2- Handouts Lecture-31 32

COURSE:COURSE: CE 201 (STATICS)CE 201 (STATICS)

LECTURE NO.:LECTURE NO.: 31 & 3231 & 32

FACULTY:FACULTY: DR. SHAMSHAD AHMADDR. SHAMSHAD AHMAD

DEPARTMENT:DEPARTMENT: CIVIL ENGINEERINGCIVIL ENGINEERING

UNIVERSITY:UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA& MINERALS, DHAHRAN, SAUDI ARABIA

TEXT BOOK:TEXT BOOK: ENGINEERING MECHANICSENGINEERING MECHANICS--STATICS STATICS by R.C. HIBBELER, PRENTICE HALLby R.C. HIBBELER, PRENTICE HALL

LECTURE NO. 31 & 32LECTURE NO. 31 & 32INTERNAL FORCES (2INTERNAL FORCES (2--D & 3D & 3--D)D)

Objectives:Objectives: To show how to use the method of sections for To show how to use the method of sections for

determining the internal forces in a member at a determining the internal forces in a member at a specified location.specified location.

INTERNAL FORCES DEVELOPED IN MEMBERSINTERNAL FORCES DEVELOPED IN MEMBERSWhen a structural member is subjected to the external loadings usuallyfollowing three types of internal forces are developed at any section of themember: Normal or axial force, N Shear force, V Bending moment, M

For example, the internal forces at point C of a simply supported beamsubjected to external forces are shown in the following figure:

C

INTERNAL FORCES IN 2INTERNAL FORCES IN 2--DD

The internal forces in two-dimensions (2-D)are shown below with their positive senses:

C

V

N

M

= Shear force

= Normal force

= Bending moment

C

V

N

M

C

V

N

M

= Shear force

= Normal force

= Bending moment

INTERNAL FORCES IN 3INTERNAL FORCES IN 3--DD

The internal forces in three-dimensions (3-D) are shown below with their positive senses:

DETERMINATION OF THE INTERNAL FORCESDETERMINATION OF THE INTERNAL FORCES

For determining the internal forces at any section of amember following procedure may be adopted: Determine the support reactions by considering the

F.B.D. of the entire structure.

Pass a vertical section at the desirable point and indicatethe internal forces (i.e., N, V, and M) on isolated parts of the member.

Select either left part or right part and apply theequilibrium conditions: Fx = 0, Fy = 0, and M about a point = 0 in such a way that most direct solutions of N, V, and M could be obtained.

UTILITY OF INTERNAL FORCESUTILITY OF INTERNAL FORCES

The analysis of the internal forces, are required forcalculation of various types of stresses and deformationsin structural members subjected to various types ofexternal forces.

The stresses and deformations are used in analysis and design of the structural and mechanical members.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1

Determine the internal forces at quarter-point (C) and mid-point (D) of the beam AB, Shown the following figure.

A BC D

4m

20 kN/m

A BC D

4m

20 kN/m


A BC D

4m

20 kN/m

A BC D

4m

20 kN/m

Support reaction at A can be determined by applying equilibriumconditions to the FBD of the entirebeam, as follows:

AB

4m

2m 2m

AyBy

FR = 204=80kN

AB

4m

2m 2m

AyBy

FR = 204=80kN

M about B = 0 4Ay 80 2 = 0 Ay = 40 kN.

Internal forces at quarter-point Ccan be determined by applyingequilibrium conditions to FBD of theportion AC of the beam, as follows:

AC

40kNVC

FR = 201=20kN

m144 =

MC

NCA C

40kNVC

FR = 201=20kN

m144 =

MC

NC

Fx = 0 Nc = 0 Ans. Fy = 0 40 20 VC = 0 VC = 20 kN () Ans. Mabout C = 0 401 20 MC = 0 MC = 30 kN-m ( ) Ans.


AB

4m

2m 2m

AyBy

FR = 204=80kN

AB

4m

2m 2m

AyBy

FR = 204=80kN

Internal forces at quarter-point D can be determined by applyingequilibrium conditions to FBD of the portion AD of the beam, as follows:

A

40kNVD

FR = 202=40kNMD

ND

2m

1mA

40kNVD

FR = 202=40kNMD

ND

2m

1m

Fx = 0 ND = 0 Ans. Fy = 0 40 40 VD = 0 VD = 0 Ans. M about D = 0 40 2 40 1 MD = 0 MD = 40 kN-m ( ) Ans.


Determine the internal forces at mid-points (D) of supports A and B and also at point just to the right of support B of the beam ABC, shown in the followingfigure.

A B C

3m 1m

30N/m

A B CD

3m 1m

30N/m


A B C

3 m 1m

30N/m

A B CD

3 m 1m

30N/m

Reactions at supports A and B can be determined by applyingequilibrium conditions to the FBD ofthe entire beam, as follows:

AB C

3m

1.33m

Ay By

m67.2432 =

NFR 602430 ==

AB C

3m

1.33m

Ay By

m67.2432 =

NFR 602430 ==

M about A = 0 60 2.67 3 By = 0 By = 53.33 N Fy = 0 Ay 60 + 53.33 = 0 Ay = 6.67 N

Internal forces at quarter-point D can be determined by applying equilibriumconditions to FBD of the portion AD of the beam, as follows:

m35.1

NDD

1.5m6.67 N VD

NFR 44.825.125.11 ==

MD

m35.1

NDD

1.5m6.67 N VD

NFR 44.825.125.11 ==

MD

Fx = 0 ND = 0 Ans. Fy = 0 6.67 8.44 VD = 0 VD = 1.77 N = 1.77N ( ) Ans. Mabout D = 0 6.667 1.5 8.44 1.53 MD = 0 MD = 5.78 N-m ( )


AB C

3m 1 m

30N/m

AB CD

3m 1 m

30N/m

Internal forces at point just to the right of support B can be determined by applying equilibrium conditions to FBD of the portion AB of the beam, as follows:

m133 =

NBB

3m6.67 N

VB

NFR 75.33235.22 ==

MB53.33 N

A

m133 =

NBB

3m6.67 N

VB

NFR 75.33235.22 ==

MB53.33 N

A

Fx = 0 NB = 0 Ans. Fy = 0 6.67 33.75 + 53.33 VB = 0 VB = 26.25 N Ans. M about B = 0 6.67 3 33.75 1 MB = 0 MB = 13.75 N-m = 13.75 N-m ( )

Ans.


Determine the internal N, V, and M in the beam at points C and D. Assume the support at B is a roller. Point C is located just to the right of the 8-kip load.


Reactions at supports A and B can be determined by applyingequilibrium conditions to the FBD ofthe entire beam, as follows:

A

C D

Ax

Ay By

8kip40kipft

AC D

Ax

Ay By

8kip40kipft

8 ft 8 ft8 ft

Fx = 0 Ax = 0

Fy = 0 Ay + By 8 = 0 Ay = 7 kip.

M about A = 0 8 8 40 24 By = 0 By = 2424 = 1 kip

Internal forces at point C can be determined by applying equilibriumconditions to FBD of the portion AC of the beam, as follows:

AC

Ay=7kip

8kip

VC

NCMC

AC

Ay=7kip

8kip

VC

NCMC8 ft

Fx = 0 Nc = 0 Ans. Fy = 0 7 8 Vc = 0 Vc = 1 kip = 1 kip () Ans. M about C = 0 7 8 Mc = 0 Mc = 56 kip-ft ( ) Ans.


AC D

Ax

Ay By

8kip40kipft

AC D

Ax

Ay By

8kip40kipft

8 ft 8 ft8 ft

Internal forces at point D can be determined by applying equilibriumconditions to FBD of the portion AD of the beam, as follows:

D

Ay=7kip VD

8kip

NDMD

D

Ay=7kip VD

8kip

NDMD

8 ft8 ft

Fx = 0 ND = 0 Ans. Fy = 0 7 8 VD = 0 VD = 1 kip = 1 kip () Ans. M about D = 0 7 16 8 8 MD = 0 MD = 48 kip-ft. ( ) Ans.

PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4The column is fixed to the floor and is subjected to the loadsshown. Determine the internal N, V, and M at points A and B.


Internal forces at point A can be determined by applying equilibriumconditions to FBD at point A of the column, as follows:

NA

1

5

0

m

m

1

5

0

m

m

6kN 6kN

MA

VA A

NA

1

5

0

m

m

1

5

0

m

m

6kN 6kN

MA

VA A

Fx = 0 VA = 0 Ans. Fy = 0 NA 6 6 = 0 NA = 12 kN Ans. Mabout A = 0 MA + 6 150 6 150 = 0 MA = 0 Ans. Internal forces at point A can be determined by applying equilibriumconditions to FBD at point B of the column, as follows:

Fx = 0 VB = 0 Ans. Fy = 0 NB 8 6 6 = 0 NB = 20 kN Ans. Mabout B = 0 MB1506+6150+8150= 0 MB = 1200 kN-mm ( ) Ans.


Determine the internal N, V, and M at point E of the frame shown below.


Vertical reaction at support C can be determined by applyingequilibrium conditions to the FBDof the entire frame, as follows:

M about A=0 3 1 4.5 1.5 3 Cy = 0 Cy = 1.25 kN = 1.25 kN ()

A

Ay Cy

CAx Cx1m

1.5m

E

DB

FR1=3kN

FR2=4.5kN

A

Ay Cy

CAx Cx1m

1.5m

E

DB

FR1=3kN

FR2=4.5kN

3 m

Horizontal reaction at support C can be determined by applying equilibriumconditions to the FBD of the portion CBof the frame, as follows:

M about B = 0 3Cx+ 4.51.5 = 0 Cx = 2.25 kN = 2.25 kN ()

By

BxB

4.5 kN

Cx

Cy

By

BxB

4.5 kN

Cx

Cy

1.5 m

1.5 m


A

Ay Cy

CAx Cx1m

1.5m

E

DB

FR1=3kN

FR2=4.5kN

A

Ay Cy

CAx Cx1m

1.5m

E

DB

FR1=3kN

FR2=4.5kN

3 m

Internal forces at point E can be determined by applying equilibriumconditions to FBD at point E of the frame, as follows:

VE

FR3 =2.25 kN

Cx=2.25 kN

Cy=1.25 kN

EVE

NE

ME

C

0.75m

VE

FR3 =2.25 kN

Cx=2.25 kN

Cy=1.25 kN

EVE

NE

ME

C

0.75m

0.75 m

Fx = 0 2.25 2.25 VE = 0 VE = 0 Ans.

Fy = 0 NE 1.25 = 0 NE = 1.25 kN Ans.

M about E = 0 2.25 1.5 2.25 0.75 + ME = 0 ME = 1.6875 kN-m Ans.


Determine the x, y, z components of force and moment at point C in the pipe assembly. Neglect the weight of the pipe. Take F1 = {350i 400j}lb and F2 = {300j+150k}lb.


Nc i

Vcz kF1

Vcy j

C

Mcy j

Mcx i

Mcz k

D

B

F2F1 = {350i 400j}lbF2 = {300j+150k}lb

The coordinates of points B, C, and D are as follows:

B (3.5, 0, 3) ft; C (1.5, 0, 0) ft, D (3.5, 0, 0) ft

rCB = {2i + 3k} ft; rCD = {2i} ft

F = 0 F1 + F2 + Nci + VcyJ + Vcjk = 0{350i 400j}+{300j + 150k}+ Nci + Vcyj +Vcjk = 0{Nc+350}i + {Vcy700}j + {Vcz+150}k = 0Nc + 350 = 0 Nc = 350 lb Ans.Vcy 700 = 0 Vcy = 700 lb Ans.Vcz+ 150 = 0 Vcz = 150 lb Ans.Mabout C = 0rCB F1 + rCD F2 + Mcxi + Mcyj + Mczk = 0{2i + 3k}{350i 400j} + {2i}{300j+150k} + Mcxi + Mcyj + Mczk = 0


{2i + 3k}{350i 400j} + {2i}{300j+150k} + Mcxi + Mcyj + Mczk = 0{800k + 1050j + 1200i} + {600k 300j} + Mcxi + Mcyj + Mczk = 0{Mcx + 1200}i + {Mcy + 750}j + {Mcz 1400}k = 0Mcx + 1200 = 0 Mcx = 1200 lb-ft Ans.Mcy + 750 = 0 Mcy = 750 lb-ft Ans.Mcz 1400 = 0 Mcz = 1400 lb-ft Ans.

Multiple Choice Problems1. For the beam shown below, the normal

force, shear force, and bending moment acting just to the left of the 6kN (i.e., at point B) are

(a) 0, 5 kN, and 15 kN-m, respectively (b) 5 kN, 0, and 15 kN-m, respectively (c) 15 kN, 5 kN, and 0, respectively (d) 5 kN, 15 kN, and 0, respectively

Ans: (a) Feedback:

Reaction at support A can be determined by applying equilibriumconditions to the FBD of the entirebeam, as follows:

M about D = 09 6 6 9 0 5y yA A = = Internal forces at point B can be determined by applying equilibrium conditions to FBD at point B of the beam, as follows:

Fx = 0 NB = 0 Fy = 0 5 VB = 0 VB = 5 kN M about B = 0 5 x 3 MB = 0 MB = 15 kN-m

Multiple Choice Problems2. For the beam shown below the normal

force, shear force, and bendingmoment acting just to the right of the 6 kN (i.e., at point C) are

(a) 1 kN, 15 kN, and 0, respectively (b) 1 kN, 0, and 15 kN-m, respectively

(c) 15 kN, 1 kN, and 0, respectively (d) 0, 1 kN, and 15 kN-m, respectively

Ans: (d)

Feedback:

M about D = 0 9 6 6 9 0 5y yA A = =

Internal forces at point C can be determined by applying equilibriumconditions to FBD at point C of the beam, as follows:

Note: The 6 kN load is considered in the FBD at point C because point C is just after the load

Fx = 0 NC = 0

Fy = 0 5 + VC 6= 0 VC = 1 kN

M about B = 0 5 x 3 MC = 0 MC = 15 kN-m

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