Related Variational and Energy Principles - Mechanics of Structures: Variational and Computational Methods - page 148

FIGURE 3.6
Frame.

The remaining reactions are found by applying the conditions of equilibrium to the free-
body diagram of Fig. 3.5b. This gives
R Ha

= 5 . 795 kN , = 14 . 346 kN , = 17 . 275 kN · m (11)

Va a

EXAMPLE 3.9 Combined Effects on Bar Systems
Determine the horizontal displacement of point d on the frame of Fig. 3.6a.
This structure, along with the structures of Fig. 3.7, is loaded such that more than either
simple extension or simple bending is generated. Apart from shear deformation, which can
always be included in structures built of beam members, a portion of the frame of Fig. 3.6a,
i.e., element 2, is subject to both extension and bending. The displacement at point b due
to the extension of element 2 is considered to be insignificant relative to the displacement
caused by bending and is sometimes ignored, but it will be considered here. Some sections
of the structures in Fig. 3.7 are also subjected to extension (or compression) and bending,
e.g., element 1 in Fig. 3.7b and all of the bar in Fig. 3.7c, while others are subjected to both
torsion and bending, e.g., element 1 in Fig. 3.7a and all of the bar of Fig. 3.7d. Member 1 in
Fig. 3.7e is deformed by bending, torsion, and compression.
First the internal bending moment and axial force in each bar of th e frame of Fig. 3.6a
will be determined. The equilibrium equations yield (Fig. 3.6b) M

3 3

| = Px, N | = 0 ,

2 2 1

Px ,

1

M | =− Pa, N | = P, M | =− Pa + N | = 0

where M N bar 2 , etc.

To find the displacement of point d in the d ire ction of P, take the de rivative of the total
complementary strain energy with respect to P : V d horizontal = ∂

|

3

= M bar 3 ,N |

2

=

U i /∂

P . Since this leads to

integrals of the form +

M

EI ∂

M

EA ∂

N N

P dx P dx compute ∂ ∂

∂ M |

3

∂ N |

3

∂ M |

2

P =

x,

P =

0 ,

P =−

a,

∂ ∂ ∂

(1)

2 1 1

∂ N | ∂ M | ∂ N |

x , = 1 ,

P =−

a + = 0

∂ P ∂ ∂ P