Solving Oneaxial Shear Test: Find Shear Force

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SUMMARY

The discussion focuses on calculating the shear force for a oneaxial shear test with a specimen height of 100mm and diameter of 54mm, subjected to a maximum force of 144.6 N and a deformation of 5mm. The user calculated the strain (ε) as 0.05 and the stress (σz) as 0.025 N/mm², leading to a shear force of 0.0127 N/mm² or 12.7 kN/m². However, the expected answer is 30 kN/m², indicating a discrepancy in the calculations that requires further clarification regarding the specimen's shape and force application points.

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Zasha
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I´m really struggling to find the shear force here.
1. Homework Statement

height h of a specimen = 100mm and diameter is 54mm.
Max force F=144,6 N and deformation is δ=5mm

Homework Equations


I have to find ε=δ/h
Area of specimen before deformation=54mm*100mm=5400mm
σz =(F(1-ε))/A
Shear force = σz /2

The Attempt at a Solution


ε=δ/h=5/100=0.05
A=5400mm
σz =(F(1-ε))/A=(144,6(1-0.05))/5400=0,025N/mm2
Shear Force=0,025/2=0,0127N/mm2 which gives 12,7kN/m2. The right answer is 30kN/m2.

Any thoughts?
 
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I, for one, would need more description. What is the shape of the specimen? Where are the forces applied? Where is the deformation measured?
 

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