Problem ...
x

A simply supported beam is subject to a distributed load w = −wo * cos 2π 
L

Figure shown below
a) Determine reaction forces at the ends
b) Plot shear-force and bending moment diagrams
c) Model the distributed load as a set of four concentrated loads determined by
integrating over each quarter length. Locate the load at the mid length of each
segment. I.e.
L
2
F2 = ∫ w( x )dx located at x =
L
4
3
L
8
Comment on the static equivalency of this model of the distributed load (is it equivalent
in force and moment?)
d) Plot shear-force and bending moment diagram for this loading. Comment on the
comparison with b) above.
Solution ....
wo := 1
L := 10
x
w( x) := −wo⋅cos 2⋅π⋅ 
L
x := 0, 0.1 .. L

as shown, w is positive up but negative in value
1
w ( x)
0
0
1
0
2
4
6
8
10
x
R0
RL

L
⌠
sum_forces :=  w( x) dx
⌡
sum_forces = 0
R0 + RL = 0
0
L
⌠
moment_wrt_R0 :=  x⋅ w( x) dx
⌡
moment_wrt_R0 = 0
0
a)
RL⋅L − moment_wrt_R0 = 0
b)
equilibrium
RL := 0
R0 := 0
x
⌠
R0 +  w( ξ ) dξ − V = 0
⌡
w(x)
0
M
V
R0
x
⌠
shear_force( x) :=  w( ξ ) dξ
⌡
0
2
hold this as
shear_force( x)
shear_force1(x) := shear_force( x)
2
w1( x) := −wo⋅cos
or ... analytically
0

0
5
10
x
x
x
⌠
⌠
L
x
x
shear_force( x) =  w1( ξ ) dξ = −wo  cos 2⋅π⋅  dξ = −wo⋅
⋅sin 2⋅π⋅ 

⌡
L
L
2⋅π


0
⌡
0
2
curves overlap
shear_force( x)
− w o⋅
L
2⋅π


⋅sin2⋅π⋅
x
0
L
2
0
5
10
x
x
equilibrium
⌠
R0 +  w( ξ ) dξ − V = 0
⌡
0
x
⌠
bending_moment( x) :=  shear_force( ξ ) dξ
⌡
0
0
bending_moment( x)
5
0
2
4
6
8
10
x
hold this as :
bending_moment1(x) := bending_moment( x)
or ... analytically:
x


⌠
x
L ⌠
bending_moment( x) =  shear_force( ξ ) dξ = −wo⋅
⋅  sin 2⋅π⋅  dξ
⌡
L
2⋅π  

0
⌡
x


0
2
L   
x
bending_moment( x) = wo⋅ 
⋅  cos 2⋅π⋅  − 1
L

 2⋅π   
the 1 is from the lower limit
0
bending_moment( x)
2
2
 L  ⋅ cos2⋅π⋅ x  −1
 
L 
 2⋅π   
w o⋅ 
4
0
2
4
6
8
10
x
c) model in segments:
i := 0 .. 3
ξ i , 0 := i⋅
following the notation in the notes:
L = 10
L
ξ i , 1 := (i + 1)⋅
4
ξi, 1
⌠
f := 
i
⌡ξ
w( x) dx
i,0
 −1.592 
1.592 
f= 
 1.592 
 −1.592


located at
L
4
0

2.5
ξ=
 5
 7.5

xx :=
i
2.5 
5

7.5 
10 
ξi, 1 + ξi, 0
2
1.25 

3.75 
xx = 
 6.25 
 8.75


x := 0, 0.1 .. L
ll := 0
ul := 3
ul
shear( x) :=
∑ fi⋅(x ≥ xxi)
2
i = ll
plotted with value from distributed
shear( x)
shear_force1( x)
0
2
0
5
10
x
ul
bending_moment( x) :=
∑ fi⋅(x − xxi)⋅(x ≥ xxi)
i = ll
0
bending_moment( x)
bending_moment1( x)
5
10
0
5
10
x
comment ... shear at the quarter points right on! bending moment a bit underestimated
bending_moment
L
 2  = 0.785
L
bending_moment1 
 2
about 20 % low .. but with just 4 segments
do the same but with 8 segments: (not expected)
i := 0 .. 7
ξ i , 0 := i⋅
L
8
L = 10
ξ i , 1 := (i + 1)⋅
L
8
 0
 1.25

 2.5
 3.75
ξ=
 5
 6.25
 7.5

 8.75
1.25 
2.5

3.75 


6.25

7.5 
8.75 
5
10

 −1.125 
 −0.466


 0.466 
 1.125 
f= 

 1.125 
 0.466 
 −0.466 

 −1.125 
ξi, 1
⌠
f := 
i
⌡ξ
w( x) dx
i,0
x := 0, 0.1 .. L
ll := 0
located at
xx :=
i
ξi, 1 + ξi, 0
2
 0.625 
 1.875


 3.125 
 4.375 
xx = 

 5.625 
 6.875 
 8.125 

 9.375 
ul := 7
ul
shear( x) :=
∑ fi⋅(x ≥ xxi)
2
i = ll
plotted with value from distributed
shear( x)
shear_force1( x)
ul
bending_moment( x) :=
∑ i(
i) (
f ⋅ x − xx ⋅ x ≥ xx
i = ll
i)
0
2
0
5
10
x
0
bending_moment( x)
bending_moment1( x)
5
10
0
5
x
comment ... shear at the quarter points right on! bending moment a bit underestimated
bending_moment
L
 2  = 0.948
L
bending_moment1 
 2
about 5 % low .. with 8 segments
10