5.Analysis of Structures Trusses .15

Highway and Transport. Engineering Department First Class College of Engineering Engineering Mechanics

Mustansiriyah University 2020-2021 Lec.Rana Hashim

م.رنا هاشم 103

قسم الطرق والنقل

كلية الهندسة .الجامعة المستنصرية

5.Analysis of Structures

Trusses .15

A truss is a structure composed of slender members joined together at their end points by

joint. In particular, planer trusses lie in a single plane and are often used to support roofs and

bridges.






The calculations for the internal force in the members of truss are based on the following

assumptions:-

1-The members of the truss are joined together by means of smooth pins at their ends.

3.The weight of the individual members can be neglected.

Each truss member will act as a two force member and the force acting at each end of the

member will be directed along the axis of the member.

Tension force:-The force tends to elongate the member.

Compression force:- The force tends to shorten the member.

.2Analysis of Truss5

There are two methods for the force analysis of the simple trusses

A. Method of joints

This method is based on the fact that if the entire truss is in equilibrium, then each of its

joints is also in equilibrium.






∑ 𝐹𝑥 = 0

∑ 𝐹𝑦 = 0

:NoteImportant

at least one known load exists and where not more Start the analysis with any joint where

( if the joint is at one of the support , then it may be than two unknown forces are present

).necessary first to calculate the external reactions at the support

F.B.D →

1.Calculate R1,R2

2.Start with joint A.

3.Joint B

2.Joint F

1.Joint A

6.Joint D

5.Joint E

4.Joint C






Examples

-:Example(1)

Compute the force in each member of the loaded cantilever truss by the method of joints.

-:Solution From the F.B.D of whole truss.

∑ ME = 0

5T − 20(5) − 30(10) = 0 T = 80kN

∑ Fx = 0

80 cos 30 − Ex = 0 Ex = 69.3kN

∑ Fy = 0

80sin30 + Ey − 20 − 30 = 0 Ey = 10kN

Joint A

∑ Fy = 0

0.866AB − 30 = 0 AB = 34.6kN T

∑ Fx = 0

AC − 0.5(34.6) = 0 AC = 17.32kN C Joint B

∑ Fy = 0

0.866BC − 0.866(34.6) = 0 BC = 34.6kN C

∑ Fx = 0

BD − 2(0.5)(34.6) = 0 BD = 34.6kN T

Joint C

∑ Fy = 0

0.866CD − 0.866(34.6) − 20 = 0 CD = 57.7kN T

∑ Fx = 0

F.B.D

Joint A

Joint B

Joint C






Joint E

CE − 17.32 − 0.5(34.6) − 0.5(57.7) = 0 CE = 63.5kN C

Joint E

∑ Fy = 0

0.866DE = 10 DE = 11.55kN C

-:Zero Force Members

1. If only two non collinear members form a truss joint and no external load or support

reaction is applied to the joint, the two members must be zero force members.

∑ Fy = 0 DCsinθ = 0 DC = 0since sinθ ≠ 0

∑ Fx = 0 DE + 0 = 0 DE = 0

∑ Fx = 0 AB = 0

∑ Fy = 0 AF = 0

2.If three members form a joint for which two of the members are collinear, the third

member is a zero force member provided no external force or support reaction has a

component that acts along this member.

∑ Fx = 0 CAsinθ = 0 CA = 0since sinθ ≠ 0

∑ Fy = 0 CB = CD

∑ Fx = 0 DA = 0

∑ Fy = 0 DC = DE






-:Example(2)

Neglect any horizontal reactions at the support and solve for the forces in all members by

joint method.

-:Solution

By symmetry RA=RE

∑ 𝐹𝑦 = 0

5 + 2RA = 0 RA = RE = 2.5kN ↑

𝛼 = tan−1(2

4) = 26.6

Joint A

∑ Fy = 0

2.5 − 1 − ABsinα = 0 AB = 3.35kN C

∑ Fx = 0

−3.35 cosα + AH = 0 AH = 3kN T

By inspection (see page 107)

BH=0 GH=AH

Joint B

∑ Fy = 0

−1 + 3.35sinα + BGsinα − BCsinα = 0

∑ Fx = 0

3.35sinα − BCcosα − BGcosα = 0

BC = 1.39kN C BG = 0.28kN C

Joint G

∑ 𝐹𝑦 = 0

CG − 2(0.28)sinα = 0 CG = 0.25 kN T

Joint A

Joint B

Joint G






By symmetry

DE = AB = 3.35kN C

CD = BC = 1.39kN C

EF = AH = 3kN T

DF = BH = 0 FG = GH = 3kN T DG = BG = 0.28kN C






sSection. Method of B

The forces in the members of a truss can be determined by drawing a free body diagram of a

portion of the truss involving one or more of the unknown forces and applying the equations

of equilibrium.

∑ Fx = 0 ∑ Fy = 0 ∑ Ma = 0

where a is a joint

Compression

Tension






Examples

-:Example(1)

Determine the force in members GE,GC and BC of the truss shown below.

-:Solution Section aa has been chosen

From F.B.D

∑ Fx = 0

400 − Ax = 0 Ax = 400N ←

∑ MA = 0

−1200(8) − 400(3) + Dy(12) = 0

∑ Fy = 0

Ay − 1200 + 900 = 0 Ay = 300N ↑

From the F.B.D of left portion

∑ MG = 0

−300(4) − 400(3) + BC(3) = 0

BC = 800N T

∑ MC = 0

−300(8) + GE(3) = 0

GE = 800N C

∑ Fy = 0

300 −3

5GC = 0

GC = 500N T

F.B.D

F.B.D of left portion

x45

Pencil






-:Example(2)

Determine the load(P) which can be supported by the truss shown below and product a force

of 10000N compression in member CE.

-:Solution

Joint C

∑ Fy = 0

Cy − 10000 (2.4

3) = 0 Cy = 8000N ↑

For all structure

∑ MA = 0

P(5.4) − Cy(3.6) = 0

p =8000(3.6)

5.4= 5333.33N ↓

x45

Pencil






-:(3)eExampl

Determine the force in member EB of the roof truss shown in figure. Indicate

whether the member is in tension or compression.

-:Solution

Section aa

∑ MB = 0

1000(4) + 3000(2) − 4000(4) + EDsin30(4)= 0 ED = 3000N C

Joint E(Section bb)

∑ Fx = 0

EF cos30 − 3000cos30 = 0 EF = 3000N C

∑ 𝐹𝑦 = 0

2(3000sin30) − 1000 − EB = 0 EB = 2000N T

x45

Pencil






-:Example(4)

Determine the forces in members BC and FG of the loaded symmetrical truss.

-:Solution Section aa

∑ MC = 0

−300(6) + FG(12) = 0 FG = 150N C

∑ MF = 0

−300(6) + BC(12) = 0 BC = 150N T

-:Example(5)

Calculate the forces in members BC,CD and CG of the loaded truss.

-:Solution

From F.B.D

∑ MJ = 0

−3(16) + D(28) − (5sin30)(48) = 0 D = 6kN

From the right portion of section

∑ Fy = 0

CGsin60 + 6 − 5sin30 = 0 CG = −4.041kN C

∑ MG = 0

F.B.D

x45

Pencil






Right portion of section

CD(8sin60) + 6(4) − (5sin30)(24) = 0

T CD=5.2kN

∑ MC = 0

GH(8sin60) + 6(8) − (5sin30)(16) = 0 T GH=-1.155kN

From the left portion of section

∑ MH = 0

6(12) − (5sin30)(32) − BC(8sin60) = 0

T BC=-1.155kN

Left portion of section

x45

Pencil

5.Analysis of Structures Trusses .15

Documents

Transcript of 5.Analysis of Structures Trusses .15