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ENGR-1100 Introduction to Engineering Analysis. Lecture 16. Today Lecture Outline. Structures: Trusses Frames Trusses analysis- method of joints Stability criteria. Trusses : Structures composed entirely of two force members.

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slide2

Today Lecture Outline

  • Structures:
          • Trusses
          • Frames
  • Trusses analysis- method of joints
  • Stability criteria.
two important structures types
Trusses: Structures composed entirely of two force members.

Frames: Structures containing at least one member acted on by forces at three or more points.

Two important structures types
plane trusses
Plane trusses: lie in a single plane.

Space trusses: not contained in a single plane and/or loaded out of the structure plane.

Plane Trusses
slide5

Assumptions

1) Truss members are connected together at their ends only.

2) Truss members are connected together by frictionless pins.

3) The truss structure is loaded only at the joints.

4) The weight of the member may be neglected.

slide6

The Four Assumptions

Truss members are two-force members

F

F

Truss member

F

F

slide7

Compressive forces tend to shorten the member.

Tensile forces tend to elongate the member.

F

F

F

F

Straight Members

Forces act along the axis of the member

slide8

“Rigid” trusses

“Rigid”- the truss will retain its shape when removed from its support

Simple truss- constructed by attaching several triangles together.

Allows a simple way to check rigidity.

slide9

What are we looking for?

The support reaction .

The force in each member.

How many equations are available? How many unknowns?

Each joint- 2 equations

Unknowns- number of members+ support reaction.

slide10

Stability Criteria

m=2j-3

2j- number of equations to be solved.

m- number of members.

3- number of support reaction

m<2j-3

Truss unstable

m>2j-3

Statically indeterminate

slide11

m=2j-3

Example

m (Number of members) =

13

j (Number of joints) =

8

Number of supports=

3

slide12

Method of Joints

Separate free-body diagrams for:

each member

each pin

Equilibrium equations for each pin:

SF=0

no moment equation

slide13

Example 7-3

Use the method of joints to determine the force in each member of the truss shown in Fig. P7-3. State whether each member is in tension or compression.

slide14

Joint B

Joint D

---------

---------

y

y

From a free-body diagram on joint D:

x

TBD

Fy = TBD - 3000 = 0

x

TAD

TCD

TBD = 3000 lb = 3000 lb (T)

From a free-body diagram on joint B:

TAB = - 1500.0 lb = 1500 lb (c)

TBC = - 2598 lb  2600 lb (c)

TAB

TBC

TBD

Solution

Fx = TBC sin 30º - TAB sin 60º = 0

Fy = - TBC cos 30º - TAB cos 60º - TBD = 0

slide15

Joint D

Joint C

---------

---------

y

y

x

TBC

TBD

x

From a free-body diagram on joint C:

TAD

TCD

TCD

Cy

From a free-body diagram on joint D:

known

Fx = - TCD - TBCsin 30

= - TCD -(2598) sin 30 = 0

TCD = 1299 lb = 1299 lb (T)

Fx = -TAD + TCD = - TAD + 1299 = 0

TAD = 1299 lb = 1299 lb (T)

slide16

Class Assignment: Exercise set 7-4

please submit to TA at the end of the lecture

Answer:

TBC=14.1 kN (T)

TAC=5.13 kN (T)

TAB=28.2 kN (C)

slide17

Class Assignment: Exercise set 7-6

please submit to TA at the end of the lecture

Answer:

TBD=3.23 kN (C)

TCD=4.62 kN (T)

TAD=0.567 kN (T)

TBC=4 kN (C)

TAB=2.38kN (C)

slide18

Free body diagram on joint C

y

FBC

x

FCD

FCD=0

FBC=0

SFx=0

Member BC and DC are zero force members

Zero Force Members

SFy=0

slide19

Free body diagram on joint B

y

y

FCB

FAB

x

x

FBD

FBD=0

Free body diagram on joint D

FDC

FAD

FAD=0

FED

SFy=0

slide20

Example 7-19

The truss shown in figure P7-19 support one side of a bridge; anidentical truss supportsthe other side. Floor beams carry vehicle loads to the truss joints. Calculate the forces inmembersBC, BG, andCGwhen a truck weighing 7500 lb is stopped in the middle of the bridge as shown. The center of gravity of the truck is midway between the frontand rear wheels.

slide21

Free body diagram on floor beams FH

Fy = 2FG - 2(3750) + 4(1125) = 0

FG= 1500 lb

Solution

From symmetry: FH = FF

Free body diagram on floor beams

GF

MG = 2FF (10) – 3750 (6) = 0

FF = 1125 lb

slide22

f= tan-1 8/5 = 57.99

= tan-1 5/10 =26.57

= tan-1 8/10 =38.66

Free body diagram on complete truss

Fx = Ax = 0

Ax = 0

ME= Ay (30) -1125 (25) - 1500(15) - 1125(5) = 0

Ay = 1875 lb = 1875 lb

slide23

Free body diagram on joint A

Fx = TAB cos 57.99 + TAH = 0

Fy = TAB sin 57.99 + 1875= 0

TAB = - 2211 lb  2210 lb (c)

Free body diagram on joint H

Fy = TBH - 1125 = 0

TBH = 1125 lb = 1125 lb (T)

slide24

Free body diagram on joint B

Fx = TBC cos 26.57 + TBG cos 38.66

+ 2211cos 57.99 = 0

Fy = TBC cos 26.57 - TBG sin 38.66

+ 2211 sin 57.99 - 1125 = 0

TBC = -1451.2 lb  1451 lb (C)

TBG = 161.31 lb  161.3 lb (T)

Free body diagram on joint C

Fx = TCD cos 26.57+ 1451.2 cos 26.57 = 0

Fy = TCD sin 26.57 - TCG + 1451.2 sin 26.57 = 0

TCD = -1451.2 lb  1451 lb (C)

TCG = 1298.2 lb  1298 lb (T)

slide25

Class Assignment: Exercise set 7-23

please submit to TA at the end of the lecture

ad