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## Homework Statement

**1.Explain how one with the help of the free body diagram can calculate forces and moments on the vertical cut line through the beam in a distance 1 m to the right from the point C.**

**2. Explain how one calculate size and distribution normal stress of the cut line. The beam is a I profile with moment of inerti I= 22,2*106 mm4 and height = 162 mm.**## Homework Equations

Part 1.

∑Fx = 0 ∑Fy = 0 ∑Fz = 0

∑Mx = 0 ∑My = 0 ∑Mz = 0.

Part 2.

The elastic flexure formula:

σx = - (Mr y / I)

## The Attempt at a Solution

Part 1.

I have applied the equations of equilibrium to the full beam to determine the unknown forces acting on it.

+ ∑Ma = 0: -20 kN(2 m) + Cy(5.5 m) = 0 → 5.5Cy = 40 + 112.5 → Cy = 27,7 kN

↑ + ∑Fy = 0: -Ay – 20 kN + 27,7 kN = 0 → Ay = -20 + 27,7 → Ay = 7.7 kN

→ + ∑Fx = 0: Ax = 0 kN

This might be overdoing it however, since we are only asked about the forces and moments on the vertical cut line through the beam in a distance 1 m to the right from the point C.

Here I have attempted

+ ∑Ma = 0: -20 kN(2 m) + Cy(4 m) = 0 → 4Cy = 40 → Cy = 10 kN

↑ + ∑Fy = 0: -Ay – 20 kN + 10 kN = 0 → Ay = -20 + 10 → Ay = -10 kN

→ + ∑Fx = 0: Ax = 0 kN

Is this correct.

Part 2.

We are given:

I = 22,2*106 mm4

H = 162 mm.

So do we simply put the numbers in

σx = - (Mr y / I)

= - 27,7 kN x 0,162 m / 22,2*106 mm4 .

This part confuses me. Help would be appriciated.