**Truss member forces by method of joints**

**Problem 3-1**

Use method of joints to determine the forces in all the members of pin-jointed plane truss shown in figure 3-1(a).

**Figure 3-1(a)**

**
Solution:**

In the given truss the support at A is roller and C is hinged. First we will find whether this truss is determinate or indeterminate.

Condition of determinacy of plane truss: m = 2j - 3

In this truss j = 6, which requires 2×6 - 3 members for the truss to be determinate. It is confirmed from the figure that there are 9 members in this truss.

Therefore the the given truss is statically determinate.

__Reactions at the support:__

Support A is on the roller, therefore it will have only vertical reaction and no horizontal reaction. Support C being hinged will experience both horizontal and vertical reactions. (refer to figure 3-1(b).

Considering horizontal reaction at C to be in the +ve x direction and. Applying the conditions of static equilibrium, we get;

(i) Σ F_{x} = 0; therefore C_{x} - 15
= 0;
eq (1)

Solving the equation we get **C _{x}**
=15 kN.
The positive sign of this value indicates that our assumption in the
direction of C

(ii) Σ F_{y} = 0; yields A_{y}+ C_{y}
- 25 -10 - 20 = 0;

A_{y}+ C_{y} = 55;
eq. (2)

(iii) Σ M_{z} = 0; Considering
z-axis perpendicular to the plane and passing through joint A. Take
moment of all the forces about z-axis (taking clock-wise negative and anticlock-wise positive);

A_{y }x 0 + C_{y } x 4 - C_{x }
x 0 - 20 x 2 + 15 x 2 - 10 x 2 + 25 x 0 =0;

we get **C _{y} **= 7.5 kN;

Therefore **A**_{y }=
47.5 kN;

**
Figure 3-1(b)**

__Calculation of member forces__

We use *method of joints* to find all the
forces in the members of the given truss.

First of all look for the joint which does not have more than 2 unknown forces. In this truss we find that joint D and F have only two unknown forces.

Let's start with joint D; In the beginning assume all the unknown forces as tensile. Tensile forces are shown with an outward arrow whereas compressive forces are shown with an inward arrow at the joint. The forces acting in the +ve directions of axes are taken as +ve whereas those acting in the -ve directions of axes are taken as -ve.

__Equilibrium of joint D__

Σ F_{x} = 0;

- F_{DE} - 15 kN = 0;

**F _{DE}** = - 15 kN
The -ve sign of F

Σ F_{y} = 0; => - F_{DC} = 0;
therefore **F _{DC}** = 0

__Equilibrium of joint F__

Σ F_{x} = 0; => F_{FE}
= 0

Σ F_{y} = 0; => - 25 - F_{FA} =
0;

**F _{FA}** = - 25 kN

The -ve sign of F_{FA} indicates that the
assumed direction of (tension)
for this forces is not correct, therefore the actual
direction of force in member FA is to be compressive.

Now the joints A and C have only two unknown forces whereas joint B and E have three unknown forces. So we come to joint A.

__Equilibrium
of joint A__

The force in
member AE is making an angle of 45 degree with +x-axis,
therefore it has to be resolved into its rectangular
components along x-axis
(F_{AE}cos45) and along y-axis (
F_{AE} sin45).

Σ F_{x} = 0; => F_{AE}
cos45 + F_{AB} = 0
(i)

Σ F_{y} = 0; => F_{AE} sin45 + A_{y}
- F_{AF} = 0
(ii)

substituting the magnitudes of A_{y}
and F_{AF } (equal to F_{FA})
into eq.(i) we get

F_{AE} sin45 + 47.5 - 25 = 0;

Therefore **F _{AE}** = -
22.5/sin45= - 31.82 kN; negative sign of F

Substituting this value of F_{AE} in eq. (i) yields.

**F _{AB}** = 22.5 kN;

__Equilibrium
of joint C__

Σ F_{y} = 0;

F_{CE}
sin45 + C_{y}
+ F_{CD} = 0 (iii)

F_{CE} sin45 + 7.5 + 0 = 0;
Therefore
**F _{CE}**
= - 7.5/sin45 = - 10.61 kN; again the negative sign of this
force indicates that it should be in the reverse direction,
hence compressive 10.61 kN.

Σ F_{x} = 0; =>
- F_{CE} cos45 - F_{CB} + C_{x} = 0
(iv)

Substituting the
value of F_{CE}
and C_{x} in eq. (iv) we get;

** F _{CB}** = + 7.5 + 15 = +
22.5 kN

__Equilibrium
of joint B__

Σ F_{y} = 0; => **F _{BE}** = 20
kN

From the previous calculations about
joint A and joint C, it is also evident that Σ F_{x}
= 0; because F_{BA} =
- F_{AB} and F_{BC} = - F_{CB} ;
This confirms our calculations.

Result of Member
forces calculations |
||

Member |
Force (kN) |
Nature of force |

AB | 22.5 | Tensile |

AF | 25 | Compressive |

AE | 31.82 | Compressive |

BC | 22.5 | Tensile |

BE | 20 | Tensile |

CD | 0 | |

CE | 10.61 | Compressive |

DE | 15 | Compressive |

EF | 0 |

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