Examples
The following example represents 3-segment beam loaded by couple, linearly distributed load and two concentrated forces.

Input:
Length of the beam L: 2.5 Number of couples (external moments): 1
Number of concentrated loads: 2 Number of distributed loads: 1

M1 - value of the couple # 1 of 1: -2 a1 - coordinate of this couple: 2
F1 - value of the concentrated load # 1 of 2: -4 b1 - coordinate of this concentrated load: 1
qL1 - left value of the lineary distributed load # 1 of 1: 3 qR1 - right value of the lineary distributed load # 1 of 1: 2
c1 - left coordinate of this distributed load: 0 d1 - right coordinate of this distributed load: 2

F2 - value of the concentrated load # 2 of 2: -1 b2 - coordinate of this concentrated load: 2.5

Output:

               Maximum shear force V is -2.75 at x=1.0
               Maximum bending moment M is 3.8333333 at x=0.0
               Maximum deflection EIv is 9.508333 at x=2.5
     Reactions:
R=F1+F2+qL1(d1-c1)+(qR1-qL1)(d1-c1)/2=
-4.0-1.0+3.0(2.0-0.0)+(2.0-3.0)(2.0-0.0)/2=0.0
MR=-M1-F1b1-F2b2-qL1(c1-d1)(c1+d1)/2-(qR1-qL1)(d1-c1)/2(c1+2(d1-c1)/3)=
+2.0+4.0*1.0+1.0*2.5-3.0(2.0-0.0)(0.0+2.0)/2.0+(2.0-3.0)(2.0-0.0)/2.0(0.0+(2.0-0.0)2/3)=3.833333333333333

               The beam has 3 segments.
     Segment #1 (0.0≤x≤1.0).
Shear force:
V=R-qL1(x-c1)-(qR1-qL1)(x-c1)2/(d1-c1)/2=
0.0-3.0(x-0.0)-(2.0-3.0)(x-0.0)2/(2.0-0.0)/2=
0.0-3.0x+0.25x2
Bending moment:
M=MR+Rx-qL1(x-c1)2/2-(qR1-qL1)(x-c1)3/(d1-c1)/6=
3.8333333+0.0x-3.0(x-0.0)2/2-(2.0-3.0)(x-0.0)3/(2.0-0.0)/6=
3.8333333+0.0x-1.5x2+0.083333336x3
Deflection:
EIv=MRx2/2+Rx3/6-qL1(x-c1)4/24-(qR1-qL1)(x-c1)5/(d1-c1)/120=
3.833333333333333x2/2+0.0x3/6-3.0(x-0.0)4/24-(2.0-3.0)(x-0.0)5/(2.0-0.0)/120=
1.9166666x2+0.0x3-0.125(x-0.0)4+0.004166667(x-0.0)5
At x=0.0 V=-7.5E-9; M=3.8333333; EIv=1.1979167E-17;
At x=1.0 V=-2.75; M=2.4166667; EIv=1.7958333;

     Segment #2 (1.0≤x≤2.0).
Shear force:
V=R-F1-qL1(x-c1)-(qR1-qL1)(x-c1)2/(d1-c1)/2=
0.0+4.0-3.0(x-0.0)-(2.0-3.0)(x-0.0)2/(2.0-0.0)/2=
4.0-3.0x+0.25x2
Bending moment:
M=MR+Rx-F1(x-b1)-qL1(x-c1)2/2-(qR1-qL1)(x-c1)3/(d1-c1)/6=
3.8333333+0.0x+4.0(x-1.0)-3.0(x-0.0)2/2-(2.0-3.0)(x-0.0)3/(2.0-0.0)/6=
3.8333333+0.0x+4.0(x-1.0)-1.5x2+0.083333336x3
Deflection:
EIv=MRx2/2+Rx3/6-F1(x-b1)3/6-qL1(x-c1)4/24-(qR1-qL1)(x-c1)5/(d1-c1)/120=
3.833333333333333x2/2+0.0x3/6+4.0(x-1.0)3/6-3.0(x-0.0)4/24-(2.0-3.0)(x-0.0)5/(2.0-0.0)/120=
1.9166666x2+0.0x3+0.6666667(x-1.0)3-0.125(x-0.0)4+0.004166667(x-0.0)5
At x=1.0 V=1.25; M=2.4166667; EIv=1.7958333;
At x=2.0 V=-1.0; M=2.5; EIv=6.4666667;
There is maximum bending moment 2.740448 on this segment at x=1.53

     Segment #3 (2.0≤x≤2.5).
Shear force:
V=R-F1-qL1(x-c1)-(qR1-qL1)(x-c1)2/(d1-c1)/2+qR1(x-d1)+(qR1-qL1)(x-d1)2/(d1-c1)/2=
0.0+4.0-3.0(x-0.0)-(2.0-3.0)(x-0.0)2/(2.0-0.0)/2+2.0(x-2.0)-(2.0-3.0)(x-2.0)2/(2.0-0.0)/2=
4.0-3.0x+0.25x2+2.0(x-2.0)-0.25(x-2.0)2
Bending moment:
M=MR+M1+Rx-F1(x-b1)-qL1(x-c1)2/2-(qR1-qL1)(x-c1)3/(d1-c1)/6+qR1(x-d1)2/2+(qR1-qL1)(x-d1)3/(d1-c1)/6=
3.8333333-2.0+0.0x+4.0(x-1.0)-3.0(x-0.0)2/2-(2.0-3.0)(x-0.0)3/(2.0-0.0)/6+2.0(x-2.0)2/2+(2.0-3.0)(x-2.0)3/(2.0-0.0)/6=
1.8333334+0.0x+4.0(x-1.0)-1.5x2+0.083333336x3+1.0(x-2.0)2-0.083333336(x-2.0)3
Deflection:
EIv=MRx2/2+M1(x-a1)2/2+Rx3/6-F1(x-b1)3/6-qL1(x-c1)4/24-(qR1-qL1)(x-c1)5/(d1-c1)/120+qR1(x-d1)4/24+(qR1-qL1)(x-d1)5/(d1-c1)/120=
3.833333333333333x2/2-2.0(x-2.0)2/2+0.0x3/6+4.0(x-1.0)3/6-3.0(x-0.0)4/24-(2.0-3.0)(x-0.0)5/(2.0-0.0)/120+2.0(x-2.0)4/24-(2.0-3.0)(x-2.0)5/(2.0-0.0)/120=
1.9166666x2-1.0(x-2.0)2+0.0x3+0.6666667(x-1.0)3-0.125(x-0.0)4+0.004166667(x-0.0)5+0.083333336(x-2.0)4-0.004166667(x-2.0)5
At x=2.0 V=-1.0; M=0.5; EIv=6.4666667;
At x=2.5 V=-1.0; M=2.499999E-9; EIv=9.508333;