|
支撑梁强度计算书 |
项目号 ITEM NO. |
|
注： |
|
|
1.黄色的背景是需要输入的项，蓝色为结果，红色字体为不合格。 |
CALCULATION OF SUPPORT BEAM
|
文件号 DOC. NO. |
|
2. 最大弯矩为M=PL/8，但考虑到符合的分布是不均匀的，并且操作时， |
|
|
|
湿填料重量有所增加，为安全起见M=PL/6 |
|
|
|
输入数据 Input Data |
|
|
|
|
|
3.承载宽度t可以是等效宽度，如果只有一个型钢 |
t=πD^2/4L |
6344.699 |
|
|
|
|
|
|
|
|
容器直径Diameter of Column |
D= |
7600 |
mm |
 |
|
4.型钢惯性矩级抗弯模量需去除腐蚀余量。 |
|
|
|
填料堆积密度Packing density |
ρp |
2000 |
kg/m3 |
|
5.型钢惯性矩和抗弯模量可以查型钢表或者按照下图计算 |
|
|
|
填料高度Height of Packing |
H |
860 |
mm |
|
型钢抗弯模量计算 |
 |
|
I1=ab3/12 |
590951.67 |
y1=b/2 |
18.5 |
|
|
填料类型Type of packing |
|
散堆填料 |
|
|
a |
140 |
mm |
|
I2=dc3/12 |
1.207E+09 |
y2=c/2 |
380 |
|
|
液体密度Density of liquid |
ρL |
0 |
kg/m3 |
|
b |
37.00 |
mm |
|
I3=ef3/12 |
0 |
y3=f/2 |
0 |
|
|
其它重量Other weight |
G |
2500 |
kg |
|
c |
760 |
mm |
|
A1=ab |
5180 |
y'1 |
18.5 |
|
|
支撑梁长度Length of beam |
L |
7150 |
mm |
|
d |
33 |
mm |
|
A2=cd |
25080 |
y'2 |
417 |
|
|
支撑梁承载宽度Width of load |
t |
800 |
mm |
|
e |
0 |
mm |
|
A3=ef |
0 |
y'3 |
797 |
|
|
支撑梁型号Type of Beam |
T型钢 |
160*40-750*26 |
|
f |
0 |
mm |
|
s1 |
330.28354 |
s2*A |
565071792 |
|
|
支撑梁惯性矩Inertia moment |
I |
1.89E+09 |
mm4 |
|
惯性矩I |
1.89E+09 |
mm4 |
|
s2 |
-68.21646 |
|
116709405 |
|
|
支撑梁抗弯模量The flexural modulus of beam |
W |
4215722.4 |
mm3 |
|
中性轴y |
348.78354 |
mm |
|
s3 |
-448.2165 |
|
0 |
|
|
压力降Pressure drop |
Δp |
20000 |
Pa |
|
抗弯模量W |
4215722.4 |
mm3 |
|
总高 |
797 |
|
|
支撑梁材料Material of Beam |
|
S30409 |
|
|
|
此表适用于角钢，工字钢，H型钢，槽钢，T型钢。如当为角钢时e，f=0 |
|
|
设计温度下许用应力 Allowa-nce stress in design temp. |
[σ]t |
64 |
Mpa |
常温下许用应力[σ] Allowance stress |
137 |
|
6. 许用挠度[f]推荐值按照下表之一 |
|
|
|
|
设计温度下弹性模量Elasticity modulus in design temp. |
E |
151000 |
MPa |
许用的挠度[f] The
Allowance deflection |
10 |
|
 |
|
|
|
|
设计计算 Checking strength |
|
. |
|
|
| a) |
操作工况 Operation |
|
|
|
筒体截面积Shell across Area |
A=πD^2/4= |
45364597.9 |
mm2 |
|
|
支撑梁上填料重量产生的载荷
The load of packing |
Pp=HLtρpg×10-9= |
96416.3 |
N |
|
78027.108 |
kg |
|
|
系数k，散堆填料0.35，丝网填料0.05;Factor K |
0.35 |
|
|
|
填料层的持液量The load of liquid |
PL=kHLtρLg×10-9= |
0.0 |
N |
|
 |
|
|
压力降产生的载荷
The load of pressure drop |
Py=Δp×Lt×10-6= |
114400.0 |
N |
|
|
其他重量产生的载荷
The load of other weight |
Pg=G×g= |
24500.0 |
N |
|
|
|
|
|
|
支撑梁上的总载荷
The total load on the beam |
P=Pp+PL+Py+Pg= |
235316.3 |
N |
|
|
|
|
|
|
支撑梁的单位均布载荷Load per unit length |
q=P/L= |
32.9 |
N/mm |
|
|
|
|
|
|
最大弯矩 The Max. bending moment |
M=PL/6= |
2.8042E+08 |
N.mm |
|
|
|
|
|
|
支撑梁应力The stress of Beam |
σ=M/W= |
66.5 |
Mpa |
|
|
|
|
|
|
支撑梁的挠度The actual deflection |
f=5qL4/(384EI)= |
3.9 |
mm |
|
|
|
|
|
|
Since f≤[f] ,挠度合格, It is safe. |
|
|
7. L/a= |
51.071429 |
L/a>11,时需要考虑梁的整体稳定性 |
|
|
|
Since σ>[σ]t , 支撑梁强度不合格.It is not safe |
|
|
| b) |
安装和检修 Installation and maintenance |
|
|
|
|
安装和检修时载荷
The load when installation and maintenance |
P1=Pp+Pg= |
120916.3 |
N |
|
|
支撑梁的单位均布载荷
Load per unit length |
q'=P1/L= |
16.9114 |
N/mm |
|
DN≤2m，在梁中点处作用1350N载荷。When DN≤2m，There are 1350N on the middle of beam. So: |
|
|
DN>2m,在距离梁两端各1/3梁长度出分别作用1000N的集中载荷。When DN>2m，There are 1000N on the both 1/3 length of beam. So: |
|
|
DN≤2m，在梁中点处作用1350N载荷。When DN≤2m，There are 1350N on the middle of beam. So: |
|
DN>2m,在距离梁两端各1/3梁长度出分别作用1000N的集中载荷。When DN>2m，There are 1000N on the both 1/3 length of beam. So: |
|
集中载荷concentrated load |
P'= |
1000.0 |
N |
|
M'=q'L^2/8+P'L/4= |
109856461 |
|
|
最大弯矩 The Max. bending moment |
M'=q'L^2/8+P'L/3= |
1.1045E+08 |
N.mm |
|
M'=q'L^2/8+P'L/3= |
110452294 |
|
108068961 |
2383333.3 |
|
|
支撑梁应力The stress of Beam |
σ'=M/W= |
26.2 |
Mpa |
|
f=5q'L^4/(384EI)+P'L^3/(48EI)= |
2.0436857 |
|
|
|
|
支撑梁的挠度The actual deflection |
f=5q'L^4/(384EI)+25P'L^3/(1944EI)= |
2.0335 |
mm |
|
f=5q'L^4/(384EI)+25P'L^3/(1944EI)= |
2.0334712 |
|
|
|
|
Since f≤[f] ,挠度合格, It is safe. |
|
|
|
|
|
|
|
Since σ'≤[σ] ,支撑梁强度合格, It is safe. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
XlsxShop
