Question 3.6: The rigid, weightless beam AC in Fig. 1 is supported at end ......
The rigid, weightless beam AC in Fig. 1 is supported at end A by a vertical column, and at end C it is supported by a vertical rod CD that is attached at D to a leveling jack, that is, a jack that can support the required load in rod CD but can also move the vertical position of D upward or downward.
Point B, the point of application of load P, can be anywhere along the beam AC, that is, anywhere in the range 0 < a < 1. The purpose of the leveling jack is to keep beam AC perfectly level, regardless of the location of B.
(a) Determine the axial stresses σ_1 in the column and σ_2 in the rod when the load P is on the beam. (b) Determine the downward displacement u_A of end A when the load is on the beam. (c) Finally, determine the displacement u_D required at the leveling jack for the beam AC to be level under the given loading, that is, to make u_C = u_A. The relevant dimensions and material properties are:
P = 2 kips, L_1 = 10 ft, L_2 = 5 ft, L = 10 ft, a = 0.4
A_1 = 2 in^2, A_2 = 0.8 in^2, E_1 = E_2 = 30 = 10^3 ksi
Plan the Solution A free-body diagram of beam AC may be used in solving for the axial forces in the column and support rod. Equation 3.14 can be used to express the elongation of each element in terms of its axial force. Finally, the elongations of the two elements can be expressed in terms of the displacements u_A = u_C and u_D.
e = fF, \text{where} f ≡\frac{L}{AE} (3.14)
(a) Axial stresses σ_1 and σ_2.
Equilibrium: The free-body diagram of beam AC in Fig. 2 will enable us to write two equilibrium equations that relate the axial force F_1 in the column and F_2 in the support rod to the external load P.
+\circlearrowleft \left(\sum{M} \right) _C = 0: F_1L + PL(1 – a) = 0
+\circlearrowleft \left(\sum{M} \right) _A = 0: F_2L – PaL = 0
or
F_1 = -P(1 – a) = -2 kips (0.6)
= -1.2 kips Equilibrium (1a,b)
F_2 = Pa = 2 kips (0.4) = 0.8 kips
Therefore, the axial stresses are
σ_1 = \frac{F_1}{A_1} = \frac{-1.2 Kips}{2 in^2} = -0.6 ksi = 0.6 ksi C (a)
σ_2 = \frac{F_2}{A_2} = \frac{0.8 Kips}{0.8 in^2} = 1.0 ksi = 1.0 ksi T
(b) The downward displacement u_A. To determine the displacement u_A, we need to determine the change in the length of the column due to the compressive force acting on it. At the same time, we can determine the elongation of the rod due to the force F_2 acting on it.
Element Force-Deformation Behavior: For the two uniform elements, the force-deformation equation, Eq. 3.14. gives
e_i = f_iF_b where f_i = \frac{L_i}{A_i E_i} i = 1, 2 Element Force- Deformation (2)
Inserting numerical values into Eq. (2) gives the two element flexibility coefficients
f_1 = \frac{L_1}{A_1 E_1} = \frac{(120 in.)}{(2 in^2)(30 \times 10^3 ksi)} = 2.00(10^{-3}) in./kip
f_2 = \frac{L_2}{A_2 E_2} = \frac{(60 in.)}{(0.8 in^2)(30 \times 10^3 ksi)} = 2.50(10^{-3}) in./kip
Note that these two elements are almost equally flexible, even though one is twice as long as the other. Inserting these flexibility coefficients, together with the element axial forces from Eqs. (1a,b), into Eq. (2) gives the two elements elongations
e_1 = f_1F_1 = -2.40(10^{-3}) in. (2′a,b)
e_2 = f_2F_2 = 2.00(10^{-3}) in.
Thus, the column is shortened due to its compressive force while the rod elongates due to its tensile force.
Geometry of Deformation: The elongation e_i of each element is the difference between the displacements of its two ends:
e_1 = -u_A Geometry of Deformation (3a,b)
e_2 = u_C – u_DTherefore, from Eqs. (2′a) and (3a), the vertical displacement at A is
u_A = 2.40(10^{-3}) in. ↓ (b) (4)
(c) The displacement u_D at the top of the support rod. The beam BC is level whenever
u_C = u_A (5)
Therefore, from Eqs. (2′b), (3b), (4), and (5),
u_D = u_C – e_2 = 2.40(10^{-3}) in. – 2.00(10^{-3}) in.
so
u_D = 0.40(10^{-3}) in. ↓ (c)
Thus, to keep the beam AC level for the given position of the load, it is necessary for the leveling jack to support a load of 0.8 kips while allowing rod-attachment point D to move downward the small amount 0.40(10^{-3}) in.
Review the Solution For statically determinate problems, like this one, the element forces can be determined from equilibrium alone, independent of the member sizes or the materials from which the members are made. However, wherever displacements are required, the additional equations come from the element force-deformation equations and the geometry-of-deformation equations.
