In Example 4.1, add a member BD of the same material and cross section as members AD and C...

In Example 4.1, add a member BD of the same material and cross section as members AD and CD. Determine the loads P required to cause deflections u = 2.00, 4.00, 4.481, and 8 mm. Plot the load–deflection curve for the structure.

Example 4.1

The stress–strain diagram for an isotropic metal at room temperature is approximated by two straight lines (Figure 4.4b). Pan AB has slope E and part BF has slope βE, where E is the modulus of elasticity and β is the strain-hardening factor for the metal. The intersection of the two lines defines the yield stress Y and yield strain ϵY = Y/E. The stress–strain relations in the region AB and BF are, respectively,

σ = Eϵ,   ϵ ≤ ϵY (elastic stress–strain)   (a)

σ = (1 − β)Y + βEϵ, ϵ > ϵY (elastic stress–strain)   (b)

(a) Determine the constants β, Y, E, and ϵY for the annealed high-carbon steel of Figure 4.1.

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(b) Consider the pin-joined structure in Figure E4.1a. Each member has a cross-sectional area 645 mm2 and is made of the steel of Figure 4.1. A load P = 170 kN is applied. Compute the deflection u.

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(c) Repeat part (b) for P = 270 kN and P = 300 kN.

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(d) Use the results of parts (b) and (c) to plot a load–deflection graph for the structure.

FIGURE E4.1: (a) Pin-joined structure. (b) Free-body diagram.

Solution:

(a) By Figure 4.1, E = 211.4 GPa, Y = 252.6 MPa, βE = 16.9 GPa (or β = 0.0799), and ϵY = Y/E = 0.001195. Equations (a) and (b) become

σ = 211,400ϵ,  ϵ ≤ ϵY (elastic stress–strain)   (c)

σ = 232.4 + 16,900ϵ, ϵ > ϵY (elastic stress–strain)   (d)

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(b) By the geometry of the structure shown in Figure E4.1a, cos θ = 0.8. By equilibrium of joint D (Figure E4.1b), we find the force F in members AD and CD to be F = P/(2cos θ)= 106.25 kN. Therefore,the stress in members AD and CD is σ = 164.73 MPa<252.6 MPa. Thus, for this load, themembers are deformed elastically. By Eq. (c), the strain in members AD and CD is ϵ = 0.000779.Since the length of members AD and CD is L = 3.0 m, their elongation is e = ϵL = 2.338 mm. Hence, u = e/cos θ = 2.922 mm.

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(c) For P = 270 kN, F = 168.75 kN and σ = 261.63 MPa (> 252.6 MPa). Hence, bars AB and CD are strained inelastically. Therefore, by Eq. (d), ϵ = 0.001730 and e = ϵL = 5.189 mm. Hence, u = e/cos θ = 6.486 mm. For P = 300 kN, F = 187.5 kN and σ = 290.7 MPa (> 252.6 MPa). By Eq. (d), the strain in the bars is ϵ = 0.003450, the elongation of the bars is e = ϵL = 10.349 mm, and the deflection is u = e/cos θ = 12.936 mm.

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(d) A summary of the load deflection data is given in Table E4.1, and the data are plotted in Figure E4.1c. The yield load (point F)is located by the intersection of the extensions of lines 0E and GH. Note that the ratio of the slope of line FGH to the slope of line 0EF is β = 0.0799.

TABLE E4.1: Load–Deflection Data

FIGURE E4.1: (c) Load–deflection plot.

FIGURE 4.4 Idealized stress–strain curves. (a) Elastic–perfectly plastic response. (b) Elastic–strain hardening response.

FIGURE 4.1 Tension and compression stress–strain diagrams for annealed high-carbon steel for initial and reversed loading. (From Sidebottom and Chang, 1952.)