Internal forces in the frame

Calculation of internal forces in frame constructions is analogous to beam structures, with the difference that we will also have vertical bars, which does not make a big problem. The frame that I will calculate looks as follows.

Before calculating the internal forces, calculate the support reactions and forces in the joint. The next step will be to plan how we will divide the construction that we count. Everything is clearly visible in the picture below.

I divided the frame into 5 sections. The forces in the joint will be visible when I calculate parts C and D. Now we can start calculating the internal forces.

Section A-A, distance: 0 ≤ x ≤ 2

\Sigma X = 0
T (x) - 4,50kN = 0
T (x) = 4,50kN

\Sigma Y = 0
N (x) + 1kN = 0
N (x) = -1kN

\Sigma M = 0
-M (x) + 4,50x = 0
M (x) = 4,50kN * x
x =0 \; \; \; \; \; \; \; M (0) = 0kNm
x = 2\: \; \; \; \; \; \; M (2) = 9kNm

Section B-B, distance: 2 ≤ x ≤ 4

\Sigma X = 0
T (x) + 5kN - 4,50kN = 0
T (x) = -0,50kN

\Sigma Y = 0
N (x) + 1kN = 0
N (x) = -1kN

\Sigma M = 0
-M (x) - 5kN * (2-x) + 4,50kN * x = 0
M (x) = -5kN * (2-x) + 4,50kN * x
x = 2\; \; \; \; \; \; \; M (2) = 9kNm
x = 4\; \; \; \; \; \; \; M (4) = 8kNm

Section C-C, distance: 0 ≤ x ≤ 2

\Sigma X = 0
-N (x) - 0,50kN = 0
N (x) = -0,50kN

\Sigma Y = 0
T (x)-1kN = 0
T (x) = 1kN

\Sigma M = 0
M (x) + 1kN * x = 0
M (x) = -1kN * x
x = 0\; \; \; \; \; \; \; M (0) = 0kNm
x = 2\; \; \; \; \; \; \; M (2) = -2kNm

Section D-D, distance: 0 ≤ x ≤ 2

\Sigma X = 0
N (x) + 0,50kN = 0
N (x) = - 0,50kN

\Sigma Y = 0
-T (x) + 1kN - 2kN * x = 0
T (x) = 1kN - 2kN * x
x = 0\; \; \; \; \; \; \; T (0) = 1,00kN
x = 2\; \; \; \; \; \; \; T(2) = -3,00kN

\Sigma M = 0
-M (x) + 1kN * x - 2kN * x * x * 0.5
M (x) = 1kN * x - 2kN * x * x * 0.5
x = 0\; \; \; \; \; \; \; M (0) = 0kNm
x = 2\; \; \; \; \; \; \; M (2) = -2kNm

Section E-E, distance: 0 ≤ x ≤ 4

\Sigma X = 0
T (x) - 0,50kN = 0
T (x) = 0,50kN

\Sigma Y = 0
N (x) + 3kN = 0
N (x) = -3kN

\Sigma M = 0
M (x) + 0,50kN * x = 0
M (x) = -0,50kN * x
x = 0\; \; \; \; \; \; \; M (0) = 0kNm
x = 4\; \; \; \; \; \; \; M (4) = -2kNm

We already know the values of all internal forces, we can make their graphs.

 

Bending moment

Moment

 

Shear forces

 

Normal forces

 

We can see from the cutting force diagram that there will be some extreme value in part D. Let’s calculate it, we equate the equation of the cutting force from section D to zero.

-2 * x + 1 = 0
x = 0,50m

We substitute the distance to the bending moment equation from section D.

M (0,50) = 1 * 0,50 - 2 * 0,50 * 0,50 * 0,50
M (0,50) = 0,25 kNm

The extreme value is 0,25kNm.

In this way, we have reached the end of the task, I hope that everything is understandably explained and as usual I encourage you to take a few examples to consolidate knowledge. If everything is clear, we can move on!