Solving Beam Bending Problems using Singularity Functions¶
To make this document easier to read, we are going to enable pretty printing.
>>> from sympy import *
>>> x, y, z = symbols('x y z')
>>> init_printing(use_unicode=True, wrap_line=False)
Beam¶
A Beam is a structural element that is capable of withstanding load primarily by resisting against bending. Beams are characterized by their second moment of area, their length and their elastic modulus.
In Sympy, we can construct a 2D beam objects with the following properties :
Length
Elastic Modulus
Second Moment of Area
Variable : A symbol that can be used as a variable along the length. By default, this is set to
Symbol(x).- Boundary Conditions
bc_slope : Boundary conditions for slope.
bc_deflection : Boundary conditions for deflection.
Load Distribution
We have following methods under the beam class:
apply_load
solve_for_reaction_loads
shear_force
bending_moment
slope
deflection
Examples¶
Let us solve some beam bending problems using this module :
Example 1¶
A beam of length 9 meters is having a fixed support at the start. A distributed constant load of 8 kN/m is applied downward from the starting point till 5 meters away from the start. A clockwise moment of 50 kN-m is applied at 5 meters away from the start of the beam. A downward point load of 12 kN is applied at the end.
\\\\| 8 KN/m
\\\\|_____________
\\\\|| | | | | | | ⭯ 50 KN-m 12 KN
\\\\|V V V V V V V | |
\\\\|________________|_______________V
\\\\|________________|_______________|
\\\\| :
\\\\|----------------|---------------|
5.0 m 4.0 m
Note
Since a user is free to choose its own sign convention we are considering the upward forces and clockwise bending moment being positive.
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E, I = symbols('E, I')
>>> R1, M1 = symbols('R1, M1')
>>> b = Beam(9, E, I)
>>> b.apply_load(R1, 0, -1)
>>> b.apply_load(M1, 0, -2)
>>> b.apply_load(-8, 0, 0, end=5)
>>> b.apply_load(50, 5, -2)
>>> b.apply_load(-12, 9, -1)
>>> b.bc_slope.append((0, 0))
>>> b.bc_deflection.append((0, 0))
>>> b.solve_for_reaction_loads(R1, M1)
>>> b.reaction_loads
{M₁: -258, R₁: 52}
>>> b.load
-2 -1 0 -2 0 -1
- 258⋅<x> + 52⋅<x> - 8⋅<x> + 50⋅<x - 5> + 8⋅<x - 5> - 12⋅<x - 9>
>>> b.shear_force()
-1 0 1 -1 1 0
- 258⋅<x> + 52⋅<x> - 8⋅<x> + 50⋅<x - 5> + 8⋅<x - 5> - 12⋅<x - 9>
>>> b.bending_moment()
0 1 2 0 2 1
- 258⋅<x> + 52⋅<x> - 4⋅<x> + 50⋅<x - 5> + 4⋅<x - 5> - 12⋅<x - 9>
>>> b.slope()
3 3
1 2 4⋅<x> 1 4⋅<x - 5> 2
- 258⋅<x> + 26⋅<x> - ────── + 50⋅<x - 5> + ────────── - 6⋅<x - 9>
3 3
─────────────────────────────────────────────────────────────────────
E⋅I
>>> b.deflection()
3 4 4
2 26⋅<x> <x> 2 <x - 5> 3
- 129⋅<x> + ─────── - ──── + 25⋅<x - 5> + ──────── - 2⋅<x - 9>
3 3 3
─────────────────────────────────────────────────────────────────
E⋅I
Example 2¶
There is a beam of length 30 meters. A moment of magnitude 120 Nm is applied in the clockwise direction at the end of the beam. A pointload of magnitude 8 N is applied from the top of the beam at the starting point. There are two simple supports below the beam. One at the end and another one at a distance of 10 meters from the start. The deflection is restricted at both the supports.
|| 8 N ⭯ 120 Nm
\/______________________________________________|
|_______________________________________________|
/\ /\
|------------|---------------------------------|
10 m 20 m
Note
Using the sign convention of upward forces and clockwise moment being positive.
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E, I = symbols('E, I')
>>> R1, R2 = symbols('R1, R2')
>>> b = Beam(30, E, I)
>>> b.apply_load(-8, 0, -1)
>>> b.apply_load(R1, 10, -1)
>>> b.apply_load(R2, 30, -1)
>>> b.apply_load(120, 30, -2)
>>> b.bc_deflection.append((10, 0))
>>> b.bc_deflection.append((30, 0))
>>> b.solve_for_reaction_loads(R1, R2)
>>> b.reaction_loads
{R₁: 6, R₂: 2}
>>> b.load
-1 -1 -2 -1
- 8⋅<x> + 6⋅<x - 10> + 120⋅<x - 30> + 2⋅<x - 30>
>>> b.shear_force()
0 0 -1 0
- 8⋅<x> + 6⋅<x - 10> + 120⋅<x - 30> + 2⋅<x - 30>
>>> b.bending_moment()
1 1 0 1
- 8⋅<x> + 6⋅<x - 10> + 120⋅<x - 30> + 2⋅<x - 30>
>>> b.slope()
2 2 1 2 4000
- 4⋅<x> + 3⋅<x - 10> + 120⋅<x - 30> + <x - 30> + ────
3
─────────────────────────────────────────────────────────
E⋅I
>>> b.deflection()
3 3
4000⋅x 4⋅<x> 3 2 <x - 30>
────── - ────── + <x - 10> + 60⋅<x - 30> + ───────── - 12000
3 3 3
──────────────────────────────────────────────────────────────
E⋅I
Example 3¶
A beam of length 6 meters is having a roller support at the start and a hinged support at the end. A clockwise moment of 1.5 kN-m is applied at the mid of the beam. A constant distributed load of 3 kN/m and a ramp load of 1 kN/m is applied from the mid till the end of the beam.
ramp load = 1 KN/m
constant load = 3 KN/m
|------------------------|
⭯ 1.5 KN-m
______________________|________________________
|_______________________________________________|
o | /\
|----------------------|-----------------------|
3.0 m 3.0 m
Note
Using the sign convention of upward forces and clockwise moment being positive.
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E, I = symbols('E, I')
>>> R1, R2 = symbols('R1, R2')
>>> b = Beam(6, E, I)
>>> b.apply_load(R1, 0, -1)
>>> b.apply_load(1.5, 3, -2)
>>> b.apply_load(-3, 3, 0)
>>> b.apply_load(-1, 3, 1)
>>> b.apply_load(R2, 6, -1)
>>> b.bc_deflection.append((0, 0))
>>> b.bc_deflection.append((6, 0))
>>> b.solve_for_reaction_loads(R1, R2)
>>> b.reaction_loads
{R₁: 2.75, R₂: 10.75}
>>> b.load
-1 -2 0 1 -1
2.75⋅<x> + 1.5⋅<x - 3> - 3⋅<x - 3> - <x - 3> + 10.75⋅<x - 6>
>>> b.shear_force()
2
0 -1 1 <x - 3> 0
2.75⋅<x> + 1.5⋅<x - 3> - 3⋅<x - 3> - ──────── + 10.75⋅<x - 6>
2
>>> b.bending_moment()
2 3
1 0 3⋅<x - 3> <x - 3> 1
2.75⋅<x> + 1.5⋅<x - 3> - ────────── - ──────── + 10.75⋅<x - 6>
2 6
>>> b.slope()
3 4
2 1 <x - 3> <x - 3> 2
1.375⋅<x> + 1.5⋅<x - 3> - ──────── - ──────── + 5.375⋅<x - 6> - 15.6
2 24
───────────────────────────────────────────────────────────────────────
E⋅I
>>> b.deflection()
4 5
3 2 <x - 3> <x - 3> 3
-15.6⋅x + 0.458333333333333⋅<x> + 0.75⋅<x - 3> - ──────── - ──────── + 1.79166666666667⋅<x - 6>
8 120
──────────────────────────────────────────────────────────────────────────────────────────────────
E⋅I
Example 4¶
An overhanging beam of length 8 meters is pinned at 1 meter from starting point and supported by a roller 1 meter before the other end. It is subjected to a distributed constant load of 10 KN/m from the starting point till 2 meters away from it. Two pointloads of 20KN and 8KN are applied at 5 meters and 7.5 meters away from the starting point respectively.
---> x
|
v y
10 KN/m
_____________ 20 KN 8 KN
| | | | | | | | |
V V V V V V V V V
_______________________________________________
|_______________________________________________|
/\ O
|-----|------|-----------------|----------|--|--|
1m 1m 3m 2m .5m .5m
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E,I,M,V = symbols('E I M V')
>>> b = Beam(8, E, I)
>>> E,I,R1,R2 = symbols('E I R1 R2')
>>> b.apply_load(R1, 1, -1)
>>> b.apply_load(R2, 7, -1)
>>> b.apply_load(10, 0, 0, end=2)
>>> b.apply_load(20, 5, -1)
>>> b.apply_load(8, 7.5, -1)
>>> b.solve_for_reaction_loads(R1, R2)
>>> b.reaction_loads
{R₁: -26.0, R₂: -22.0}
>>> b.load
0 -1 0 -1 -1 -1
10⋅<x> - 26.0⋅<x - 1> - 10⋅<x - 2> + 20⋅<x - 5> - 22.0⋅<x - 7> + 8⋅<x - 7.5>
>>> b.shear_force()
1 0 1 0 0 0
10⋅<x> - 26.0⋅<x - 1> - 10⋅<x - 2> + 20⋅<x - 5> - 22.0⋅<x - 7> + 8⋅<x - 7.5>
>>> b.bending_moment()
2 1 2 1 1 1
5⋅<x> - 26.0⋅<x - 1> - 5⋅<x - 2> + 20⋅<x - 5> - 22.0⋅<x - 7> + 8⋅<x - 7.5>
>>> b.bc_deflection = [(1, 0), (7, 0)]
>>> b.slope()
3 3
5⋅<x> 2 5⋅<x - 2> 2 2 2
────── - 13.0⋅<x - 1> - ────────── + 10⋅<x - 5> - 11.0⋅<x - 7> + 4⋅<x - 7.5> + 28.2916666666667
3 3
───────────────────────────────────────────────────────────────────────────────────────────────────
E⋅I
>>> b.deflection()
4 4 3 3
5⋅<x> 3 5⋅<x - 2> 10⋅<x - 5> 3 4⋅<x - 7.5>
28.2916666666667⋅x + ────── - 4.33333333333333⋅<x - 1> - ────────── + ─────────── - 3.66666666666667⋅<x - 7> + ──────────── - 28.7083333333333
12 12 3 3
────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
E⋅I
Example 5¶
A cantilever beam of length 6 meters is under downward distributed constant load with magnitude of 4.0 KN/m from starting point till 2 meters away from it. A ramp load of 1 kN/m applied from the mid till the end of the beam. A point load of 12KN is also applied in same direction 4 meters away from start.
---> x .
| . |
v y 12 KN . | |
| . | | |
V . | | | |
\\\\| 4 KN/m . | | | | |
\\\\|___________ . 1 KN/m | |
\\\\|| | | | | | . V V V V V V V
\\\\|V V V V V V |---------------|
\\\\|________________________________
\\\\|________________________________|
\\\\| : : :
\\\\|----------|-----|----|----------|
2.0 m 1m 1m 2.0 m
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E,I,M,V = symbols('E I M V')
>>> b = Beam(6, E, I)
>>> b.apply_load(V, 0, -1)
>>> b.apply_load(M, 0, -2)
>>> b.apply_load(4, 0, 0, end=2)
>>> b.apply_load(12, 4, -1)
>>> b.apply_load(1, 3, 1, end=6)
>>> b.solve_for_reaction_loads(V, M)
>>> b.reaction_loads
{M: 157/2, V: -49/2}
>>> b.load
-2 -1
157⋅<x> 49⋅<x> 0 0 1 -1 0 1
───────── - ──────── + 4⋅<x> - 4⋅<x - 2> + <x - 3> + 12⋅<x - 4> - <x - 6> - <x - 6>
2 2
>>> b.shear_force()
-1 0 2 2
157⋅<x> 49⋅<x> 1 1 <x - 3> 0 1 <x - 6>
───────── - ─────── + 4⋅<x> - 4⋅<x - 2> + ──────── + 12⋅<x - 4> - <x - 6> - ────────
2 2 2 2
>>> b.bending_moment()
0 1 3 2 3
157⋅<x> 49⋅<x> 2 2 <x - 3> 1 <x - 6> <x - 6>
──────── - ─────── + 2⋅<x> - 2⋅<x - 2> + ──────── + 12⋅<x - 4> - ──────── - ────────
2 2 6 2 6
>>> b.bc_deflection = [(0, 0)]
>>> b.bc_slope = [(0, 0)]
>>> b.slope()
1 2 3 3 4 3 4
157⋅<x> 49⋅<x> 2⋅<x> 2⋅<x - 2> <x - 3> 2 <x - 6> <x - 6>
──────── - ─────── + ────── - ────────── + ──────── + 6⋅<x - 4> - ──────── - ────────
2 4 3 3 24 6 24
──────────────────────────────────────────────────────────────────────────────────────
E⋅I
>>> b.deflection()
2 3 4 4 5 4 5
157⋅<x> 49⋅<x> <x> <x - 2> <x - 3> 3 <x - 6> <x - 6>
──────── - ─────── + ──── - ──────── + ──────── + 2⋅<x - 4> - ──────── - ────────
4 12 6 6 120 24 120
──────────────────────────────────────────────────────────────────────────────────
E⋅I
Example 6¶
An overhanging beam of length 11 meters is subjected to a distributed constant load of 2 KN/m from 2 meters away from the starting point till 6 meters away from it. It is pinned at the starting point and is resting over a roller 8 meters away from that end. Also a clockwise moment of 5 KN-m is applied at the overhanging end.
2 KN/m ---> x
_________________ |
| | | | | | | | | v y
V V V V V V V V V ⭯ 5 KN-m
____________________________________________________|
O____________________________________________________|
/ \ /\
|--------|----------------|----------|---------------|
2m 4m 2m 3m
>>> R1, R2 = symbols('R1, R2')
>>> E, I = symbols('E, I')
>>> b = Beam(11, E, I)
>>> b.apply_load(R1, 0, -1)
>>> b.apply_load(2, 2, 0, end=6)
>>> b.apply_load(R2, 8, -1)
>>> b.apply_load(5, 11, -2)
>>> b.solve_for_reaction_loads(R1, R2)
>>> b.reaction_loads
{R₁: -37/8, R₂: -27/8}
>>> b.load
-1 -1
37⋅<x> 0 0 27⋅<x - 8> -2
- ──────── + 2⋅<x - 2> - 2⋅<x - 6> - ──────────── + 5⋅<x - 11>
8 8
>>> b.shear_force()
0 0
37⋅<x> 1 1 27⋅<x - 8> -1
- ─────── + 2⋅<x - 2> - 2⋅<x - 6> - ─────────── + 5⋅<x - 11>
8 8
>>> b.bending_moment()
1 1
37⋅<x> 2 2 27⋅<x - 8> 0
- ─────── + <x - 2> - <x - 6> - ─────────── + 5⋅<x - 11>
8 8
>>> b.bc_deflection = [(0, 0), (8, 0)]
>>> b.slope()
2 3 3 2
37⋅<x> <x - 2> <x - 6> 27⋅<x - 8> 1
- ─────── + ──────── - ──────── - ─────────── + 5⋅<x - 11> + 36
16 3 3 16
────────────────────────────────────────────────────────────────
E⋅I
>>> b.deflection()
3 4 4 3 2
37⋅<x> <x - 2> <x - 6> 9⋅<x - 8> 5⋅<x - 11>
36⋅x - ─────── + ──────── - ──────── - ────────── + ───────────
48 12 12 16 2
───────────────────────────────────────────────────────────────
E⋅I
Example 7¶
There is a beam of length l, fixed at both ends. A concentrated point load
of magnitude F is applied in downward direction at mid-point of the
beam.
^ y
|
---> x
\\\\| F |\\\\
\\\\| | |\\\\
\\\\| V |\\\\
\\\\|_____________________________________|\\\\
\\\\|_____________________________________|\\\\
\\\\| : |\\\\
\\\\| : |\\\\
\\\\|------------------|------------------|\\\\
l/2 l/2
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E, I, F = symbols('E I F')
>>> l = symbols('l', positive=True)
>>> b = Beam(l, E, I)
>>> R1,R2 = symbols('R1 R2')
>>> M1, M2 = symbols('M1, M2')
>>> b.apply_load(R1, 0, -1)
>>> b.apply_load(M1, 0, -2)
>>> b.apply_load(R2, l, -1)
>>> b.apply_load(M2, l, -2)
>>> b.apply_load(-F, l/2, -1)
>>> b.bc_deflection = [(0, 0),(l, 0)]
>>> b.bc_slope = [(0, 0),(l, 0)]
>>> b.solve_for_reaction_loads(R1, R2, M1, M2)
>>> b.reaction_loads
⎧ -F⋅l F⋅l F F⎫
⎨M₁: ─────, M₂: ───, R₁: ─, R₂: ─⎬
⎩ 8 8 2 2⎭
>>> b.load
-2 -2 -1 -1 -1
F⋅l⋅<x> F⋅l⋅<-l + x> F⋅<x> l F⋅<-l + x>
- ───────── + ────────────── + ─────── - F⋅<- ─ + x> + ────────────
8 8 2 2 2
>>> b.shear_force()
-1 -1 0 0 0
F⋅l⋅<x> F⋅l⋅<-l + x> F⋅<x> l F⋅<-l + x>
- ───────── + ────────────── + ────── - F⋅<- ─ + x> + ───────────
8 8 2 2 2
>>> b.bending_moment()
0 0 1 1 1
F⋅l⋅<x> F⋅l⋅<-l + x> F⋅<x> l F⋅<-l + x>
- ──────── + ───────────── + ────── - F⋅<- ─ + x> + ───────────
8 8 2 2 2
>>> b.slope()
2
l
1 1 2 F⋅<- ─ + x> 2
F⋅l⋅<x> F⋅l⋅<-l + x> F⋅<x> 2 F⋅<-l + x>
- ──────── + ───────────── + ────── - ──────────── + ───────────
8 8 4 2 4
────────────────────────────────────────────────────────────────
E⋅I
>>> b.deflection()
3
l
2 2 3 F⋅<- ─ + x> 3
F⋅l⋅<x> F⋅l⋅<-l + x> F⋅<x> 2 F⋅<-l + x>
- ──────── + ───────────── + ────── - ──────────── + ───────────
16 16 12 6 12
────────────────────────────────────────────────────────────────
E⋅I
Example 8¶
There is a beam of length 4*l, having a hinge connector at the middle. It
is having a fixed support at the start and also has two rollers at a distance
of l and 4*l from the starting point. A concentrated point load P is also
applied at a distance of 3*l from the starting point.
---> x
\\\\| P |
\\\\| | v y
\\\\| V
\\\\|_____________________ _______________________
\\\\|_____________________O_______________________|
\\\\| /\ : /\
\\\\| oooo : oooo
\\\\|----------|-----------|----------|-----------|
l l l l
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E, I = symbols('E I')
>>> l = symbols('l', positive=True)
>>> R1, M1, R2, R3, P = symbols('R1 M1 R2 R3 P')
>>> b1 = Beam(2*l, E, I)
>>> b2 = Beam(2*l, E, I)
>>> b = b1.join(b2, "hinge")
>>> b.apply_load(M1, 0, -2)
>>> b.apply_load(R1, 0, -1)
>>> b.apply_load(R2, l, -1)
>>> b.apply_load(R3, 4*l, -1)
>>> b.apply_load(P, 3*l, -1)
>>> b.bc_slope = [(0, 0)]
>>> b.bc_deflection = [(0, 0), (l, 0), (4*l, 0)]
>>> b.solve_for_reaction_loads(M1, R1, R2, R3)
>>> b.reaction_loads
⎧ -P⋅l 3⋅P -5⋅P -P ⎫
⎨M₁: ─────, R₁: ───, R₂: ─────, R₃: ───⎬
⎩ 4 4 4 2 ⎭
>>> b.load
-2 -1 -1 -1
P⋅l⋅<x> 3⋅P⋅<x> 5⋅P⋅<-l + x> -1 P⋅<-4⋅l + x>
- ───────── + ───────── - ────────────── + P⋅<-3⋅l + x> - ──────────────
4 4 4 2
>>> b.shear_force()
-1 0 0 0
P⋅l⋅<x> 3⋅P⋅<x> 5⋅P⋅<-l + x> 0 P⋅<-4⋅l + x>
- ───────── + ──────── - ───────────── + P⋅<-3⋅l + x> - ─────────────
4 4 4 2
>>> b.bending_moment()
0 1 1 1
P⋅l⋅<x> 3⋅P⋅<x> 5⋅P⋅<-l + x> 1 P⋅<-4⋅l + x>
- ──────── + ──────── - ───────────── + P⋅<-3⋅l + x> - ─────────────
4 4 4 2
>>> b.slope()
⎛ 2 2 2 2⎞ ⎛ 1 2 2 2⎞ ⎛ 1 2 2 2⎞
⎜5⋅P⋅l P⋅<-2⋅l + x> P⋅<-3⋅l + x> P⋅<-4⋅l + x> ⎟ 0 ⎜ P⋅l⋅<x> 3⋅P⋅<x> 5⋅P⋅<-l + x> P⋅<-2⋅l + x> ⎟ 0 ⎜ P⋅l⋅<x> 3⋅P⋅<x> 5⋅P⋅<-l + x> P⋅<-2⋅l + x> ⎟ 0
⎜────── - ───────────── + ───────────── - ─────────────⎟⋅<-2⋅l + x> ⎜- ──────── + ──────── - ───────────── + ─────────────⎟⋅<x> ⎜- ──────── + ──────── - ───────────── + ─────────────⎟⋅<-2⋅l + x>
⎝ 48 4 2 4 ⎠ ⎝ 4 8 8 4 ⎠ ⎝ 4 8 8 4 ⎠
──────────────────────────────────────────────────────────────────── + ──────────────────────────────────────────────────────────── - ───────────────────────────────────────────────────────────────────
E⋅I E⋅I E⋅I
>>> b.deflection()
⎛ 2 3 3 3⎞ ⎛ 2 3 3 3⎞ ⎛ 3 2 3 3 3⎞
⎜ P⋅l⋅<x> P⋅<x> 5⋅P⋅<-l + x> P⋅<-2⋅l + x> ⎟ 0 ⎜ P⋅l⋅<x> P⋅<x> 5⋅P⋅<-l + x> P⋅<-2⋅l + x> ⎟ 0 ⎜7⋅P⋅l 5⋅P⋅l ⋅(-2⋅l + x) P⋅<-2⋅l + x> P⋅<-3⋅l + x> P⋅<-4⋅l + x> ⎟ 0
⎜- ──────── + ────── - ───────────── + ─────────────⎟⋅<x> ⎜- ──────── + ────── - ───────────── + ─────────────⎟⋅<-2⋅l + x> ⎜────── + ───────────────── - ───────────── + ───────────── - ─────────────⎟⋅<-2⋅l + x>
⎝ 8 8 24 12 ⎠ ⎝ 8 8 24 12 ⎠ ⎝ 24 48 12 6 12 ⎠
────────────────────────────────────────────────────────── - ───────────────────────────────────────────────────────────────── + ────────────────────────────────────────────────────────────────────────────────────────
E⋅I E⋅I E⋅I
Example 9¶
There is a cantilever beam of length 4 meters. For first 2 meters
its moment of inertia is 1.5*I and I for the other end.
A pointload of magnitude 20 N is applied from the top at its free end.
---> x
\\\\| |
\\\\| 20 N v y
\\\\|________________ |
\\\\| |_______________V
\\\\| 1.5*I _______I_______|
\\\\|________________|
\\\\| :
\\\\|----------------|---------------|
2.0 m 2.0 m
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E, I = symbols('E, I')
>>> R1, R2 = symbols('R1, R2')
>>> b1 = Beam(2, E, 1.5*I)
>>> b2 = Beam(2, E, I)
>>> b = b1.join(b2, "fixed")
>>> b.apply_load(20, 4, -1)
>>> b.apply_load(R1, 0, -1)
>>> b.apply_load(R2, 0, -2)
>>> b.bc_slope = [(0, 0)]
>>> b.bc_deflection = [(0, 0)]
>>> b.solve_for_reaction_loads(R1, R2)
>>> b.load
-2 -1 -1
80⋅<x> - 20⋅<x> + 20⋅<x - 4>
>>> b.shear_force()
-1 0 0
80⋅<x> - 20⋅<x> + 20⋅<x - 4>
>>> b.bending_moment()
0 1 1
80⋅<x> - 20⋅<x> + 20⋅<x - 4>
>>> b.slope()
⎛ 1 2 2 ⎞
⎜80⋅<x> - 10⋅<x> + 10⋅<x - 4> 120 ⎟
⎜─────────────────────────────── - ─── ⎟ ⎛ 1 2 2⎞ 0 ⎛ 1 2 2⎞ 0
⎜ I I 80.0⎟ 0 0.666666666666667⋅⎝80⋅<x> - 10⋅<x> + 10⋅<x - 4> ⎠⋅<x> 0.666666666666667⋅⎝80⋅<x> - 10⋅<x> + 10⋅<x - 4> ⎠⋅<x - 2>
⎜───────────────────────────────────── + ────⎟⋅<x - 2> + ──────────────────────────────────────────────────────── - ────────────────────────────────────────────────────────────
⎝ E E⋅I ⎠ E⋅I E⋅I
Example 10¶
A combined beam, with constant flexural rigidity E*I, is formed by joining
a Beam of length 2*l to the right of another Beam of length l. The whole beam
is fixed at both of its both end. A point load of magnitude P is also applied
from the top at a distance of 2*l from starting point.
---> x
|
\\\\| P v y |\\\\
\\\\| | |\\\\
\\\\| V |\\\\
\\\\|____________ ________________________|\\\\
\\\\|____________O________________________|\\\\
\\\\| : : |\\\\
\\\\| : : |\\\\
\\\\|------------|------------|-----------|\\\\
l l l
>>> from sympy.physics.continuum_mechanics.beam import Beam
>>> from sympy import symbols
>>> E, I = symbols('E, I')
>>> l = symbols('l', positive=True)
>>> b1 = Beam(l ,E,I)
>>> b2 = Beam(2*l ,E,I)
>>> b = b1.join(b2,"hinge")
>>> M1, A1, M2, A2, P = symbols('M1 A1 M2 A2 P')
>>> b.apply_load(A1, 0, -1)
>>> b.apply_load(M1, 0 ,-2)
>>> b.apply_load(P, 2*l, -1)
>>> b.apply_load(A2, 3*l, -1)
>>> b.apply_load(M2, 3*l, -2)
>>> b.bc_slope=[(0, 0), (3*l, 0)]
>>> b.bc_deflection=[(0, 0), (3*l, 0)]
>>> b.solve_for_reaction_loads(M1, A1, M2, A2)
>>> b.reaction_loads
⎧ -5⋅P -13⋅P 5⋅P⋅l -4⋅P⋅l ⎫
⎨A₁: ─────, A₂: ──────, M₁: ─────, M₂: ───────⎬
⎩ 18 18 18 9 ⎭
>>> b.load
-2 -2 -1 -1
5⋅P⋅l⋅<x> 4⋅P⋅l⋅<-3⋅l + x> 5⋅P⋅<x> -1 13⋅P⋅<-3⋅l + x>
─────────── - ────────────────── - ───────── + P⋅<-2⋅l + x> - ─────────────────
18 9 18 18
>>> b.shear_force()
-1 -1 0 0
5⋅P⋅l⋅<x> 4⋅P⋅l⋅<-3⋅l + x> 5⋅P⋅<x> 0 13⋅P⋅<-3⋅l + x>
─────────── - ────────────────── - ──────── + P⋅<-2⋅l + x> - ────────────────
18 9 18 18
>>> b.bending_moment()
0 0 1 1
5⋅P⋅l⋅<x> 4⋅P⋅l⋅<-3⋅l + x> 5⋅P⋅<x> 1 13⋅P⋅<-3⋅l + x>
────────── - ───────────────── - ──────── + P⋅<-2⋅l + x> - ────────────────
18 9 18 18
>>> b.slope()
⎛ 1 2 2⎞ ⎛ 1 2 2⎞ ⎛ 2 1 2 2 2⎞
⎜5⋅P⋅l⋅<x> 5⋅P⋅<x> 5⋅P⋅<-l + x> ⎟ 0 ⎜5⋅P⋅l⋅<x> 5⋅P⋅<x> 5⋅P⋅<-l + x> ⎟ 0 ⎜P⋅l 4⋅P⋅l⋅<-3⋅l + x> 5⋅P⋅<-l + x> P⋅<-2⋅l + x> 13⋅P⋅<-3⋅l + x> ⎟ 0
⎜────────── - ──────── + ─────────────⎟⋅<x> ⎜────────── - ──────── + ─────────────⎟⋅<-l + x> ⎜──── - ───────────────── - ───────────── + ───────────── - ────────────────⎟⋅<-l + x>
⎝ 18 36 36 ⎠ ⎝ 18 36 36 ⎠ ⎝ 18 9 36 2 36 ⎠
──────────────────────────────────────────── - ───────────────────────────────────────────────── + ───────────────────────────────────────────────────────────────────────────────────────
E⋅I E⋅I E⋅I
>>> b.deflection()
⎛ 2 3 3⎞ ⎛ 2 3 3⎞ ⎛ 3 2 2 3 3 3⎞
⎜5⋅P⋅l⋅<x> 5⋅P⋅<x> 5⋅P⋅<-l + x> ⎟ 0 ⎜5⋅P⋅l⋅<x> 5⋅P⋅<x> 5⋅P⋅<-l + x> ⎟ 0 ⎜5⋅P⋅l P⋅l ⋅(-l + x) 2⋅P⋅l⋅<-3⋅l + x> 5⋅P⋅<-l + x> P⋅<-2⋅l + x> 13⋅P⋅<-3⋅l + x> ⎟ 0
⎜────────── - ──────── + ─────────────⎟⋅<x> ⎜────────── - ──────── + ─────────────⎟⋅<-l + x> ⎜────── + ───────────── - ───────────────── - ───────────── + ───────────── - ────────────────⎟⋅<-l + x>
⎝ 36 108 108 ⎠ ⎝ 36 108 108 ⎠ ⎝ 54 18 9 108 6 108 ⎠
──────────────────────────────────────────── - ───────────────────────────────────────────────── + ─────────────────────────────────────────────────────────────────────────────────────────────────────────
E⋅I E⋅I E⋅I
