Question 3.20: A (statically determinate) two-bar planar truss has the conf......
A (statically determinate) two-bar planar truss has the configuration shown in Fig. 1 when it is assembled. If a downward load P = 20 kN is applied to the pin at C, and, at the same time, element (1) is cooled by 20°C, (a) what are the stresses σ_1 and σ_2 in elements (1) and (2), respectively? (b) What are the horizontal and vertical displacements, u_C and ν_C) respectively?
(a) Determine the element stresses σ_1 and σ_2.
Equilibrium: We should use a free-body diagram of joint C shown in Fig. 2, summing forces in the x and y directions.
\underrightarrow{+}\sum{F_y} = 0: \frac{4}{5}F_1 – 20 kN = 0 F_1 = 25 kN
F_1 = 25 kN Equilibrium (1)
F_2 = -15 kN
Thus, the stresses are:
σ_1 = \frac{F_1} {A_1} = \frac{25000 N}{0.001 m^2} = 25 MPa
σ_2 = \frac{F_2} {A_2} = \frac{-15000 N}{0.001 m^2} = -15 MPa
σ_1 = 25 MPa (T), σ_2 = 15 MPa (C) Ans. (a)
(b) Determine the joint displacement components u_C and ν_C.
Element Force-Temperature-Deformation Behavior: From Eq. 3.27,
e = fF + αLΔT, f = \frac{L}{AE} (3.27)
e_i = f_iF_i + α_iL_iΔT_i Element Force-Temp.-Deformation Behavior (2)
where
f_1 = \left(\frac{ L}{AE}\right)_1 = \frac{2.5 m}{ (0.001 m^2)(200 × 10^9 N/m^2)} = 1.250(10^{-8}) m/N
f_2 = \left(\frac{ L}{AE}\right)_2 = \frac{1.5 m}{(0.001 m^2)(100 × 10^9 N/m^2)} = 1.500(10^{-8}) m/N
Then, from Eq. (2),
e_1 = (1.250 × 10^{-8} m/N)(25000 N) + (20 × 10^{-6}/°C)(2.5 m)(-20°C)
= -0.000 688 m = -0.688 mm
e_2 = (1.500 × 10^{-8} m/N)(-15000 N) = -0.225 mm
Geometry of Deformation: From Eq. 3.29,
e = e_u + e_ν = u cos θ + ν sin θ (3.29)
e_i = u_C \cos θ_i + ν_C \sin θ_i Geometry of Deformation (3)
\cos θ_1 = \frac{3}{5}, \sin θ_1 = -\frac{4}{5}, \cos θ_2 = 1, \sin θ_2 = 0
e_1 = \frac{3}{5}u_C – \frac{4}{5} ν_C = -0.688 mm
e_2 = u_C = -0.225 mm
Then,
u_C = -0.225 mm, ν_C = 0.691 mm Ans. (b)
Review the Solution The displacement of joint C, upward and to the left, seems a bit strange, since the applied force P pulls downward.
However, we must remember that since element (1) is cooled, it will tend to pull joint C upward and to the left, apparently overriding the downward displacement due to load P.
Note that Eq. (3), the geometry-of-deformation equation, is the only really “new” feature that distinguishes this truss problem from the axial-deformation problems treated in Section 3.5.
