Question

Theory of Structures

Determine ALL the SUPPORT REACTIONS and determine the forces in each member of the truss shown. Redundant/s can be freely chosen based on preference 100 KN 100 KN (4000 mm) 75 kN — D (5000 mm) (4000 mm) (4000 mm) 3 m (5000 mm) (4000 mm) 4 m E = 200 GPa

Answer 1

The above truss is statically indeterminat of a degree Choose member BC as the redundant member. So, Relased member for P-forces: 100 KN 1100 75 kN ac | 156.25 1156.25 '143.15 for force in member AD, FAD Coso - 75 2) FADX 4/5=75 7 FAD 2 93.75

redundant force. Structure with for k- value, 315 || 315 / 1 415 1 2050 = FAB = 1 sino = FBD = FAB= 415 FBD = 3/5 is given as The ñ value E - E Pke AE 5 к?. AE we can cancel E since it is same for all members consider Tensim(t) & Comp. (-)

I ke 0.6 -156.25 Member Prvalue / K-value / Length / Area okela IA I Faptkal AS 6.4810" + 56.28241 co 1-75 1-0.8 0.06 6.4x10 |- 29.97 kn 1-100 10.ous |27X10 1-57.788KH 0.07 2.7810 (-114.04KN AD 93.75 5000/0.09375 0.001 | 23.4 kN 5000 10 10.001 - 70.353 KN {/0.26875 3.82510 Hence, ã =- -0-26875 3.82X163 -70.353 KN s nam w

FAB = PAB+ KAB (20 + (-0.6)(-70.353) = 56.2824 KN similarly all forces are calculated SOROS RB = A rea Roth as = FBD - FBC sino 2 114.04- 70.353% 315 RB = 71.83 KN HA = 75 KN RA= 128.17 kn
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