Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

F 1 F 1 F 2
Since the axial members and the 220 kN load are arranged
symmetrically relative to midpoint B of the rigid
D E F
beam, the forces in the two aluminum bars (1) must be
identical. The internal forces in the axial members are
related to their deformations by Equation (5.2). Because
members (1) and (2) are connected to rigid beam
2 m 2 m
ABC, they are not free to deform independently of each
other. On the basis of this observation and considering
the symmetry of the structure, we can assert that the
deformations in members (1) and (2) must be equal.
(1) (2)
(1) This fact can be combined with the relationship between
A B C
the internal force in a member and the member’s
deformation [Equation (5.2)] to derive another equation,
which is expressed in terms of the unknown member
220 kN
forces F 1 and F 2 . This equation is called a compat-
ibility equation. The equilibrium and compatibility
0.75 m 0.75 m
equations can be solved simultaneously to calculate the
member forces. Then, after F 1 and F 2 have been determined,
the normal stresses in each bar and the deflection
of rigid beam ABC can be calculated.
(1) (2)
(1)
A B C
104
220 kN
0.75 m 0.75 m
SOLUTION
Step 1 — Equilibrium Equations: An FBD of
rigid beam ABC is shown. From the overall symmetry
of the structure and the loads, we know that
the forces in members AD and CF must be identical;
therefore, we will denote the internal forces in
each of these members as F 1 . The internal force in
member BE will be denoted F 2 .
From this FBD, equilibrium equations can
be written for (a) the sum of forces in the vertical
direction (i.e., the y direction) and (b) the sum of moments about joint A:
∑ F = 2F + F − 220 kN = 0
(a)
y 1 2
∑ M = (1.5 m) F + (0.75 m) F − (0.75 m)(220 kN) = 0
A 1 2 (b)
Two unknowns appear in these equations (F 1 and F 2 ), and at first glance it seems as
though we should be able to solve them simultaneously for F 1 and F 2 . However, if
Equation (b) is divided by 0.75 m, then Equations (a) and (b) are identical. Consequently,
a second equation that is independent of the equilibrium equation must be
derived in order to solve for F 1 and F 2 .
Step 2 — Geometry of Deformation: By symmetry, we know that rigid beam ABC
must remain horizontal after the 220 kN load is applied. Consequently, joints A, B, and
C must all displace downward by the same amount: v A = v B = v C . How are these rigidbeam
joint displacements related to member deformations δ 1 and δ 2 ? Since the members
are connected directly to the rigid beam (and there are no other considerations,
such as gaps or clearances in the pin connections),
v = v = δ and v = δ
(c)
A C 1 B 2