Post on 21-Jul-2015
οΏ½ οΏ½πΌπππππππππππ
πΈπ
π πππππ
http://www.elrincondelingeniero.com/ For the structure given in the figure, determine:
a) Degree of static indeterminacy and its implications.
b) Reaction forces at the supports.
c) Force laws in the whole structure (normal and shear forces as well as and bending).
d) Force laws diagrams obtained in part d) of the exercise.
a) Stability
EDSI = 4 β 3 = 1
IDOF = 3(2 β 1) = 3
IL = 2(2 β 1) = 2
IDSI = 2 β 3 = β1
DSI = EDSI + IDSI = 1 β 1 = 0
The structure is entirely linked and the reactions exerted by the supports can be determined just by using the three scalar equations of equilibrium.
b) Reaction calculation
οΏ½MB = 2Ha β 2.1 = 0 βππ = π π€π (left)
οΏ½ Fx = 0 β HD = 2 β 1 β ππ = π π€π (left)
οΏ½Me = 0 β 3Va + 2.1 β12
. 2.3.23β 2.1 = 0
ππ = πποΏ½ π€π
οΏ½ Fy = 0 β Ve =12
. 3.2 β23
ππ = πποΏ½ π€π
c) Force laws
{(0 β€ y β€ 1) β© (x = 0)}
ππ = βπ π π€π
ππ = π π€π
ππ = π² π€π.π¦
{(1 β€ y β€ 2) β© (x = 0)}
ππ = βπ π
π€π
V2 = 1 β 2 β ππ = βπ π€π
M2 = y β 2(y β 1)
ππ = βπ²+ π π€π.π¦
{(0 β€ x β€ 1) β© (y = 2)}
ππ = βπ π€π
ππ =π π
π€π
M3 =2 3
x β 2.1 + 1.2 β ππ =π π π± π€π.π¦
οΏ½ οΏ½πΌπππππππππππ
πΈπ
π πππππ
http://www.elrincondelingeniero.com/ {(1 β€ x β€ 3) β© (y = 2)}
q(x) =qxL
=3x2
kNm
VT(y) = οΏ½3ΖΊ2dΖΊ =
xβ1
0οΏ½3ΖΊ2
4οΏ½0
xβ1
VT =34
(x β 1)2
MT(y) = οΏ½3ΖΊ2
4dΖΊ =
xβ1
0οΏ½ΖΊ3
4οΏ½0
xβ1
MT =(x β 1)3
4
ππ =π π βππ
(π± β π)π π€π
ππ =ππ± π
β(π± β π)π
ππ€π.π¦
Mmax βdM(x)
dx= V(x)
V(x) = β0,75x2 + 1,5x β 0,083 kN = 0
x = 1,93 m
ππ¦ππ± = M4(x = 1,93 ) = π,ππ π€π.π¦
{(x = 3) β© (0 β€ y β€ 2)}
ππ = βπ π π€π
ππ = π π€π
ππ = π π€π.π¦
d) Force laws diagrams
At this point there is enough information to completely define the force laws and bending moment diagrams that are shown in the next page: