Review: Trusses
• Each joint consists of a single pin to which the
respective members are connected individually.
• No member extends beyond a joint.
• Loads are applied only at joints.
• Each Member is a 2-force member and carries
only axial load
Internal forces in 2-force members
W
T
T
A
W
W
The internal force in a straight 2-force member consists of a
force acting normal to the surface of internal surfaces that are
perpendicular to the member. Force can be positive or negative.
Positive force is tension, negative force is compression.
1
Frames
• Structures consisting of structural members
connected in arbitrary ways.
Truss vs Frames
P
P
P
P
4
2
Calculating forces in structures made of 2force members
Find the internal force in each member of the idealized
bike frame. Assume all joints are hinges; all members
have equal length.
B
C
W
A
D
Method of Joints Net force at each joint is zero.
B
FAB
FBC
FAB
Ax
A
C
FCD
FBD
W
FAD
FBC
Cy
FBD
FCD
FAD
All angels 60o
D
Ay
8 Unknowns: Force in each member (5): FAB, FBC, FAD, FCD, FBD
reaction forces (3) Ax, Ay, Cy
8 Equations: Fx= Fy=0 (2 at each of the 4 jonts)
6
3
B
Equations
FAB
FBC
FAB
FAD
Ax
A
C
FCD
FBD
W
All angles 60o
FBC
Cy
FBD
FCD
FAD
D
Ay
F
F
F
F
Ax
Bx
Cx
Dx
 FAD  FAB cos 60  Ax  0,  FAy  FAB sin 60  Ay  0
 FBC  FBD cos 60  FAB cos 60  0,  FBy  W  FAB sin 60  FBD sin 60  0
  FBC  FCD cos 60  0,  FCy   FCD cos 60  C y  0
  FAD  FBD cos 60  FDC cos 60  0,  FDy  FBD sin 60  FCD sin 60  0
w
2w
2 3w
2 3w
4 3w
{ ax 0, cy , ay
, fcd 
, fad 
, fab 
, fbc 
3
3
9
9
9
3w
9
, fbd 
2
3w
9
}
7
8
4
3D problem:  Fx   Fy   Fz  0
At each joint
D
C
Dz
k
a
3a
Dy
j
a
Dx
Cy
i
A
5a
Cx
2a
FAD
Cz
FAC
FAD
W
B
FAC
Equations:Fx=Fy=Fz=0 at each joint (12)
Unknowns: Total of 12:
Member forces FAC, FAD, FAB, (3)
Reactions :Bx,By,Bz, Cx,Cy,Cz, Dx,Dy,Dz (9)
By
FAB
FAB
W
Bz
Bx
9
Forces act parallel to the members
C
3a
j
FAD
D
j
k
a
FAC
k
i
i
A
a
5a
2a
A
FAB
W
W
B
Member
FABn AB  FAC n AC  FADn AD  Wi  0
Unit Vector pointing away from Joint A.
Member AB
n AB  (2ai  5ak ) / 29a 2  (2i  5k ) / 29
Member AC
n AC  (3i  j  5k ) 35
Member AD
n AD  (3i  j  5k ) 35
10
5
j
TAD
Joint A
k
i
TAC
A
TAB
W
FABn AB  FAC n AC  FAD n AD  Wi  0
FAB (2i  5k ) / 29  FAC (3i  j  5k ) / 35  FAD (3i  j  5k ) / 35  Wi  0
FAx  2 FAB / 29  3FAC / 35  3FAD / 35  W  0
FAy   FAC / 35  FAD / 35  0
FAz  5FAB / 29  5FAC / 35  5FAD / 35  0
FAD  FAC  ( 35 /10)W
FAB  ( 29 / 5)W
11
D
C
Joints B,C,D
3a
j
k
a
i
A
a
5a
2a
W
B
By
Dy
Cy
FAB
Dz
Cz
Bz
Cx
Bx
Dx
FAC
FAD
FAB n BA  Bx i  By j  Bz k  0
FAC nCA  Cx i  C y j  Cz k  0
FAD n DA  Dx i  Dy j  Dz k  0
 FAB n BA  Bx i  By j  Bz k  0
 FAC n AC  Cx i  C y j  Cz k  0
 FAD n AD  Dx i  Dy j  Dz k  0
F
F
F
F
F
F
F
F
F
Bx
  FAB 2 / 29  Bx  0
By
 By  0
Bz
 5FAB / 29  Bz  0
Cx
  FAC 3/ 35  Cx  0
Cy
 FAC / 35  C y  0
Cz
 5FAC / 35  Cz  0
Dx
 FAD 3/ 35  Dx  0
Dy
 FAD / 35  Dy  0
Dz
 5FAD / 35  Dz  0
12
6