6.3 Trusses: Method of Sections

6.3 Trusses: Method of Sections

6.3 Trusses: Method of Sections xample 1, page 1 of 2 1. etermine the force in members,, and, and state whether the force is tension or compression. 5 m 4 kn 4 kn 5 m 1 Pass a section through the three members whose forces are to be 6 kn determined. 6 kn 5 m

+ 6.3 Trusses: Method of Sections xample 1, page 2 of 2 2 ree-body diagram of portion of truss above the section 4 quations of equilibrium for the portion of the truss (Note that moments are summed about point, even though point is not part of the free body): 5 m + x = 0: + 4 kn + = 0 y = 0: = 0 3 t each cut through a member, a force is shown to represent the effect of the portion of the member on one side of the section pulling on the portion on the other side. t is convenient to always assume the force to be tension. 4 kn 5 m 5 M = 0: () ()(5 m + 5 m) (4 kn)(5 m) = 0 Solving simultaneously gives = 13.33 kn (T) = 6.0 kn = 6.0 kn () = 13.33 kn = 13.33 kn () ns. ns. ns. We had assumed member to be in tension. alculations showed that is negative, so our assumption was wrong: must be in compression. Similarly must be in compression.

6.3 Trusses: Method of Sections xample 2, page 1 of 3 2. etermine the force in members,, and, and state whether the force is tension or compression. 8 ft 800 lb 800 lb 800 lb 12 ft 12 ft 12 ft 12 ft 1 Pass a section through the three members whose forces are to be determined. 800 lb 800 lb 800 lb

+ 6.3 Trusses: Method of Sections xample 2, page 2 of 3 2 ree-body diagram of portion of truss to right of section 4 quations of equilibrium for the portion of the truss: 8 ft + x = 0: sin = 0 (1) y = 0: cos 800 lb + y = 0 (2) 800 lb M = 0: (8 ft) + y (12 ft) = 0 (3) y 12 ft 5 eometry 3 t each cut through a member, a force is shown. = tan -1 12 ft = 56.31 8 ft 8 ft 12 ft 6 Three equations but four unknowns, so another equation is needed.

6.3 Trusses: Method of Sections xample 2, page 3 of 3 7 ree-body diagram of entire truss. 8 ft x y 800 lb 800 lb 800 lb y 12 ft 12 ft 12 ft 12 ft 8 quilibrium equation for entire truss. This will give the needed fourth equation. M = 0: (800 lb)(12 ft) (800 lb)(2 12 ft) (800 lb)(3 12 ft) + y (4 12 ft) = 0 Solving gives y = 1,200 lb. 9 Substituting y = 1,200 lb into qs. 1, 2, and 3 and solving simultaneously gives = 1,800 lb (T) = 721 lb (T) = 2,400 lb = 2,400 lb () ns. ns. ns.

6.3 Trusses: Method of Sections xample 3, page 1 of 3 3. The diagonal members are not connected to each other where they cross. etermine the force in members,, and, and state whether the force is tension or compression. 4 m 4 m 4 m 2.5 m 3 kn 3 kn 3 kn 1 Pass a section through the three members whose forces are to be determined. 3 kn 3 kn 3 kn

6.3 Trusses: Method of Sections xample 3, page 2 of 3 2 ree-body diagram of portion of truss to right of section 4 m 3 kn y 2.5 m 5 eometry = tan -1 4 m = 58.0 2.5 m 4 m 3 t each cut through a member, a force is shown 4 quations of equilibrium for the portion of the truss: + + x = 0: sin sin = 0 (1) y = 0: cos + cos + y 3 kn = 0 (2) M = 0: sin (2.5 m) + y (4 m) = 0 (3) 6 Three equations but four unknowns, so another equation is needed. 2.5 m

6.3 Trusses: Method of Sections xample 3, page 3 of 3 7 ree-body diagram of entire truss (This will give the needed fourth equation). y 4 m 4 m 4 m x 2.5 m 3 kn 3 kn 3 kn y 8 quilibrium equation for entire truss. M = 0: (3 kn)(4 m) (3 kn)(2 4 m) + y (3 4 m) = 0 Solving gives y = 3.0 kn. Then substituting simultaneously gives = 5.66 kn (T) = 5.66 kn (T) = 9.6 kn = 9.6 kn () = 58.0 and y = 3.0 kn into qs. 1, 2, and 3 and solving ns. ns. ns.

6.3 Trusses: Method of Sections xample 4, page 1 of 4 4. etermine the force in members,,, and, and state whether the force is tension or compression. 1.5 m 6 kn 6 kn J J 4 kn 4 kn 1 Pass a section through at least some of the members whose forces are to be determined. The general idea is to choose as few members as possible --three in this instance-- because each time a member is cut by a section, an additional unknown is introduced into the equilibrium equations. 6 m 6 m

+ 6.3 Trusses: Method of Sections xample 4, page 2 of 4 2 ree-body diagram of portion of truss above section (Using the upper portion of the truss rather than the lower eliminates the need to calculate the reactions at the bottom of the truss). L J 6 kn 4 kn L 3 quations of equilibrium for the portion of the truss: M = 0: (6 kn)() + () cos (L ) = 0 (1) M = 0: (6 kn)(2 ) 4 kn)() + cos (L ) = 0 (2) + x = 0: sin + + sin + 4 kn + 6 kn = 0 (3)

6.3 Trusses: Method of Sections xample 4, page 3 of 4 4 eometry 5 = tan -1 ' = tan -1 1.5 m = 7.125 J' 4 (6 m) tan () tan J L = + () tan + () tan = 3.75 m L = + (6 m) tan + (6 m) tan = 4.50 m 6 Substituting these values for, L, and L into qs. 1, 2, and 3 and solving simultaneously gives: = 10.75 kn ns. = kn = 7.33 kn () ns. = 10.75 kn = 10.75 kn () ns. 1.5 m ' 1.5 m

6.3 Trusses: Method of Sections xample 4, page 4 of 4 7 ree-body diagram of joint. This free body will enable us to calculate the remaining unknown force the force in member. = 10.75 kn () 8 quilibrium equations for joint + x = 0: cos + sin 7.125 + (10.75 kn)(sin 7.125 ) 7.33 kn = 0 (4) y = 0: sin cos 7.125 (10.75 kn)(cos 7.125 ) = 0 (5) = 7.33 kn () 9 eometry () tan L = 4.50 m = 7.125 = tan -1 = 31.61 () tan 7.125 + 4.50 m 10 Substituting = 31.608 into qs. 4 and 5 and solving simultaneously gives: = 14.97 kn = 14.97 kn () = 7.99 kn (T) ns.

6.3 Trusses: Method of Sections xample 5, page 1 of 4 5. etermine the force in members RS, LS, L, and, and state whether the force is tension or compression. N O P Q R S T J K L M 2 m 2 m 4 kn 4 kn 4 kn N O P Q R S 1 Pass a section through the four members whose forces are to be T determined. t does not appear possible to find a section that cuts only three of these members. J K L M 4 kn 4 kn 4 kn

6.3 Trusses: Method of Sections xample 5, page 2 of 4 2 ree body diagram of truss portion to right of section line RS S T LS M 2 m L 2 m y 3 quations of equilibrium for the portion of the truss: + + M S = 0: (2 2 m) + y () = 0 (1) M = 0: RS (2 2 m) + y() = 0 (2) y = 0: L LS + y = 0 (3) 4 Three equations with five unknowns so two more equations are needed.

6.3 Trusses: Method of Sections xample 5, page 3 of 4 5 ree-body diagram of entire truss (This free body will enable us to calculate the reaction at ). N O P Q R S T J K L M 2 m x 2 m y 4 kn 4 kn 4 kn y 6 quation of equilibrium for the entire truss. M = 0: (4 kn)(2 ) (4 kn)(3 ) (4 kn)(4 ) + y (18 m) = 0 (4) Solving gives y = 6 kn Substituting y = 6 kn into qs. 1 and 2 and solving gives: = 4.5 kn (T) RS = 4.5 kn = 4.5 kn () ns. ns.

6.3 Trusses: Method of Sections xample 5, page 4 of 4 7 ree-body diagram of joint S. This free body will enable us to calculate the force in member LS. RS = 4.5 kn () S ST 10 ree body diagram of joint T ST T MS MT LS 8 quations of equilibrium for joint S. Note that there are three unknowns but only two equations. + x = 0: 4.5 kn + ST + MS cos = 0 (5) y = 0: LS MS sin = 0 (6) 11 Two members meet at joint T, they are not collinear and no external force acts at joint T, so members ST and MT are zeroforce members. Substituting ST = 0 in q. 5 and solving qs. 5 and 6 simultaneously gives: MS = 5.41 kn = 5.41 kn () 9 eometry S T LS = 3.0 kn (T) ns. 2 m M = tan -1 2 m = 33.69 12 Substituting LS = 3.0 kn and y = 6 kn into q. 3 and solving gives: L = 3.0 kn = 3.0 kn () ns.

6.3 Trusses: Method of Sections xample 6, page 1 of 4 6. etermine the force in members TU,, and U. State whether the force is tension or compression. T U X V W 5 m N O P Q R S 5 m J K L M 12 panels @ 4 m each 10 kn T U V W X N O P Q R S J K L M 1 ven though we were not asked to determine the force in member P, we have to pass the section through it because we must make the section go completely through the truss. 10 kn

+ 6.3 Trusses: Method of Sections xample 6, page 2 of 4 2 ree-body diagram of portion of the truss to the left of the section T TU 5 m 3 quations of equilibrium for the portion of the truss: N O U + x = 0: x + TU + P cos + = 0 (1) 5 m x P y = 0: y + U + P sin = 0 (2) M = 0: y (4 4 m) TU (2 5 m) = 0 (3) y 4 m 4 m 4 m 4 m 4 Three equations with six unknowns so three more equations are needed. 5 eometry P = tan -1 5 m = 51.34 5 m 4 m 4 m

6.3 Trusses: Method of Sections xample 6, page 3 of 4 6 ree-body diagram of entire truss (This free body will enable us to calculate the reactions at support ). T U V W X N O P Q R S x J K L M y 10 kn M y 12 panels @ 4 m each 7 quations of equilibrium for entire truss. Note that we only write two equations because we only need to calculate x and y, since only x and y appear in qs. 1, 2, and 3. + x = 0: x = 0 (4) M M = 0: (10 kn)(6 4 m) y (12 4 m) = 0 (5) Solving gives x = 0 and y = 5 kn. 8 onsideration of joint shows that member P is a zero-force member, so P = 0. ut if member P is removed (because it is a zero-force member), consideration of joint P shows that member P is also a zero-force member, so P = 0.

6.3 Trusses: Method of Sections xample 6, page 4 of 4 9 Substituting = 51.34, x = 0, y = 5 kn, and P = 0 into qs. 1, 2, and 3, and solving gives: TU = 8 kn = 8 kn () = 8 kn (T) U = 5 kn = 5 kn () ns. ns. ns.

6.3 Trusses: Method of Sections xample 7, page 1 of 6 7. etermine the force in members KM, LM, and K. State whether the force is tension or compression. M J K L 6 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m M J K 1 We choose a section that cuts at least some of the members whose forces are to be determined. ut the section should cut as few other members as possible, since each time a member is cut, an additional unknown appears in the equilibrium equations. L

+ 6.3 Trusses: Method of Sections xample 7, page 2 of 6 2 ree-body diagram of portion of truss to right of section. t is not essential but we can save some work if we use the principle of transmissibility as shown in Step 3. KM ML 3 Same free body as in Step 2, but now the force KM has been moved along its line of action to joint (principle of transmissibility) and then expressed in terms of vertical and horizontal components. Similarly ML is moved to joint. K L K KM sin L ML sin KM cos ML cos y y 4 quations of equilibrium for free body in Step 3. Note that because we were not asked to determine, we choose two moment equations in which does not appear. 2 m 2 m 2 m 2 m M = 0: ()(2 m) + ()(3 2 m) KM sin (4 2 m) = 0 (1) M = 0: ()(2 m) ()(3 2 m) (2kN)(4 2 m) + y (4 2 m) + ML sin (4 2 m) = 0 (2) 5 Two equations but three unknown forces, so another equilibrium equation is needed.

6.3 Trusses: Method of Sections xample 7, page 3 of 6 6 eometry M 6 m 8 m 2 m = tan -1 6 m ( 2 m ) = 71.56 = tan -1 6 m ( 2 m + 8 m ) = 30.96

6.3 Trusses: Method of Sections xample 7, page 4 of 6 7 ree-body diagram of entire truss. This free body will enable us to calculate the reaction at support. M J K L x y y 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m 8 quilibrium equation for entire truss M = 0: ()(2 m) ()(3 2 m) ()(5 2 m) ()(7 2 m) ()(9 2 m) ()(10 2 m) + y (10 2 m) = 0 (3) 9 Solving gives y = 7 kn. Substituting y = 7 kn, = 71.56, and = 30.96 in qs. 1 and 2, and solving simultaneously gives: KM = 2.11 kn (T) ML = 5.83 kn = 5.83 kn () ns. ns.

6.3 Trusses: Method of Sections xample 7, page 5 of 6 10 ree-body diagram of joint K (This free body will enable us to calculate the force in member K). y KM = 2.11 kn (T) x K KL 11 Since there are only two unknown forces, KL and K, we could write force-equilibrium equations in the x and y directions and then solve them simultaneously. owever, we can save work by noticing that a zero-force member is present. K 12 ree- body diagram of entire truss 13 onsideration of joint K shows that KL must be a zero-force member, so KL = 0. M J K L x y y

6.3 Trusses: Method of Sections xample 7, page 6 of 6 14 ree-body diagram of joint K (repeated) y KM = 2.11 kn (T) K x KL K y = 2.11 kn K = 0 (4) Solving gives K = 2.11 kn (T) ns.

6.3 Trusses: Method of Sections xample 8, page 1 of 3 8. etermine the force in members,, and. State whether the force is tension or compression. lso, find the tension in the cable. able 30 J 3 ft 5 kip 5 ft 5 ft 5 ft 5 ft 5 ft able 1 The section must pass through the cable. Otherwise the portion of the truss to the left of the section could not be isolated as a free body. 30 J 5 kip

6.3 Trusses: Method of Sections xample 8, page 2 of 3 2 ree-body diagram of portion of truss to left of section T 30 3 The tension in the cable is one of the unknowns. 3 ft 5 ft 5 ft 5 kip 4 quations of equilibrium for the portion of the truss: + + + M = 0: T cos 30 (3 ft) (3 ft) + (5 kip)(2 5 ft) = 0 (1) M = 0: (5 kip)(2 5 ft) + (3 ft) + sin (3 ft) = 0 (2) y = 0: 5 kip + T sin 30 + cos 0 (3) 5 eometry 5 ft 3 ft = tan -1 5 ft ( 3 ft ) = 59.04 6 Three equations but four unknown forces, so another equilibrium equation is needed.

6.3 Trusses: Method of Sections xample 8, page 3 of 3 7 ree-body diagram of entire truss T 30 J Jx 3 ft J y 5 kip 5 ft 5 ft 5 ft 5 ft 5 ft 8 quation of equilibrium for the entire truss. Only one equation is used because we need to calculate T only; the reactions at J are not needed. M J = 0: (5 kip)(5 5 ft) T sin 30 (3 5 ft) = 0 (4) 9 T = 16.67 kip ns. Substituting = 59.04 and T = 16.67 kip into qs. 1, 2, and 3 and solving simultaneously gives: = 2.23 kip (T) = 11.11 kip = 11.11 kip () = 6.48 kip = 6.48 kip () ns. ns. ns.