Geomechanics and Ground Control · Engineering Mechanics · 2 marks · Numerical answer
Three collinear steel-bar segments have lengths 250, 500, 250 mm and equal area 0.01 m². E=200 GPa. Axial forces at successive nodes are 400 kN right, 500 kN left, 300 kN right, 200 kN left. Find overall length-change magnitude in mm, to three decimals.
Taking tension positive, the internal segment forces are−400, +100, −200 kN. Divide the sum of N×L by AE, with A=10000 mm² and E=200000 N/mm². Contributions are−0.050, +0.025, −0.025 mm, totalling−0.050 mm. The bar therefore shortens by 0.050 mm. The official positive interval grades the magnitude; a signed extension would be negative.
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