Any help with this question please!! Axialh Loaded members A steel bar AD (Fig.

Question

Any help with this question please!!

Axialh Loaded members A steel bar AD (Fig. 1) has a cross-sectional area of 260 mm2 and is loaded by force P1=12 kN, P2=8 kN, and P3=6 kN. The lengths of the segments of the bar are a=1.5 m, b=0.6 m, and c=0.9 m. Assuming that the modulus of elasticity E=210 Gpa, calculate the change in length delta of the bar. Does the bar elongate or shorten? By what amount P should the load P3 be increased so that the bar does not change in length when the three loads are applied? Fig. 1 Steel bar AD loaded by three forces P1, P2, and P3.

Explanation / Answer

elongation = PL/AE

so for the three forces

P1

dl1 = P1L1/AE = 12000 x 1.5 / 2.6 x 10^-4 x 210 x 10^9

dl1 = 3.296 x 10^-4 m ( elongation)

due to P2

dl1 = P2L2/AE = 8000 x 2.1 / 2.6 x 10^-4 x 210 x 10^9

dl1 = 3.0769 x 10^-4 m ( elongation)

due to P3

dl1 = P3L3/AE = 6000 x 3 / 2.6 x 10^-4 x 210 x 10^9

dl1 = 3.296 x 10^-4 m ( compression)

so net elongation

dl1 + dl2 - dl3

= 3.0769 x 10^-4 m ( elongation)

now inorder to counter the elongation

extra force of compression required will be

P = dlnet x AE / L

= 3.0769 x 10^-4 x 2.6 x 10^-4 x 210 x 10^9 / 3

  

= 5.599 kN

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