Question

Member BC is composed of two members, one on each side of member AB. Find the minimum combined cross-sectional area of member BC if the ultimate normal stress is 400 MPa and the factor of safety is 3.5.

Answer 1

Answer:

$A = 0.1237mm^2$

Explanation:

Let's consider the moment equilibrium condition about point named A.

$summation MA = 0$

$-20 * 10^3 (1100 * 10^-3) -Fbc cos45° (2200 * 10^3) = 0$

Fbc = -141.42KN

The formula for minimum combined cross-sectional area is

$(Φu / factor of safety) = ( Fbc / Area)$

Let's make Area subject of the formula, we have;

Area = ( Fbc * factor of safety) /( Φu)

$Area = (- 141.42 * 10^3 *3.5) / (400 * 10^6)$

$Area = 0.1237mm^2$

Note: I couldn't add the diagram