TRUSSES
Plane Trusses Trusses- It is a system of uniform bars or members (of various circular section, angle section, channel section etc.) joined together at their ends by riveting or welding and constructed to support loads.
Practical Application A truss is mainly used to support roofs of building, television towers, bridges etc.
FRAMES Frames It is a structure consisting of several bars or members pinned together and in which one or more than one of the members is subjected to more than two forces. They are designed to support loads are stationary structures. In case of frames forces may act anywhere on the beam.
Perfect Truss A perfect truss is made of members just sufficient to keep it in equilibrium when loaded, without destroying its shape. A perfect truss is just rigid, hence it is also known as rigid truss. Mathematical condition of perfect or rigid truss:- m + 3 = 2j Where m = number of members Ex:-1 1 j = number of joints in the truss 1 2 2 3 4 5 4 3 m = 5, j = 4 L.H.S = m + 3 = 8 R.H.S = 2j = 8, non-collapsable
Ex:-2 2 3 m = 6, j = 6 m + 3 = 2 j L.H.S = 9 + 3 = 12 1 6 5 4 R.H.S = 2j = 12, Non-Collapsable Imperfect truss:- A truss which does not satisfy the above design equation is known as imperfect truss. m + 3 2 Rigid or Collapsible Truss (Deficient Truss):- The truss is said to be collapsible if the number of members in a frame are less than (2j - 3) then it is known as deficient truss. m + 3 < 2 j
Ex:-1 2 3 This figure shows the typical example of deficient truss 1 6 5 4 Redundant Truss:- A redundant truss is another type of imperfect truss in which the number of members are more than that required as per the equation. m + 3 > 2j 2 3 1 6 5 4
Basic Assumptions of Perfect Truss 1. The joints of a simple truss are assumed to be pin connections and frictionless. 2. The loads on the truss are applied at the joints only. 3. The self weight of the members are negligible. 4. The truss is statically in determinant. 5. The members of the truss are straight two force members with the forces acting collinear with the centre line of the members.
Statically In determinant A truss is statically determinant if the equations of static equilibrium alone are sufficient to determine the axial forces in the members without the need of considering their deformation.
Determination of Axial forces in the Members The various members are : 1. Method of Joints. 2. Method of Sections. 3. Graphical Method Method of Joints The procedure is as follows:- 1. Draw the FBD of the entire truss and compute the support reactions. Determination of the support reactions may not be necessary in the case of cantilever type of truss. 2. Assume and mark the direction of the axial forces in the members.
If in the solution, the magnitude of forces comes out to be negative the assumed direction of the force should be simply reversed. Start form the joint where not more than two unknown forces appear. Special Condition 1. When the members meeting at a joint are not collinear and there is no external force acting at a joint, then the force in both the members is zero. E E C D P F 1 = 0 F 2 = 0 A B
When there are three members meeting at a joint, of which two are collinear and third be at an angle and if there is no load at the joint the force in the third member is zero. C D E R A A F H G P B R B F FD FFD = 0 F GE FGE = 0 F FA F F FH F GH G F GB
A F AB B 3 m F AE F AD F BD F BC 6 KN E F DE D F DC C 8 KN 4 m 4 m
R AY Step-1: Support Reactions R AX A F AB B F x = 0 F y = 0, M A = 0 M A = 0, R D x 4 = 8 x 8, 3 m F AE F AD F BD F BC R D = 16 N F y =R D 8 6 R AY = 0 R AY = 2 KN E D C 6 KN F DE R D F DC 8 KN R D = 16 N R AY = 2 KN R AX = 0 4 m 4 m F x = 0, R AX = 0
R AX R AY A F AB B Step-2: Start form the joint where not more than two unknown forces appear F BC 3 m F AE 6 KN E F DE F AD F BD D R D F DC F BC C 8 KN F DC C 8 KN F y = 0, F BC sin 36.87-8 = 0 F BC = 13.33 KN (T) 4 m 4 m F x = 0 F y = 0 - F DC - F BC cos 36.87 = 0 F BC = 13.33 KN (T) - F DC - 13.33 cos 36.87 = 0 F DC = - 10.67 KN (C) F DC = - 10.67 KN (C)
R AX R AY A F AB B 90 36.87 = 53.13 F x = 0 F y = 0 F BC = 13.33 KN (T) F DC = - 10.67 KN (C) R D = 16 N R AY = 2 KN 3 m F AE F AD F BD F BC B 6 KN E F DE F x = 0, F BC sin 53.13 - F AB = 0 D R D F DC 4 m C 8 KN F AB F BD 53.13 F BC F y = 0 - F BD - F BC cos 53.13 = 0 13.33 sin 53.13 - F AB = 0 - F BD - 13.33 cos 53.13 = 0 F AB = 10.66 KN (T) F BD = - 8 KN (C)
R AX R AY A F AB B 90 36.87 = 53.13 F BC = 13.33 KN (T) F DC = - 10.67 KN (C) F BD = - 8 KN (C) F AE F AD F AB = 10.66 KN (T) R D = 16 N 3 m F BD F BC R AY = 2 KN R AX = 0 E 6 KN F y = 0, F AE - 6 = 0 F DE F AE = 6 KN D R D F DC 4 m C 8 KN F AE E F DE F x = 0, F DE = 0 6 KN
R AX F AE R AY 3 m 6 KN A F AB F AD 53.13 B F F BD BC E D F DE R D F x = 0, - R AY - F AE F AD cos 53.13 = 0-2 - 6 F AD cos 53.13 = 0 F AD = -13.33 KN (C) F DC 4 m 4 m C 8 KN F BC = 13.33 KN (T) F DC = - 10.67 KN (C) F BD = - 8 KN (C) F AB = 10.66 KN (T) R D = 16 N F AE = 6 KN R AY = 2 KN F DE = 0 R AX = 0 R AX R AY F AE A 53.13 F AB F AD
Step-1: Support Reactions R D = 16 N R AY = 2 KN R AX = 0 Step-2: Start form the joint where not more than two unknown forces appear F DC = - 10.67 KN (C) F BC = 13.33 KN (T) F AB = 10.66 KN (T) F BD = - 8 KN (C) F AE = 6 KN F AD = -13.33 KN (C) F DE = 0
Member Force Magnitude (KN) Nature AB F AB 10.66 T AD F AD 13.32 C AE F AE 6 T BC F BC 13.33 T BD F BD 8 C CD F CD 10.67 C DE F DE 0 -
Q.2 Use method of joints and obtain forces in the members of truss. 500 N 500 N A 50 cm C 50 cm 30 0 D E B
R AX R AY A F AE F AC 50 cm 30 0 500 N 500 N C F CE E F CD 50 cm D 30 0 F DE F BD F BE 60 0 B Step-1: Support Reactions F x = 0 F y = 0, M A = 0 R B x AB 500 x 50-500 x 100 = 0 R B = 649.51 N F y = 0 R B = 649.51 N R AY = 216.5 N R AX = - 500 N R B F y =R B + R AY 2 x 500sin 60 = 0 R AY = 216.5 N F x = - R AX 2 x 500cos 60 = 0 R AX = - 500 N
Joint A F y = 0, F x = 0 R AY R B = 649.51 N R AY = 216.5 N R AX = - 500 N F y = R AY sin60 - R AX sin30 - F AE sin30 = 0 F AE = 875 N (T) F x = R AY cos60 + R AX cos30 + F AE cos30 + F AC = 0 F AC = -433.01 N (C) F AE = 875 N (T) F AC = - 433.01 N (C) 60 0 R AX F AC 30 0 F AE
Joint C F y = 0, F x = 0 500 N R B = 649.51 N R AY = 216.5 N R AX = - 500 N F AE = 875 N (T) F AC = - 433.01 N (C) F AC F y = -500 F CE = 0 F CE = - 500 N (C) F x = F AC + F CD = 0 F CD = -433.01 N (C) F CE F CD F CE = -500 N (C) F CD = - 433.01 N (C)
Joint D F y = 0, F x = 0 500 N R B = 649.51 N R AY = 216.5 N R AX = - 500 N F AE = 875 N (T) F AC = - 433.01 N (C) F CE = -500 N (C) F CD = - 433.01 N (C) F DE = 500 N (T) F BD = -750 N (C) F CD 30 0 F DE F BD F x = - F CD F DE cos 30 = 0 F x = 433.01 F DE cos 30 = 0 F DE = 500 N (T) F y = - 500 F BD F DE sin 30 = 0 F y = - 500 F BD 500 sin 30 = 0 F BD = -750 N (C)
Joint B F y = 0, F x = 0 F BE F BD 60 0 30 0 R B R B = 649.51 N F AE = 875 N (T) F AC = - 433.01 N (C) F CE = -500 N (C) F CD = - 433.01 N (C) F DE = 500 N (T) F BD = -750 N (C) F BE = 375.03 N (T) F y = R B cos 30 + F BD + F BE cos 60 = 0 F y = 649.51 cos 30-750 + F BE cos 60 = 0 F BE = 375.03 N (T)
Member Force Magnitude (KN) Nature AC F AC 433.01 C AE F AE 875 T CD F CD 433.01 C CE F CE 500 C DB F DB 750 C DE F DE 500 T BE F BE 375.03 T
Q.3 A cantilever truss is loaded. Find out axial forces in all members. A 2 m 2 m 1 m C E 1 m B D 1000 N
1 m A 2 m 2 m F AC C F CE E Step-1: No need to evaluate support reactions for cantilever beam. 1 m B F BC F BD D F DE 1000 N Joint E F y = 0, F x = 0 F y = - 1000 F DE sin 26.56 = 0 F DE = - 2236.46 N (C) F x = - F CE F DE cos 26.56 = 0 F x = - F CE + 2236.46 cos 26.56 = 0 F CE = 2000.43 N (T) F CE 26.56 F DE 1000 N F DE = - 2236.46 N (C) F CE = 2000.43 N (T)
A 2 m 2 m Joint D 1 m 1 m B F AC F BC F BD F y = 0, F x = 0 C D F CE F DE F y = F DC + F DE sin 26.56 = 0 F DC 2236.46 sin 26.56 = 0 F DC = 1000 N (T) F X = F DE cos 26.56 F BD = 0-2236.46 cos 26.56 F BD = 0 F BD = - 2000.43 N (C) E 26.56 F BD F DC F DE 26.56 1000 N F DE = - 2236.46 N (C) F CE = 2000.43 N (T) F DC = 1000 N (T) F BD = - 2000.43 N (C)
A 2 m 2 m Joint C 1 m 1 m B F BC F AC F BD F y = 0, F x = 0 C D F CE F DE E 1000 N F y = F AC sin 26.56 - F BC sin 26.56 - F CD = 0 F AC = 0.006 N = 0 26.56 F AC sin 26.56 - F BC sin 26.56 = 1000 F x = - F AC cos 26.56 - F BC cos 26.56 + F CE = 0 - F AC cos 26.56 - F BC cos 26.56 = 2000.43 F BC = -2236.45 N F AC 26.56 F BC 1 26.56 F CD F CE F DE = - 2236.46 N (C) F CE = 2000.43 N (T) F DC = 1000 N (T) 2 F BD = - 2000.43 N (C)
Member Force Magnitude (N) Nature EC F EC 2000.43 C ED F ED 2236.46 C CD F CD 1000 T BD F BD 2000.43 C BC F BC 2236.46 C AC F AC 0 -