斜腿剛構計算
一、程式編制說明
本程式全部採用visual basic語言編寫，可以進行簡單杆繫結構的受力分析，能夠精確計算出結構體系的結點位移和內力，並採用文字方式輸入和輸出資料，方便實用。例子是針對斜腿剛構的計算，全橋共劃分為6個單元，計算簡圖如下。
二、子程式劃分
data：資料的準備與輸入
zg：總體剛體矩陣
dg：單元剛度矩陣
translate：座標轉換矩陣
result：杆端位移與杆端力計算
mult：矩陣相乘
mult1：矩陣與向量相乘
gas：高斯消元法
boundary：邊界條件
load：外荷載
三、識別符號說明：
1、整數型變數
nn：結點總個數
ele：單元總個數
gd：固定支座個數
jz：鉸支座個數
zc：支承個數
ms:系統總自由度數目（總剛階數）
pn：荷載作用的結點數（結點荷載個數）
2、一維陣列變數
ei：桿件抗彎剛度
ea：桿件抗拉剛度
cn：cos值
sn：sin值
ln：單元左端結點標號陣列
rn：單元右端結點標號陣列
fx：水平方向結點外力
fy：垂直方向結點外力
moment：結點彎矩
z：總體座標系下結點位移儲存
p：總體座標系下結點力儲存
3、二維陣列變數
bke：區域性座標單元剛度矩陣儲存
tke：體系總體剛度矩陣儲存
t：座標轉換矩陣儲存
四、源程式**
dim i, j as double
dim nn as integer, ele as integer, ms as integer
dim gd as integer, jz as integer, zc as integer, pn as integer
dim lrn(1 to 12, 1 to 12) as double, l(1 to 7) as double, g(1 to 21, 1 to 21) as double
dim ea(1 to 7) as double, ei(1 to 7) as double, cn as double, sn as double
dim x(1 to 7) as double, y(1 to 7) as double, p(1 to 21) as double
dim tke(1 to 21, 1 to 21) as double, tke1(1 to 21, 1 to 21) as double
dim bke(1 to 6, 1 to 6) as double, t(1 to 6, 1 to 6) as double
dim tz(1 to 6, 1 to 6) as double, z(1 to 21) as double
dim ln(1 to 6) as integer, rn(1 to 6) as integer
dim fx(1 to 7) as double, fy(1 to 7) as double, moment(1 to 7) as double
private sub form_click()
open "e:\homework\" for input as #1
open "e:\homework\" for output as #2
call data '輸入資料
call zg '總剛度矩陣
call boundary ' 輸入邊界條件
call load '輸入荷載
call result '計算結果（杆端力）
close #1
close #2
end sub
public sub data() '資料的準備與輸入
print #2, "平面杆系有限元程式——斜腿剛構計算"
print #2
print #2, "原始資料的輸入"
print #2,
print #2, "結點數"; spc(3); "單元數"; spc(3); "固定支座數"; spc(3); "鉸支座數"; spc(3); "支承數"; spc(3); "結點荷載數"; spc(3); "系統總自由度數"
input #1, nn, gd, jz, zc, pn
ms = 3 * nn: ele = nn - 1
print #2, nn; spc(6); ele; spc(6); gd; spc(6); jz; spc(6); zc; spc(6); pn; spc(6); ms
print #2,
print #2
print #2, "結點座標的輸入"
print #2,
print #2, "結點"; spc(3); "x"; spc(6); "y"; spc(3); "單位：公尺"
for i = 1 to nn
input #1, i, x(i), y(i)
print #2, i; spc(3); x(i); spc(6); y(i)
next i
print #2,
print #2
print #2, "桿件相關資訊的輸入"
print #2,
print #2, "桿件號"; spc(3); "左結點號"; spc(3); "右結點號"; spc(3); "抗彎剛度"; spc(3); "抗拉剛度"; spc(3); "桿件長度"; spc(3); "e=2.06e+11n/m2"
for i = 1 to ele
input #1, i, ln(i), rn(i), ei(i), ea(i)
next i
for i = 1 to ele
l(i) = sqr((y(rn(i)) - y(ln(i))) ^ 2 + (x(rn(i)) - x(ln(i))) ^ 2)
print #2, i; spc(8); ln(i); spc(8); rn(i); spc(8); format(ei(i), "0.000e+00"); spc(8); format(ea(i), "0.000e+00"); spc(8); (format(l(i), "0.
000"))
next i
'輸入結點編號
lrn(1, 1) = 3#: lrn(1, 2) = 3#: lrn(1, 3) = 4#
lrn(2, 1) = 1#: lrn(2, 2) = 2#: lrn(2, 3) = 3#
lrn(1, 4) = 5#: lrn(1, 5) = 7#: lrn(1, 6) = 6#
lrn(2, 4) = 4#: lrn(2, 5) = 5#: lrn(2, 6) = 5#
print #2,
print #2
print #2, "輸入結點荷載"
print #2,
print #2, "結點號"; spc(3); "fx"; spc(4); "fy"; spc(4); "moment"; spc(4); "單位：n"
for i = 1 to nn
input #1, i, fx(i), fy(i), moment(i)
next i
for i = 1 to nn
print #2, i; spc(3); format(fx(i), "0.0e+"); spc(4); format(fy(i), "0.0e+"); spc(4); format(moment(i), "0.
0e+")
next i
print #2,
print #2
print #2, "輸入約束條件"
end sub
public sub dg(k) '形成單元剛度矩陣
'for k = 1 to ele
for i = 1 to 6
for j = 1 to 6
bke(i, j) = 0#
next j
next i
bke(1, 1) = ea(k) / l(k)
bke(2, 1) = 0#
bke(2, 2) = 12 * ei(k) / l(k) * l(k) * l(k)
bke(3, 1) = 0#
bke(3, 2) = 6 * ei(k) / l(k) * l(k)
bke(3, 3) = 4 * ei(k) / l(k)
bke(4, 1) = -ea(k) / l(k)
bke(4, 2) = 0#
bke(4, 3) = 0#
bke(4, 4) = ea(k) / l(k)
bke(5, 1) = 0#
bke(5, 2) = -12 * ei(k) / l(k) * l(k) * l(k)
bke(5, 3) = -6 * ei(k) / l(k) * l(k)
bke(5, 4) = 0#
bke(5, 5) = 12 * ei(k) / l(k) * l(k) * l(k)
bke(6, 1) = 0#
bke(6, 2) = 6 * ei(k) / l(k) * l(k)
bke(6, 3) = 2 * ei(k) / l(k)
bke(6, 4) = 0#
bke(6, 5) = -6 * ei(k) / l(k) * l(k)
bke(6, 6) = 4 * ei(k) / l(k)
for i = 1 to 6
for j = 1 to 6
bke(i, j) = bke(j, i)
next j
next i
'print #2, k, "單剛"
' for i = 1 to ele
' for j = 1 to ele
' print #2, (format(bke(i, j), "0.000")); spc(2);
' next j
' print #2,
' next i
' next k
end sub
public sub translate(k) '形成座標變換矩陣
' for k = 1 to ele
cn = (x(rn(k)) - x(ln(k))) / l(k)
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