
Problem
31
Using method of joints determine the
forces in all the members of the pinjointed plane truss shown in
figure 31(a)
Figure 31(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 31(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_{x} was correct.
(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
zaxis perpendicular to the plane and passing through joint A. Take
moment of all the forces about zaxis (taking clockwise negative and anticlockwise 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; 
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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_{DE }indicates that the
assumed direction (tensile) was wrong, therefore the actual nature
of force F_{DE} will be compressive.
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 +xaxis,
therefore it has to be resolved into its rectangular
components along xaxis
(F_{AE}cos45) and along yaxis (
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_{AE}
indicates that this force should be in the opposite sense,
hence it should be a compressive force of 31.82 kN.
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. 

Problem 32 Solving Truss by method of sections
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