Answer:
axial stress in bar B = 25Mpa.
Deformation of bar A = 0.4mm.
Explanation:
PS: Kindly check the attached picture for the diagram showing the two bars that is to say the bar A and the bar B.
So, we are given the following data or information or parameters which we are going to use in solving this particular question or problem. Here they are;
The cross-sectional areas of Bars A and B = 400 mm2, the modulus of elasticity of bar A and bar B = 200 GPa, applied force = 10kN.
STEP ONE: The first step is to determine or calculate the axial stress in bar B. Therefore,
Axial stress in bar B = 10 × 10³ ÷ 400 × 10⁻⁶ = 25 Mpa.
STEP TWO: The second step here is to determine or calculate the deformation of bar A. Therefore,
The deformation of bar A = 20 × 10³ ×1.5 ÷ 400 × 10⁻⁶ × 200 × 10³ = 0.375 mm.
