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
PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 1Example # 1
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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
PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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 ( )
PROBLEM SOLVING: PROBLEM SOLVING: Example # 2Example # 2
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 3Example # 3
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 4Example # 4
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
Determine the internal N, V, and M at point E of the frame shown below.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
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
PROBLEM SOLVING: PROBLEM SOLVING: Example # 5Example # 5
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
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.
PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
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
PROBLEM SOLVING: PROBLEM SOLVING: Example # 6Example # 6
{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