Unit- Truss ,Engineering mechanics by shubham singh PDF

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Contents Chapter 1 :- Introduction Chapter 2 :- Free body diagram & equilibrium Chapter 3 :- Trusses & frames Chapter 4 :- Friction Chapter 5 :- Virtual work Chapter 6 :- Rectilinear & curvilinear motion of the particle Chapter 7 :- Kinetics of a particle (a) Work & energy

(b) Impulse & momentum Chapter 8 :- Impact : collision of elastic bodies Chapter 9 :- Kinetics of Rigid body (a) Force & acceleration (b) Work & energy

SHUBHAM SINGH

Basic Concepts of Truss :Concepts :- In The Method of Joints , Equilibrium of each joints is considered. The procedure is as follow :(1) Consider the free body diagram of the entire truss. (2) Find the reactions at the supports. (3) Consider the equilibrium of a joint , Where only 2 unknown members are meeting.

USE :-

∑Fx = 0 ∑Fy = 0

NOTE : If a member pushes a joint then member itself will be in compression with the same magnitude. Member

Compression Joint

 If a member pulls a joint then the member itself will be in tension with same magnitude.

Member

Tension Joint SHUBHAM SINGH

CASE – A If at a joint 3 members are meeting & 2 are collinear then force in 3rd member will be zero.(If there is no load or reaction at that joint)

FAB

FAB θ

θ

FAD

90-θ

FAD y

FAC

FAC ∑Fx = 0,

FAB - FAC + FADsin(90- θ) = 0

FADcos(90-θ) = 0 FADsinθ = 0 ,

x

∑Fy = 0,

∴ sinθ ≠ 0

FAD = 0

FAB - FAC + FADcosθ = 0

FAB = FAC

CASE-B If at a joint 2 members are meeting & they are non collinear then force in both the members will be zero.(if there is no load or reaction at that joint)

FAB

∑FY = 0 FABsinθ = 0

θ

FAC

∴sinθ ≠ 0 FAB = 0

 If FAB = 0 then Obvious, FAC = 0

SHUBHAM SINGH

Special case FCD

0

FBC

FAB

Fig. 1 ∑Fy = 0, FCD = 0,

 If FCD = 0 then Obvious, FAB = FBC

E

C

D Joint E

P B

A

FEC

Fig. 2 So, FEC = FED = 0

FED

Concepts :- In The Method of section :- Equilibrium of section is considered. The procedure is as follow :(1) Find the reactions at the supports. (2) Cut the member under consideration by a section 1-1 & consider equilibrium of either left hand side or right hand side of section .

(3) Use equilibrium equation for the cut section, USE :-

∑Fx = 0 , ∑Fy = 0 , ∑M = 0

Following points should be noted while using the method of section, (1) The section should be passed through the members and not through the joints.

(2) A section should divide the truss into two clearly separate and unconnected portion. (3) A section should cut only three members since only three unknowns can be determined from the three equation of equilibrium . However , in special case more than three members may be cut by a section.

(4) When using the moment equation , the moment can be taken about any convenient point which may or may not lie on the section under consideration. D FCD U FCF F

B

FEF

As you can see that when we are solving such type of question then we are using moment equation at that joint which is not in our taken section means we are using that joint which is opposite to the taken section like in this fig we are using U joint for the moment equation. i.e. meaning of underline statement. SHUBHAM SINGH

Q.A truss is loaded and supported as shown in fig. find by the method of joints the axial forces in each member of the truss. Sol.

0.375 m

1200 N

0.5 m

0.4 m

Using the concept Method of Joints,

θ’

θ RA

RB 0.4 m

So, tanθ =

0.375 m

1200 N

0.375 0.9

θ = 22.61o

0.5 m

tanθ’ =

0.375 0.5

θ’ = 36.86o

∑Fy = 0 , RA + RB = 1200 N

SHUBHAM SINGH

∑MA = 0 , 1200 x 0.9 – RB x 0.4 = 0 RB = 2700 N RA = - 1500 N Negative (-)sign means the Taken direction of RA is not correct , it will be in the reversed direction means RA direction will be in the vertically downward direction .

RA

RB

0.375 m

1200 N

0.5 m

0.4 m

Considering the Joint A, Using common sense for the direction of the forces at joint A, FAC θ

A

RA

FAB

SHUBHAM SINGH

∑Fy = 0 ,

∑Fx = 0 ,

FACsinθ = RA

FACcosθ = FAB

FACsinθ = 1500 N

3900x cos22.61 = FAB FAB = 3600 N (Comp)

FACsin22.61 = 1500 N

FAC = 3900 N (Tension)

Considering the Joint B, Using common sense for the direction of the forces at joint B, FBC θ'

FAB RB ∑FX = 0, FBCcosθ’ = FAB FBCcos36.86 = 3600 FBC = 4500 N (comp) For more understanding, click on the given link below, https://drive.google.com/drive/folders/1srHydMNehwn-J9xh6zdbbJyvCYMj8v7?usp=sharing

Another Approach for solving the same question,i.e (without common sense) Considering the Joint A, Using no common sense for the direction of the forces at joint A, means Take any direction of the forces of the members.

FAC θ

A

FAB

RA

∑Fx = 0 ,

∑Fy = 0 ,

-FACcosθ = FAB

-FACsinθ = RA

-1500 x sin22.61 = FAB

-FACsin22.61 = -1500

FAB = -3600

FAC = 1500 N (Tension)

 Negative (-) sign indicates that taken direction is wrong ,so change the direction. FAB = 3600 N (comp) Similar approach for the joint B, FBC = 4500 N SHUBHAM SINGH

Q. A truss is loaded and supported as shown . Find out the members in which the axial force are zero ? F D

H

B

A

J

30o

30o

C 1m

E 1m

G

1m

P

K

I 1m

L

1m

1m

P

Sol . Starting from left hand side to right hand side , Using the (CASE – A ) so in the member AB & BD same amount of force is there . But in the member BC there is no force as from CASE –A .

FBC = 0 FBD

0

FAB

FBC

SHUBHAM SINGH

As you can easily see that member EG & GI are collinear but FG is perpendicular to each other so from CASE – A the force in member FG is equal to zero. FFG = 0

FFG 0

FGI

FEG

As you can easily see that member HJ & JL are collinear but JK is inclined so from CASE – A the force in member JK is equal to zero.

FJK = 0 FHJ

0

FJL FJK

So the diagram after solving the problem is look like that F D

H

B

A

J

30o

30o

C 1m

E 1m

P

1m

G 1m

L

K

I 1m

1m

P

 Dotted line shows that in that dotted line member there is no force. SHUBHAM SINGH

Diagram for the joint “K”

0

FJK

FHK

θ

FKI

FKL

∑Fx = 0 ,

∑Fy = 0 , FHKsinθ = 0

FKI - FKL = 0

∴ sinθ≠0

FKI = FKL

FHK = 0

Diagram for the joint “H” FFH

θ

FHJ FHK

FHI 0

∑Fx = 0 ,

∑Fy = 0 , FHIcosθ = 0 FHI = 0

∴ sinθ≠0

FFH - FHJ = 0 FFH = FHJ

SHUBHAM SINGH

Diagram for the joint “ I ” 0

FHI

FFI

θ

FGI

FKI ∑Fx = 0 ,

∑Fy = 0 ,

∴ sinθ≠0

FFIsinθ = 0

FGI - FKI = 0 FGI = FKI

FFI = 0

Final diagram :F D

H

B

A

J

30o

30o

L E

C 1m

1m P

1m

G

K

I 1m

1m

1m

P

 Dotted line shows that in member BC , FG , FI , HI , HK , JK there is no force.

SHUBHAM SINGH

Q. A truss is loaded and supported as shown . Find out the members in which the axial force are zero ?

P B

E

C

G

A

I

J D

F

H

Sol. Try it by your own self.

Ans :- forces in members AD, CD, EF, GH, JH is zero.

SHUBHAM SINGH

Q. Determine by the method of sections the axial force in the member EB of the truss shown in fig. I 15 KN

2m F

G

H 2m

C

D

E 2m B

A 4m

Sol. By method of section , we have already located the dotted line in the fig , so remove the upper part from the lower part so we will get this fig,

 In this method don't starts from bottom portion bcoz if we started from bottom portion then we have to find the Reaction of the supports & it will make the process more inconvenient and lengthy so we are going to start from the upper portion.

 But one question arises how we will be able to know whether starts from upper portion or bottom portion so it will come from solving more question applying your common sense.

SHUBHAM SINGH

I

15 KN PART 1 H

F

FCD

C

FDE

15 KN

E

15 KN

FAC

FEB

PART 2 FEB

FAC FCD

FDE

RAX RAY

RB

SHUBHAM SINGH

I

15 KN PART 1 G

H

F

FCD

C

FAC

FDE

E

15 KN

15 KN

FEB

Using equation of equilibrium , i.e. Applying moment at joint ‘C’

∑ MC = 0 , FEB x 4 – 15 x 2 – 15 x 4 = 0 , FEB = 22.5 KN



So if we are started from lower portion we will get same value of EB but in this approach you have to calculate support reaction and then you to have the find extra force of the members to calculate the force in the member EB.

SHUBHAM SINGH

Q. A cantilever truss is loaded as shown in fig ,Find axial force in all the members. Sol. 2m

A

2m

1m C

E

1m 1000 N B

D

Applying method of joint at the joint E,

FEC

θ

FED

So from diagram, tanθ =

E

1000N

1 2

, θ = 26.56 deg

∑Fy = 0 ,

∑Fx = 0 ,

FEDsin26.56 = 1000 N

FEDcos26.56 = FEC

FED = 2236 N

FEC = 2000 N

SHUBHAM SINGH

Joint “D” FCD FED θ

FBD

∑Fy = 0 ,

∑Fx = 0 ,

FEDsin26.56 = FCD

FEDcos26.56 = FBD

FCD = 1000 N

FBD = 2000 N

Applying method of section ,

A

2m

2m

1m C

E

1m 1000 N

B

D

SHUBHAM SINGH

θ‘ =90 – 26.56

A FAC

C

FBC

E

θ‘

1000 N

FBD

D

Applying moment at joint ‘ A ’ ∑MA = 0 , FBCsin63.44 x 1 + FBCcos63.44 x 2 + 1000 x 4 – FBD x 2 = 0 FBC = 0 You can also calculate the force FBC by the method of joint at C ,

FAC θ

FCE θ‘

FBC

FCD

SHUBHAM SINGH

FAC θ

FCE θ‘

FBC

FCD

∑Fy = 0 , FACsin26.56 - FBCcos63.44 – FCD = 0 FACsin26.56 - FBCsin63.44 – 1000 = 0

Equ (1)

∑Fx = 0 ,

FACcos26.56 + FBCsin63.44 = FCE FACcos26.56 + FBCsin63.44 = 2000

Equ (2)

From solving Equ 1 & 2 we will get the value of FAC & FBC FAC = 2236 N FBC = 0 N

SHUBHAM SINGH

Q. A truss is loaded and supported as shown in fig. find axial force in the member BD,CD,CE . 1 KN

1 KN a

1 KN a

a

D

A

a

H

F

B

a

E

C

G

Sol. As we can see that we have to find only three forces in the corresponding members so we are using method of section for solving this question but you can also solve this question by method of joint.

1 KN

1 KN

a A

a

D

1 KN a

a

F

B

RA

H RH

a

C

∑ Fy = 0 , RA + RH = 3 KN , RA = 1.5 KN

E

G

∑M A = 0 , RH x 4a – 1xa – 1x2a – 1x3a = 0 RH = 1.5 KN

SHUBHAM SINGH

1 KN

1 KN

1 KN

a

a a

A

D

a

F

B

RA

RH a

C

G

E

FBD

1 KN

1 KN

D

F

45o

H RH

FCD

C

FCE ∑MC = 0 , RH x 3a – 1xa – 1x2a – FBD x a = 0 1.5 x 3a – 1xa – 1x2a – FBD x a = 0 FBD = - 1.5 KN ,

FBD = 1.5 KN (comp.)

 Means Taken direction of Force BD is wrong i.e. member BD will be in compression instead of Tension . ∑Fy = 0 , FCDsin45 + 1 + 1 - RH = 0

FCDsin45 + 1 + 1 – 1.5 = 0 FCD = - 0.707 KN FCD = 0.707 KN (Comp.)

SHUBHAM SINGH

∑Fx = 0 , FBD + FCE + FCDcos45 = 0 - 1.5 + FCE - 0.707cos45 = 0 FCE = 2 KN FCE = 2 KN (Tension)

Q. Determine the forces in bars 1 , 2 & 3 of the plane truss loaded and supported as shown in fig. C

D

1

h

2 A

B

3 a a

P

P

a

a P

a P

P

Try by your self. Similar question as we have done in above question.

Ans :- F1 =

−4𝑃𝑎 ℎ

, F2 =

−𝑃 2

, F3 =

+4𝑃𝑎 ℎ

SHUBHAM SINGH

Q. A truss is loaded and supported as shown in fig , Find axial force in the members BD,DE & EG ? F 1 KN

D

2.25 m B

H

A C 1m

E 1m

G 1m

1m

Sol. As we can see that we have to find only three forces in the corresponding members so we are using method of section for solving this question but you can also solve this question by method of joint. F 1 KN

D 2.25 m

B H

A

RA

C 1m

G

E 1m

1m

1m

∑Fy = 0 ,

∑MA = 0 ,

RA + RH = 1 KN

RH x 4 – 1 x 1 = 0

RA = 0.75 KN

RH

RH = 0.25 KN SHUBHAM SINGH

F 1 KN

D 2.25

B

H

θ

A

E

C

RA

1m

1m

G 1m

F

1m

RH

, θ = 36.86 deg  tanθ = 2.25 3

D θ

FBD FDE FEG

H

G

E 1m

1m

RH

∑ME = 0 ,

∑Fy = 0 ,

RH x 2 + FBDcos36.86 x 1.5 = 0

FBDsin36.86 +FDE – RH = 0

FBD = - 0.417 KN

-0.417sin36.86 + FDE – 0.25 = 0

FBD = 0.417 KN (comp)

FDE = 0.5 KN (Tension) SHUBHAM SINGH

∑Fy = 0 , FBDcos36.86 + FEG = 0 -0.417xcos36.86 + FEG = 0 FEG = 0.333 KN (Tension)

NOTE :- Mathematical Condition for Rigid or perfect Truss, A truss consists of a number of members which are connected together and form a certain number of joints. For a truss to be rigid or perfect, the relationship between its number of members and the number of joints is ,  m + 3 = 2j , Statically Determinate (Equation of static equilibrium alone are sufficient to determine the axial forces in the members without the need of considering their deformation.)

 m + 3 > 2j , over rigid & statically indeterminate(Truss member is more than Truss joint) Means Equation of Equilibrium is just not sufficient to find the forces at a joints, so it is solved by considering the compatibility equation.

 m + 3 < 2j , collapsible or under rigid , (there is a deficiency of internal members, and the truss is unstable and will collapse under load)

ASSUMPTIONS FOR THE PERFECT TRUSS :- (Only for IES )

 . All the members are pin-jointed & frictionless.  . The Truss is loaded only at the joints.  . The truss is a statically determinate.  . The weight of the member are negligibly small unless otherwise mentioned.

 . The members of a truss are two straight two force members with the forces acting collinear with the centre line of the members . SHUBHAM SINGH

Q. A truss is loaded and supported as shown in fig .find the axial forces in the members AB , EF & CD. B

D

0.75 m E

F 1 KN

0.75 m

A

C

0.5 m

0.5 m

0.5 m

Sol. As we can see that we have to find only three forces in the corresponding members so we are using method of section for solving this question but you can also solve this question by method of joint.

B

D

E

F 1 KN

A

RA

C

RC SHUBHAM SINGH

If you are looking to the section cut, then one question comes into your mind “ Ki kahan se chalu karein “ so the answerer of this question is that whenever such type of situation comes into your mind then starts from that section where we have to find minimum unknowns, like in this question in upper part there is less unknown than the lower part bcoz in lower part we have to find reactions at both supports so starting from upper parts.

B

D

FEF

F FCD

FAB 1 KN

∑MF = 0 , FAB x 1 - FCD x 0.5 = 0

FCD = 2 FAB ∑Fy = 0 ,

∑FX = 0 ,

FCD + FAB + 1 = 0 2FAB + FAB = -1

FEF = 0 FEF = 0

3FAB = -1

FAB = - 0.333 KN FAB = 0.333 KN (comp.) Using equation , FCD = 2 FAB FCD = 2x (-0.333) = - 0.667 KN FCD = 0.667 KN (comp.)

SHUBHAM SINGH

Q. A plane truss is supported and loaded as shown in fig . Find the axial force in members CD , EF & CF by method of section ? D C

θ

2m

3m

A

B

E

F 30 KN

2m

2m

2m

Sol. As we can see that we have to find only three forces in the corresponding members so we are using method of section...