w a truss composed by 8 nodes and 13 bars. Dimensions are provided in the metres
ID: 1714388 • Letter: W
Question
w a truss composed by 8 nodes and 13 bars. Dimensions are provided in the metres. There are pinned supports at node numbers 1 and 8. A uniformly Question 1 Figure I sho figure in distributed vertical load of 16 kN/m is applied to bars 2-4 and 4-6. The value of the modlus of elasticity CE) muliplied by the cross-sectional area (A) is the same for all bars (a) Classify the truss in Figure 1 as either a mechanism, critical, statically determinate or and given by EA = 100× 103 kN. indeterminate and state why. If the truss is statically indeterminate, provide the degree of ( 10%) indeterminacy ars. (c) Find the vertical displacement of node 4. (30%) 4 m 4 m 16 kN/m16 kN/m 6 2 3 Figure 1Explanation / Answer
a) Statically indeterminate means more unknowns than equations.
In simple words, if m+3 = 2j, it is statically determinate
if m+3>2j, it is statically indeterminate and if m+3<2j, it is unstable truss
m= no of members an j = no of joints
In the above figure, no of member= 13 and number of joints = 8
Therefore, m+3 = 13+3=16 & 2j= 2*8 = 16. Hence the structure is statically determinate.
b) Summation of Vertical = 0, Summation of Moment = 0, Summation of Horizontal = 0
Reaction at Node 1= Node 2 = (16*4 + 16*4)/2 =64 KN
Forces in each arm can be solved w.r.t to known force an resolving the unknowns with Sine or Cosine multiplication.
For example at Node 1, Vertical Reaction(upward) =64KN and Summation of vertical = 0
Since the length of bar 1-2 is 4m an 2-3 is 4m, the angle is 45 degree
At joint we also initially need to assume bar as either tension or compression,
Since, Summation of vertical = 0,
64 minus Force in bar 1-3*SIn45 = 0,
Therefore Force in bar 1-3 = 64/Sin 45 = 90.5 (Compression)
For Summation of Horizontal = 0,
Force in bar 1-2 minus 90.5 cos 45 =0
Therefore Force in 1-2 =64 (Tension)
Similarly all nodes can be resolved.
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