Question

Learning Goal To apply the method of sections to a truss and determine the unknown forces in specified members. If a truss is in equilibrium then every member in the truss must also be in equilibrium. This principle is used in analyzing a truss using the method of sections. To apply the method of sections, we use an imaginary cut through members of the truss to expose the forces in those members as external forces (Figure 1) If the free-body diagram of the joint on either side of the cut is known, we can apply the equations of equilibrium to that part of the member to determine the forces at the cut section. Since only three equilibrium equations (F 0, 2Fy 0, and Mo 0) can be applied to the free-body diagram of any segment, we should try to select a cut that passes through no more than three members of the truss in which the forces are unknown. Note that the moment equation can be written multiple times, but there are still only 3 independent equilibrium equations available For the Howe bridge truss shown, d-11 ft. F 2950 lb F2 1750 lb, F32550 lb, and F4-3850 lb (Figure 2) Figure 1 of 2 Interna T tensile forces Internal|C compressive forces Tension ▼T Compression lc

Part A - Determining the forces in three specified members of the truss

Determine the forces in members CD, DH, and GH.

Express the magnitudes of the forces in pounds to three significant figures. Separate your answers with commas.

Part B - The Sense of the Forces

State if the members considered in part A are in tension or compression.

Complete each sentenc

e by dragging the appropriate word into the blank. The words can be used more than once.

compression tension Member CD is in Member DH is in Member GH is in Part C - Determining the forces in three members on the left-hand side of the truss Determine the forces in members HI, BH, and BC. Express the magnitudes of the forces in pounds to three significant figures. Separate your answers with commas. Part D - The Sense of the Forces State if the members considered in part C are in tension or compression. Complete each sentence by dragging the appropriate word into the blank. The words can be used more than once. compression tension Member BC is in Member HI is in Member BH is in

Answer 1

in this problem first of all we have to find the support reaction at A and E

Then applying method of section to find the magnitude and nature of forces in the given member

C. RA Re Given data 1750 6 hle have to And Ges in member CO Pent CLD- Magnitude and na af foyces in me ee and BH Solation FxoRA Horizontal Teaction Tating moment a

foom -the iven diagram SCH Pant-A memberDH, Co and CH and dlivide he touss into fo 38 So.2 RE-4600 b

Taking moment about H Tabing morment aborut D ゲー-4600 사 takins moment akout c Sings fosce in me 4600 4 ) .. masnitmide et frce in membel DH10co

Pnt B -4600b C ㄩ Ftence fosce inbe GH is comessive in Hence fesce i oembe DH is Compressie i foy rme.mbot CD CDC atiroe membe cD i in Tensio Cns) Member DH is n Compressior (Ans) Membe is in Compress 、 、 S

Pant-C Consider a Section t-t Suchthaf卝cut the member , BH and BC and divide the ussinto troo part Gonsideing equwbiame aft part at 七 RA Toleng moment about

tating moment about B Taking moment about C ヲ :3550 lb (Au) се magnitude rAd) u'. r0 PortD for memberHE fosce inmeber HL io Compmess ive in nature

o545.584 46 595.5804 tb BH Hence fosce in member e is compressire in BC force in membo RC i tensile in : member e i Tension CAns : member H is in Comprsn (ng natrse

Answer 2

Free-body diagram of the entire truss, C D Moment equilibrium about point A, 514-0; R (4d)-F (3d) -F (2d)-F (d)-F (0) 0 RE 4x11)-3850 (3x11)-2550(2x1)-1750(11)-2950(0) 0 RE 4600lb Part A, B Section through the members CD, DH, and GH Moment equilibrium about point H, H FoHC 4600 (2d) -F (d) -Foc (d) 0 4600 (2x11)-3850 (11)- Foc (11)0 FDc5350 1b 5.35 kip (T) DH 45 Force equilibrium along vertical direction, 4600 FFD sin 45 0 4600 3850F sin 450 FDH1061lb-1.06 kip(C)

Force equilibrium along horizontal direction, FoH -1061) cos45o-5350 0 -FeH 4600- 0 Fo4600 1b 4.60 kip(C) Part C, D Section through the members HI, HB, and CB Moment equilibrium about point B, 4600 (3d) F (2d) -F (d) +F#r (d) 0 4600 (3x11)-3850 (2 x11)-2550 (11)+F (11) 0 FR 3550lb 3.55kip(C) Force equilibrium along vertical direction, 4600 F F - FH sin 45 0 4600 3850-2550 FH sin 45 0 1800 Fi 450 FHB -2546lb 2.55 kip(C) Force equilibrium along horizontal direction, -(-3550)-(-2546) cos45-FC0 Fc 5350 lb 5.35 kip(T)