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

`A = 0.1237mm^2`

Step-by-step 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