
−8−
20 2
=
500 −
20 − 8 −
20 2
=
462mm
αs
=
M α1 fcbh02
=
180 ×106
1.0×14.3× 250× 4622
= 0.236
查表得：ξ = 0.273 < ξb = 0.518
选 4 25+4 20 钢筋， As = 3220mm2 (满足构造要求)。
5
=
444.5mm
x=
f y As − f y′As′ α1 fcb
= 435×1473 − 410× 402 1.0×14.3× 200
= 166.41mm < ξbh0
= 0.482× 444.5 = 214mm
2as′
=
2× (c
+8+
d )
2
=
2× (35
+
8
+
16 )
2
=
102mm
<
x
Mu
= α1 fcbx(h0
ρmin
= 0.45
ft fy
= 0.45× 1.43 300
= 0.00215 > 0.002
ρminbh = 0.00215×1000×100 = 215mm2 < As = 665mm2 ，满足
1
3－3 解： fc = 31.8 N mm2 ； ft = 2.14 N mm2 ； f y = 360 N mm2 ； α1 = 0.96 ；
× 0.518× (1− − 35)
0.5× 0.518)
=
316mm2
As
=
α1 fcbξbh0 fy
+
f y′As′
=
1×19.1× 200× 0.518× 440 + 360× 316 360
=
2734mm2
As′ 选 2 16， As′ ＝402mm2； As 选 2 28+4 22， As ＝2752 mm2
−
120 )
2
= 298.6×106 N ⋅ mm
Mu2 = M − Mu1 = 650×106 − 298.6×106 =351.4×106N﹒mm
αs
=
Mu2 α1 fcbh02
= 351.4×106 1×14.3× 300× 6402
= 0.200
4
查表得ξ = 0.225 < ξb = 0.518
3－6 解：
M u1
=
f
′
y
As′
(h0
− as′) =360×941×(440-35)=137.2×106N﹒mm
Mu2 = M − Mu1 =330×106-137.2×106=192.8×106N﹒mm
αs
=
Mu2 α1 fcbh02
= 192.8×106 1×19.1× 200× 4402
=
α1
fcb′f
h′f
(h0
−
h′f 2
)
=
1.0
×14.3×
600 ×120 ×
(640
−
120 ) 2
= 597.2 ×106 N ⋅ mm < 650×106 N ⋅ mm
属于第二类 T 形截面
Mu1
= α1
fc (b′f
− b)h′f
(h0
−
h′f 2
)
=
1.0×14.3×
(600
−
300)
×120 × (640
−
x)+ 2
f y′As′(h0
− as′ )
=
1.0
×14.3×
200
×166.41×
(444.5
−
166.41 )
+
2
410× 402× (444.5 − 51)
= 236.81kN ⋅ m > 200kN ⋅ m
3
3－8 解：
安全。 fc = 14.3 N mm2 ； f y = f y′ = 360 N mm2 ；
β1 = 0.76 ；h=600mm；b=250mm
h0
=
h
−
c
−8
−
20 2
=
600
−
20
−8−
20 2
=
562mm
αs
=
M α1 fcbh02
=
270 ×106 0.96× 31.8× 250× 5622
= 0.112
查表得：ξ = 0.1199 < ξb = 0.481
As
=
α1 fcbξ h0 fy
= 0.45
ft fy
= 0.45× 1.43 360
= 0.0018 < 0.002
ρminbh = 0.002× 200 × 600 = 240mm2 < As = 2945mm2
取 As = 2945mm2
3－9 解：
fc
= 14.3 N
mm2 ； f y =
f
′
y
=
360
N
mm2 ；
h0 = h − as = 700 − 60 = 640mm
=
0.96× 31.8× 250× 0.1199× 562 360
= 1429mm2
选 3 25 钢筋， As = 1473mm2 (满足构造要求)。
ρmin
= 0.45
ft fy
= 0.45× 2.14 360
= 0.0027 > 0.002
ρminbh = 0.0027 × 250× 600 = 405mm2 < As = 1473mm2
取 As = 1256mm2
3－2 解： fc = 14.3 N mm2 ； ft = 1.43 N mm2 ； f y = 300 N mm2 ；
h0
=
h
−
c
−
10 2
=
100
−15
−
10 2
=
80mm
αs
=
M α1 fcbh02
=
14.5 ×106
1.0 ×14.3 ×1000 × 802
=
0.1584
=
73.2mm
<
ξb h0
=
0.482× 414
= 199.5mm
ρmin
= 0.45
ft fy
= 0.45× 1.71 = 0.0018 < 0.002 435
ρminbh = 0.002× 250× 450 = 225mm2 < As = 804mm2
Mu
= α1
fcbx(h0
−
x )
2
= 1.0×19.1× 250× 73.2× (414 −
73.2 ) 2
= 131.9kN
⋅m
>
M
= 127kN
⋅m
安全
3－5
解：
fc = 19.1N
mm2 ；
fy
=
f
′
y
=
360
N
mm2
；设受拉钢筋放两层，
as=c+8+20+25/2=20+8+20+12.5=60.5≈60mm，h0=h-as=500-60=440mm
αs
=
M α1 fcbh02
=
330 ×106
h0 = h − as = 600 − 60 = 540mm
α1
fcb′f
h′f
(h0
−
h′f 2
)
=
1.0
×14.3 ×1000 ×100 ×
(540
−
100 ) 2
= 700.7 ×106 N ⋅ mm > 520 ×106 N ⋅ mm
属于第一类 T 形截面
αs
=
M α1 fcb′f h02
=
取 As = 1473mm2
3－4 解： fc = 19.1N mm2 ； ft = 1.71N mm2 ； f y = 435 N mm2 ； As = 804mm2
h0
=
h
−
c
−8−
d 2
=
450 −
20 − 8 − 16 2
=
414mm
x
=
f y As α1 fcb
=
435 × 804 1.0×19.1× 250
As 2
=
α1 fcbξ h0 fy
= 1.0×14.3× 300× 0.225× 640 360
= 1716mm2
As1
=
α1
fc (b′f − b)h′f fy
= 1×14.3× (600 − 300) ×120 = 1430mm2 360
As = As1 + As2 = 1430 +1716 = 3146mm2
查表得：ξ = 0.173 < ξb = 0.55
As
=
α1 fcbξ h0 fy
= 1.0×14.3×1000× 0.173×80 300
=
660mm2
选 12@170， As = 665mm2 ，分布钢筋选φ 6@180，其截面面积
为 28.3×1000/180=157mm2>0.15×b×h=0.15×1000×100=150 mm2, 且 >15%As=0.15×665=99.75mm2。截面配筋如图
1×19.1× 200× 4402
= 0.446
2
查表得：ξ = 0.671 > ξb = 0.518
应设计成双筋截面，取 x = ξbh0 得
As′
=
M
− α1
fcbh02ξb (1− f y′(h0 − as′ )
0.5ξb )
=
330 ×106
−1×19.1× 200× 4402 360× (440
0.261
查表得：ξ = 0.309 < ξb = 0.518
x = ξ h0 =0.309×440=136mm>2 as′ =70mm
As 2
=
α1 fcbξ h0 fy
= 1.0×19.1× 200× 0.309× 440 360
=1443mm2
As1
=
As′
f
′
y
fy
= 941× 360 =941mm2 360
As=As1+As2=941+1443=2384mm2
选 5 25 钢筋， As = 2454mm2 (满足构造要求)。
3－7 解：
fc = 14.3 N mm2 ； f y = 435N / mm2; f y′ = 410 N mm2 ；
h0
=
h−
c
−8−
d 2
=
500 − 35 − 8 −
25 2
520 ×106
1×14.3×1000× 5402
= 0.1247
查表得ξ = 0.134 < ξb = 0.518
As
=
α1 fcb′f ξ h0 fy
= 1.0×14.3×1000× 0.134× 540 360
=
2874mm2
选 6 25 钢筋， As = 2945mm2 (满足构造要求)。
ቤተ መጻሕፍቲ ባይዱ
ρmin
As
=
α1 fcbξ h0 fy
= 1.0×14.3× 250× 0.273× 462 360
= 1253mm2
选 4 20 钢筋， As = 1256mm2 (满足构造要求)。
ρmin
= 0.45
ft fy
= 0.45× 1.43 360
= 0.0018 < 0.002
ρminbh = 0.002× 250× 500 = 250mm2 < As = 1256mm2
题
3－1 解： fc = 14.3 N mm2 ； ft = 1.43 N mm2 ； f y = 360 N mm2 ；
h0
=
h
−c
−8−
20 2
=
500 −
20 − 8 −
20 2
=
462mm
αs
=
M α1 fcbh02
=
180 ×106
1.0×14.3× 250× 4622
= 0.236
查表得：ξ = 0.273 < ξb = 0.518
选 4 25+4 20 钢筋， As = 3220mm2 (满足构造要求)。
5
=
444.5mm
x=
f y As − f y′As′ α1 fcb
= 435×1473 − 410× 402 1.0×14.3× 200
= 166.41mm < ξbh0
= 0.482× 444.5 = 214mm
2as′
=
2× (c
+8+
d )
2
=
2× (35
+
8
+
16 )
2
=
102mm
<
x
Mu
= α1 fcbx(h0
ρmin
= 0.45
ft fy
= 0.45× 1.43 300
= 0.00215 > 0.002
ρminbh = 0.00215×1000×100 = 215mm2 < As = 665mm2 ，满足
1
3－3 解： fc = 31.8 N mm2 ； ft = 2.14 N mm2 ； f y = 360 N mm2 ； α1 = 0.96 ；
× 0.518× (1− − 35)
0.5× 0.518)
=
316mm2
As
=
α1 fcbξbh0 fy
+
f y′As′
=
1×19.1× 200× 0.518× 440 + 360× 316 360
=
2734mm2
As′ 选 2 16， As′ ＝402mm2； As 选 2 28+4 22， As ＝2752 mm2
−
120 )
2
= 298.6×106 N ⋅ mm
Mu2 = M − Mu1 = 650×106 − 298.6×106 =351.4×106N﹒mm
αs
=
Mu2 α1 fcbh02
= 351.4×106 1×14.3× 300× 6402
= 0.200
4
查表得ξ = 0.225 < ξb = 0.518
3－6 解：
M u1
=
f
′
y
As′
(h0
− as′) =360×941×(440-35)=137.2×106N﹒mm
Mu2 = M − Mu1 =330×106-137.2×106=192.8×106N﹒mm
αs
=
Mu2 α1 fcbh02
= 192.8×106 1×19.1× 200× 4402
=
α1
fcb′f
h′f
(h0
−
h′f 2
)
=
1.0
×14.3×
600 ×120 ×
(640
−
120 ) 2
= 597.2 ×106 N ⋅ mm < 650×106 N ⋅ mm
属于第二类 T 形截面
Mu1
= α1
fc (b′f
− b)h′f
(h0
−
h′f 2
)
=
1.0×14.3×
(600
−
300)
×120 × (640
−
x)+ 2
f y′As′(h0
− as′ )
=
1.0
×14.3×
200
×166.41×
(444.5
−
166.41 )
+
2
410× 402× (444.5 − 51)
= 236.81kN ⋅ m > 200kN ⋅ m
3
3－8 解：
安全。 fc = 14.3 N mm2 ； f y = f y′ = 360 N mm2 ；
β1 = 0.76 ；h=600mm；b=250mm
h0
=
h
−
c
−8
−
20 2
=
600
−
20
−8−
20 2
=
562mm
αs
=
M α1 fcbh02
=
270 ×106 0.96× 31.8× 250× 5622
= 0.112
查表得：ξ = 0.1199 < ξb = 0.481
As
=
α1 fcbξ h0 fy
= 0.45
ft fy
= 0.45× 1.43 360
= 0.0018 < 0.002
ρminbh = 0.002× 200 × 600 = 240mm2 < As = 2945mm2
取 As = 2945mm2
3－9 解：
fc
= 14.3 N
mm2 ； f y =
f
′
y
=
360
N
mm2 ；
h0 = h − as = 700 − 60 = 640mm
=
0.96× 31.8× 250× 0.1199× 562 360
= 1429mm2
选 3 25 钢筋， As = 1473mm2 (满足构造要求)。
ρmin
= 0.45
ft fy
= 0.45× 2.14 360
= 0.0027 > 0.002
ρminbh = 0.0027 × 250× 600 = 405mm2 < As = 1473mm2
取 As = 1256mm2
3－2 解： fc = 14.3 N mm2 ； ft = 1.43 N mm2 ； f y = 300 N mm2 ；
h0
=
h
−
c
−
10 2
=
100
−15
−
10 2
=
80mm
αs
=
M α1 fcbh02
=
14.5 ×106
1.0 ×14.3 ×1000 × 802
=
0.1584
=
73.2mm
<
ξb h0
=
0.482× 414
= 199.5mm
ρmin
= 0.45
ft fy
= 0.45× 1.71 = 0.0018 < 0.002 435
ρminbh = 0.002× 250× 450 = 225mm2 < As = 804mm2
Mu
= α1
fcbx(h0
−
x )
2
= 1.0×19.1× 250× 73.2× (414 −
73.2 ) 2
= 131.9kN
⋅m
>
M
= 127kN
⋅m
安全
3－5
解：
fc = 19.1N
mm2 ；
fy
=
f
′
y
=
360
N
mm2
；设受拉钢筋放两层，
as=c+8+20+25/2=20+8+20+12.5=60.5≈60mm，h0=h-as=500-60=440mm
αs
=
M α1 fcbh02
=
330 ×106
h0 = h − as = 600 − 60 = 540mm
α1
fcb′f
h′f
(h0
−
h′f 2
)
=
1.0
×14.3 ×1000 ×100 ×
(540
−
100 ) 2
= 700.7 ×106 N ⋅ mm > 520 ×106 N ⋅ mm
属于第一类 T 形截面
αs
=
M α1 fcb′f h02
=
取 As = 1473mm2
3－4 解： fc = 19.1N mm2 ； ft = 1.71N mm2 ； f y = 435 N mm2 ； As = 804mm2
h0
=
h
−
c
−8−
d 2
=
450 −
20 − 8 − 16 2
=
414mm
x
=
f y As α1 fcb
=
435 × 804 1.0×19.1× 250
As 2
=
α1 fcbξ h0 fy
= 1.0×14.3× 300× 0.225× 640 360
= 1716mm2
As1
=
α1
fc (b′f − b)h′f fy
= 1×14.3× (600 − 300) ×120 = 1430mm2 360
As = As1 + As2 = 1430 +1716 = 3146mm2
查表得：ξ = 0.173 < ξb = 0.55
As
=
α1 fcbξ h0 fy
= 1.0×14.3×1000× 0.173×80 300
=
660mm2
选 12@170， As = 665mm2 ，分布钢筋选φ 6@180，其截面面积
为 28.3×1000/180=157mm2>0.15×b×h=0.15×1000×100=150 mm2, 且 >15%As=0.15×665=99.75mm2。截面配筋如图
1×19.1× 200× 4402
= 0.446
2
查表得：ξ = 0.671 > ξb = 0.518
应设计成双筋截面，取 x = ξbh0 得
As′
=
M
− α1
fcbh02ξb (1− f y′(h0 − as′ )
0.5ξb )
=
330 ×106
−1×19.1× 200× 4402 360× (440
0.261
查表得：ξ = 0.309 < ξb = 0.518
x = ξ h0 =0.309×440=136mm>2 as′ =70mm
As 2
=
α1 fcbξ h0 fy
= 1.0×19.1× 200× 0.309× 440 360
=1443mm2
As1
=
As′
f
′
y
fy
= 941× 360 =941mm2 360
As=As1+As2=941+1443=2384mm2
选 5 25 钢筋， As = 2454mm2 (满足构造要求)。
3－7 解：
fc = 14.3 N mm2 ； f y = 435N / mm2; f y′ = 410 N mm2 ；
h0
=
h−
c
−8−
d 2
=
500 − 35 − 8 −
25 2
520 ×106
1×14.3×1000× 5402
= 0.1247
查表得ξ = 0.134 < ξb = 0.518
As
=
α1 fcb′f ξ h0 fy
= 1.0×14.3×1000× 0.134× 540 360
=
2874mm2
选 6 25 钢筋， As = 2945mm2 (满足构造要求)。
ቤተ መጻሕፍቲ ባይዱ
ρmin
As
=
α1 fcbξ h0 fy
= 1.0×14.3× 250× 0.273× 462 360
= 1253mm2
选 4 20 钢筋， As = 1256mm2 (满足构造要求)。
ρmin
= 0.45
ft fy
= 0.45× 1.43 360
= 0.0018 < 0.002
ρminbh = 0.002× 250× 500 = 250mm2 < As = 1256mm2