chapter 6 space trusses - universiti teknologi...

CHAPTER 6 Space Trusses

INTRODUCTION

• A space truss consists of members joined together at their ends to form a stable three-dimensional structures

• A stable simple space truss can be built from the basic tetrahedral, formed by connecting six members with four joints

INTRODUCTION

• To determine the stability and determinacy of space trusses

• To determine member forces of space trusses using tension coefficient analysis

OBJECTIVES

1. Simple Space Truss

This truss is constructed from a tetrahedron. The truss can be enlarged by adding three members.

TYPE OF SPACE TRUSSES

2. Compound Space Truss This truss is constructed by combining two or more simple truss.


3. Complex Space Truss Complex truss is a truss that cannot be classified as simple truss or compound truss.


• Due to three dimensions, there will be three equations of equilibrium for each joint.

(Fx = 0; Fy = 0; Fz = 0)

• The external stability of the space truss requires that the support reactions keep the truss in force and moment equilibrium.

• Generally, the least number of required reactions for stable and externally determinate is SIX

If r < 6 Unstable

If r > 6 Externally Indeterminate

DETERMINACY & STABILITY

• For internal determinacy, if m = number of members; j = number of joints; r = number of supports; therefore:

If m = 3j + r Stable and Internally Determinate

If m < 3j + r Unstable

If m > 3j + r Internally Indeterminate

• Internal stability can sometimes be checked by careful inspection of the member arrangement.


Internally m + r = 3j Determinate Truss m + r > 3j Indeterminate Truss m + r < 3j Unstable Truss Externally r < 6 Unstable Truss r = 6 Determinate if Truss is Stable r > 6 Indeterminate Truss


TYPES OF SUPPORT

EXAMPLE 1

Ball & Socket

m = 3, j = 4, r = 9 m + r = 12 3j = 12 m + r = 3j Determinate Truss

EXAMPLE 2

m = 15, j = 10, r = 15 m + r = 30 3j = 30 m + r = 3j Determinate Truss

Ball & Socket

EXAMPLE 3

Ball & Socket

m = 13, j = 8, r = 12 m + r = 25 3j = 24 m + r = 25 3j = 24 Indeterminate Truss in the First Degree

• The members are joined together by smooth pins (no friction – cannot resist moment)

• All loadings and reactions are applied centrally at the joints

• The centroid for each members are straight and concurrent at a joint

Therefore, each truss member acts as an axial force member:

If the force tends to elongate Tensile (T)

If the force tends to shorten Compressive (C)

ASSUMPTIONS FOR DESIGN

THEOREM 1:

If all members and external force except one member at a joint, (say, joint B) lie in the same plane, then, the force in member A is zero.

The force in member A is zero

ZERO FORCE MEMBERS

P

Member A

x y

z B

THEOREM 2:

If all members at a joint has zero force except for two members, (say member A and B), and both members (A and B) do not lie in a straight line, then the force in member A and B are zero.

Both members A and B has zero force because both members do not lie in a straight line.

ZERO FORCE MEMBERS

FAy FAx

Member A

Member B

FBx

FBy

0 0

0

If members A and B lie in a straight line, then, the forces in these members MIGHT NOT be zero. In fact, referring to the example below:

FAx = -FBx

FAy = -FBy

ZERO FORCE MEMBERS

FAy FAx

Member A

Member B

FBx FBy

0 0

0

THEOREM 3:

If three members at a joint do not lie in the same plane and there is no external force at that joint, then the force in the three members is zero.

Three members connected at a

joint has zero force. A plane can

consists of two members, say

member A and B. Thus, no force

can balance the component of

member C that is normal to the

plane.

ZERO FORCE MEMBERS

Member B

Member C

Member A

Identify the members of the space truss that has zero force.

Theorem 1: Joint L F1= 0 and F2 = 0 Theorem 3: Joint K F3 = F4 = F5 = 0 and F1= 0 Theorem 3: Joint M F6 = F7 = F8 = 0 and F2= 0 Theorem 3: Joint J F9 = F10 = F11= 0 and F5 = F6= 0

EXAMPLE 4

Theorem 1: Joint F

Members FE, FC, FD lie in a plane, except member FG. Thus, member FG has zero force.

Theorem 3: Joint F

Members FE, FC, FD have zero force.

Theorem 3: Joint E

Members ED, EA, EH have zero force.

EXAMPLE 5

• To determine

member forces

• Based on 3D particle

equilibrium

Lx = L cos : Ly = L cos ; Lz = L cos

Fx = F cos : Fy = F cos ; Fz = F cos

Therefore: 𝐹𝑥

𝐿𝑥=

𝐹

𝐿

𝐹𝑦

𝐿𝑦=

𝐹

𝐿

𝐹𝑧

𝐿𝑧=

𝐹

𝐿

ANALYSIS METHOD: TENSION COEFFICIENT

• Tension Coefficient: t = F/L

Fx = tLx ; Fy = tLy ; Fz = tLz

• If t positive (tension), if t negative (compression)

• The actual force and length are given from:

𝐿 = 𝐿𝑥2 + 𝐿𝑦

2+ 𝐿𝑧2 𝐹 = 𝐹𝑥

2 + 𝐹𝑦2+ 𝐹𝑧

2

• For 3D equilibrium:

Fx = 0 ; Fy = 0 ; Fz = 0

tLx = 0 ; tLy = 0 ; tLz = 0


The component of length, L, and force, F in the x, y, z direction


y

x

z

Ly , Fy

Lz , Fz

A

L, F

Lx , Fx

Where t = F/L

t = tension coefficient


𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑥

𝐹𝑥 = 𝐹𝐿𝑥

𝐿=

𝐹

𝐿𝐿𝑥

𝐹𝑥 = 𝑡 ∙ 𝐿𝑥

y

x

z

Fx

Fy

Fz

C E

D

A

B

O

F

X


𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦

𝐹𝑦 = 𝐹𝐿𝑦

𝐿=

𝐹

𝐿𝐿𝑦

𝐹𝑦 = 𝑡 ∙ 𝐿𝑦

y

x

z

Fz

C E

D

B

O Fx

Fy

A Y F


𝐹𝑧 = 𝐹𝑐𝑜𝑠𝜃𝑧

𝐹𝑧 = 𝐹𝐿𝑧

𝐿=

𝐹

𝐿𝐿𝑧

𝐹𝑧 = 𝑡 ∙ 𝐿𝑧

y

x

z

C

D

B

Fx

Fy

A

Fz

E

O

F

Z

The space truss shown in the figure has roller and socket support at joint A, B, C and D

i) Determine the member force for all members at joint F and G

ii) Determine the reaction at support C

EXAMPLE 6

y

40 kN 10 kN

40 kN

20 kN

4 m 1 m

2 m

2 m

A

B

E

D

C

F

G

40 kN

x

z

1. Start at Joint G

Member Lx (m) Ly (m) Lz (m) L (m) t (kN/m) F (kN)

GC 0 0 -2 2 -20 -40 GE -4 -2 0 4.47 2.5 11.18 GF 0 -2 0 2 -2.5 -5

Force (kN) 10 0 -40

Fx = 0: -4tGE + 10 = 0 tGE= 2.5 kN/m

Fz = 0: -2tGC – 40 = 0 tGC = -20 kN/m

Fy = 0: -2tGE – 2tGF = 0

-2(2.5) – 2tGF = 0 tGF = -2.5 kN/m

EXAMPLE 6 – Solution

2. Move to Joint F


FC 0 2 -2 2.83 -5 -14.14 FD 0 -1 -2 2.24 -15 -33.6 FE -4 0 0 4 0 0 FG 0 2 0 -2.5

Force (kN) 0 0 -40

Fx = 0: -4tFE = 0 tFE = 0 kN/m

Fy = 0: 2tFC – tFD + 2tFG = 0 2tFC – tFD – 5 = 0 ....... (i)

Fz = 0: -2tFC – 2tFD – 40 = 0 .......(ii)

(i) + (ii): -3tFD -45 = 0 tFD = -15 kN/m

From (i): -2tFC – (-15) – 5 = 0 tFC = -5 kN/m


3. Calculate the Reaction at Joint C


CF 0 -2 2 -5

CG 0 0 2 -20

Force (kN) RCx RCy RCz

Fx = 0: RCx = 0 kN

Fy = 0: -2tCF + RCy = 0

-2(-5) + RCy = 0 RCy = -10 kN ( )

Fz = 0: 2tCF + 2tCG + RCz = 0

2(-5) + 2(50) + RCz = 0 RCz = 50 kN ( )


Determine the force in each member of the space truss shown.

EXAMPLE 7

4 m

x

z

4 m

2 m

2 m

y

A

C

D

E

Ez = 4 kN

B

Short link

Ball & socket

Slotted roller constraint in a

cylinder

1. Start with joints where there are only 3 unknowns force/reaction.

Joint B – 5 unknowns

Joint C – 5 unknowns

Joint A – 7 unknowns

Joint E – 4 unknowns

Joint D Theorem 3: Three members at a joint and no external force. Thus , all members have zero forces. tDC = tDA = tDE = 0


EXAMPLE 7 – Solution Joint E


EC -2 -4 4 4.47 0 0 EB 2 -4 4 6 1 6 ED -2 -4 0 4.47 0 0 EA 2 -4 0 4.47 -1 -4.47

Force (kN) 0 0 -4

Fz = 0: 4tEB + 4tEC – 4 = 0 .......(i)

Fx = 0: -2tEC + 2tEB – 2tED + 2tEA = 0 ....... (ii) where tED = 0

Fy = 0: -4tEC – 4tEB – 4tED – 4tEA = 0 .......(iii) where tED = 0

Solve Eq. (i), (ii) & (iii): tEC = 0 kN/m, tEA = -1 kN/m and tEB = 1 kN/m

EXAMPLE 7 – Solution Joint C

Joint C has three (3) unknowns, as tCE = tCD = 0

Therefore, by Theorem 3:

tCB = tCA = RCy = 0

x

z

y

A

C

D

E

Ez = 4 kN

B

RCy

EXAMPLE 7 – Solution Joint C


CE 2 4 -4 6 0 0 CA 4 0 -4 5.66 0 0 CD 0 0 -4 4 0 0 CB 4 0 0 4 0 0

Force (kN) 0 -RCy 0

Fz = 0: 4tCA = 0 tCA = 0 kN/m

Fx = 0: 4tCA + 4tCB = 0 tCB = 0 kN/m

Fy = 0: -RCy = 0

EXAMPLE 7 – Solution Joint B

Joint B has three (3) unknowns.

x

z

y

A

C

D

E

Ez = 4 kN

B RBy

RBx

EXAMPLE 7 – Solution Joint B


BC -4 0 0 4 0 0 BA 0 0 -4 4 -1 -4 BE -2 -4 -4 6 1 6

Force (kN) RBx -RBy 0

Fx = 0: -4tBC – 2tBE + RBx = 0 RBx = 2 kN

Fy = 0: -4tBE – RBy = 0 RBy = -4 kN

Fz = 0: -4tBA – 4tBE = 0 tBA = -1 kN/m

EXAMPLE 7 – Solution Joint A

x

z

y

A

C

D

E

Ez = 4 kN

B

RAy

RAx

RAz

EXAMPLE 7 – Solution Joint A


AB 0 0 4 4 -1 -4 AC -4 0 4 5.66 0 0 AD -4 0 0 4 0 0 AE -2 4 0 4.47 -1 -4.47

Force (kN) -RAx RAy RAz

Fx = 0: -4tAC – 4tAD – 2tAE – RAx = 0 RAx = -2 kN

Fy = 0: 4tAE + RAy = 0 RAy = -4 kN

Fz = 0: 4tAB + 4tAC + RAz = 0 RAz = 4 kN

EXAMPLE 7 – Solution Method 2

Reactions at A, B and C can also be obtained from Method 2

x

z

y

A

C

D

E

Ez = 4 kN

B

RAy

RAx

RAz

RBy

RBx

RCy

EXAMPLE 7 – Solution Method 2

My = 0

-4(2) + RBx (4) = 0

RBx = 2 kN

Mz = 0

RCy = 0 kN

Mx = 0

RBy (4) – 4(4) = 0

RBy = 4 kN

Fx = 0

2 – RAx = 0

RAx = 2 kN

Fy = 0

RAy – 4 = 0

RAy = 4 kN

Fz = 0

RAz – 4 = 0

RAz = 4 kN

A

B

C

D

x

y

z

O

2 m

3 m

7 m

2 m

1800 N

The space truss shown in the Figure consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Analyse the force in each of the members for the given loading.

CLASS EXERCISE

chapter 6 space trusses - universiti teknologi...

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