Tension and compression tasks

Model name

length of segments, m (comma separated)


- Sectional areas

On-point loads, kN (up - positive, down - negative)

distribution load kN/m (up - positive, down - negative)

Elastic modulus, MPa

weigth of material γ = kN/m3


Design model #179710

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Detailed solution.

Axial forces

N1 = + 30 = 30kN

N2 = + 30 - 20 = 10kN

N3 = + 30 - 20 - 50 = -40kN

Normal stress

σ1 = 30000/30=1000 MPa

σ2 = 10000/80=125 MPa

σ3 = -40000/50=-800 MPa

Define deformation of segments by Hooke's law, including axial force N, kN, length l, m, area А, mm2 and elastic modulus E, MPa

Δl = N×l/E×A

Δl1 = 30000 × 0.8 / (210000 × 30) = 0.00381m

Δl2 = 10000 × 2 / (210000 × 80) = 0.00119m

Δl3 = -40000 × 1.2 / (210000 × 50) = -0.004571m

Common deformation

Δl = + 0.00381 + 0.00119 - 0.004571 = 0.000429 m


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