A load W=2 kN is applied vertically to joint C of truss ABCDE as indicated. You will use the method of joints to obtain the axial forces in the bars and reactions at the supports A, E .

E4_1?
All the bars in truss ABC have constant cross section and are made of a homogeneous linear elastic material. Under the effect of a 1 kN horizontal load applied at B , the pin at A is observed to displace to the right by 6 cm.
Use the method of joints to obtain the numerical value (in kN) of the axial forces in the bars.
NAB=
kN

unanswered

NBC=
kN

unanswered

NCA=
kN

unanswered

E4_1B
Obtain the numerical value (in kN) of the reactions at the supports.
RAy=
kN

unanswered

RCx=
kN

unanswered

RCy=
kN

unanswered

E4_1C
Obtain the numerical value (in kN/m) of the stiffness of bar CA .
KCA=
kN/m

unanswered

E4_2: SOLVING TRUSSES WITH MATLAB PART 1: SELECT DEGRESS OF FREEDOM

All the bars of the truss in the figure below have a cross-sectional area of 10 mm2 . We want to determine the axial forces in each of the bars and the Cartesian components of the reactions at supports C and D using the method of joints.

We will soon use MATLAB to solve this problem, but first we need to identify the "free" and the "constrained" degrees of freedom (DOFs) of the joints of the truss. Because this is a 2D problem in the x-y plane, each joint of the truss can only have two (Cartesian) DOFs (i.e., the joint can move only along x and along y). Some joints are hinges directly attached to the wall. These joint/hinges (like joint C in the example above) cannot freely move when the truss is loaded, because the wall prevents it: these DOFs are "constrained". In contrast, a joint like A is not attached directly to the wall, so it is free to move in both Cartesian directions: its DOFs are unconstrained or "free".

For each component below, select "Free" if it is unconstrained and "Fixed" if it is constrained.

Joint A , x component
FreeFixed

Joint A , y component
FreeFixed

Joint B , x component
FreeFixed

Joint B , y component
FreeFixed

Joint C , x component
FreeFixed

Joint C , y component
FreeFixed

Joint D , x component
FreeFixed

Joint D , y component
FreeFixed

Joint E , x component
FreeFixed

Joint E , y component
FreeFixed

A load W=2 kN is applied vertically to joint C of truss ABCDE as indicated. You will use the method of joints to obtain the axial forces in the bars and reactions at the supports A, E .

We start with the classification of the degrees of freedom as "free" or "constrained".

From the geometry of the truss, we see that we have:

Free DOF x and y at B, C, and D.

Constrained DOF x and y at A and E .

Question 1: Use MATLAB to find the axial forces in the bar.

Order the vector {X} of unknown axial forces in the bars as:

I obtained,

X =

4.00
2.00
-2.82
2.00
-2.82
-2.00

but is incorrect, can someone help me please?

To solve this problem using the method of joints, we need to analyze the equilibrium of forces at each joint of the truss. We will start by identifying the unknown axial forces in the bars and the reactions at the supports.

1. Axial forces in the bars:

We can start by analyzing the joint at B. The given information states that there is a horizontal load of 1 kN applied at B, which causes a displacement of 6 cm at joint A. This displacement is a result of the elongation or compression of the bars. We can assume that the horizontal axial force in bar AB is equal to the applied load at B, which is 1 kN in this case.

Now, moving on to joint C, we have a vertical load of 2 kN applied at C. This vertical load creates a reaction at joint C, which we need to find.

To determine the axial forces in the bars, we will consider the equilibrium of forces at each joint:

Joint C: The vertical force at C is 2 kN (given load). We assume that the axial force in bar AC is equal to the vertical reaction at C, which we can calculate.

Joint C equation:
Vertical force (2 kN) - Vertical reaction at C = 0

Therefore, the vertical reaction at C is 2 kN.

Joint B: The horizontal force at B is 1 kN (given load). We assume that the axial force in bar BC is equal to the horizontal reaction at B, which we can calculate.

Joint B equation:
Horizontal force (1 kN) - Horizontal reaction at B = 0

Therefore, the horizontal reaction at B is 1 kN.

Joint A: The axial force in bar AB is already known to be 1 kN (equal to the horizontal force at B).

Now that we have obtained the axial forces in bars AB and BC, we can calculate the axial forces in the remaining bars using the method of joints.

2. Reactions at the supports:

To find the reactions at the supports A and E, we need to consider the equilibrium of forces in the vertical and horizontal directions.

For support A, we have the horizontal reaction (RAx) and the vertical reaction (RAy).

Equilibrium in the horizontal direction:
RAx - 1 kN (axial force in AB) = 0

Therefore, the horizontal reaction at A, RAx, is 1 kN.

Equilibrium in the vertical direction:
RAy + 2 kN (vertical force at C) = 0

Therefore, the vertical reaction at A, RAy, is -2 kN (negative sign indicates upward direction).

For support E, we have the vertical reaction REy.

Equilibrium in the vertical direction:
REy + 2 kN (vertical force at C) = 0

Therefore, the vertical reaction at E, REy, is -2 kN (negative sign indicates upward direction).

To summarize:

Axial forces in the bars:
NAB = 1 kN
NBC = 1 kN
NCA = 2 kN

Reactions at the supports:
RAx = 1 kN
RAy = -2 kN
REy = -2 kN

Please note that these values are calculated based on the given information and assumptions made.