Subject
Year
: S0695 / STATICS
: 2008
Internal Loading Developed in Structural Members
Session 13 - 18
What are the Internal Loads used for ?
To determine the forces and moment that act
within it.
Stresses are determined by the bending moment
and
shear stresses are determined by the maximum shear force
and the maximum torsional moments
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5.1. Internal loading at specified Point
Cross
section Method
Æ can be determined by using
Internal force act along the
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axis
5.1. Internal loading at specified Point
Internal Loading for Coplanar structure will consist :
• Normal/Axial Force ( N )
• Shear Force ( V / D )
• Bending Moment ( M )
These loads actually represent the RESULTANTS of
the STRESS DISTRIBUTION acting over the member of
cross section
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5.1. Internal loading at specified Point
Normal /Axial ForceÆ
x-section
•
An internal force at a given
•
It acts
•
It is a longitudinal component of a reactant internal force
along the longitudinal axis of a member
vector
•
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A reactant internal force vector is a vector equal in magnitude,
but oppositely directed, to the resultant force vector of all external
loads applied on the left or right side of a given x-section
5.1. Internal loading at specified Point
Sign Convention Æ
Normal / Axial Force Æ
N +
N +
Longitudinal axis
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5.1. Internal loading at specified Point
Shear ForceÆ
x-section
•
An internal force at a given
•
It acts
•
It is a
•
A reactant internal force vector is a vector equal in magnitude, but
oppositely directed, to the resultant force vector of all external loads applied on
the left or right side of a given x-section
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normal to longitudinal axis
of a member
transverse component of a reactant internal force vector
5.1. Internal loading at specified Point
Sign Convention Æ
Shear ForceÆ
D+
DLongitudinal axis
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5.1. Internal loading at specified Point
Bending MomentÆ
force couple at a given x-section
•
An internal
•
It is a normal
•
component of a reactant internal couple vector
Its vector is transverse to the longitudinal axis of a member and is
located in a horizontal plane for members loaded in the vertical plane of
symmetry
•
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A reactant internal couple vector is a vector equal in magnitude, but
oppositely directed, to the resultant couple vector of all external loads applied
on the left or right side of a given x-section
5.1. Internal loading at specified Point
Sign Convention Æ
Bending MomentÆ
M+
M+
Longitudinal axis
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5.1. Internal loading at specified Point
Sign Convention Æ
Bending MomentÆ
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5.1. Internal loading at specified Point
Torsional MomentÆ
couple at a given x-section
•
An internal force
•
Its vector is located on the longitudinal axis of a member
•
It is a longitudinal component of a reactant internal couple vector
•
A reactant internal couple vector is a vector equal in magnitude,
but oppositely directed, to the resultant couple vector of all
external loads applied on the left or right side of a given x-section
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5.1. Internal loading at specified Point
Sign Convention Æ
Torsional MomentÆ
M+
M+
Longitudinal axis
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5.2. Shear & Moment functions
the internal shear and moment function will be
discontinuous , at points where the
magnitude of the distributed load changes or
where concentrated forces or couple moments
are applied.
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5.2. Shear & Moment functions
For that reason…
shear and moment function must be
determined by section of structure
located any two discontinuities of loading
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5.2. Shear & Moment functions
Analysis Procedure :
Support reactions Æ
Determine the support reaction of the
structure, and resolve all the external forces
into component acting perpendicular and parallel
axis
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5.2. Shear & Moment functions
Analysis Procedure :
Shear and Moment Functions Æ
• Specify separate coordinate and associates
origin, extending into region between
concentrated loads, or couple moments or
where there is a discontinuity of loading
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5.2. Shear & Moment functions
Analysis Procedure :
Shear and Moment Functions Æ
2. Section the part of structure to its axis at each
distance “x “ and from the free-body diagram
determine the unknown D and M at the cutting
section.
3. On the free-body diagram , should be shown their
possitive direction ( see sign convention )
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5.2. Shear & Moment functions
Analysis Procedure :
Shear and Moment Functions Æ
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5.2. Shear & Moment functions
Analysis Procedure :
Shear and Moment Functions Æ
4. D is obtained from ΣFY = 0
5. M is obtained by summing moments about the point of
cutting location ΣMi = 0
6. The result can be checked by noting that :
dM/dx = D
dD/dx = w ( x )
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5.3. Shear & Moment Diagrams
The variation of D and M as function of “x” Æ
plotted in diagram Æ
Shear diagram
and
moment diagram
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5.3. Shear & Moment Diagrams
dM/dx = D
Æ Slope of bending moment
diagram at a given point is equal to the
shear force at this point
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5.3. Shear & Moment Diagrams
dD/dx = - w(x)
Æ Slope of shear
force
diagram at a given point is equal
minus distributed loading
intensity at this point
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5.3. Shear & Moment Diagrams
Analysis Procedure :
Shear diagrams Æ
• Establish the D and “x” axis and plot the values of Shear
for each section.
• Since dD/dx = - w , the slope of shear diagram at any point
is equal to the negative, that w is positive when acts
downward.
• Since w(x) is integrated to obtain D, if w(x) is a curve of
degree n , then D(x) is degree n+1 ( eg If w(x) is uniform ,
D(x) will be linear)
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5.3. Shear & Moment Diagrams
Analysis Procedure :
Moment diagrams Æ
• Establish the M and “x” axis and plot the values of
Moment for each section.
• Since dM/dx = D , the slope of Moment diagram at any
point is equal to the intensity of the shear at the point.
• Since D(x) is integrated to obtain M, if D(x) is a curve of
degree n , then M(x) is degree n+1 ( eg If D(x) is linear ,
M(x) will be parabolic.
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Problems (1)
Determine and draw the Shear Force and
Moment Diagrams for the structure below :
q =1 t/m
A
B
4m
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4m
Problems (1)
Step 1 :
q =1 t/m
Determine all
reaction forces
A
q =1t/m
ΣMA=0
w.2m - VB.8m = 0
4t.2m - VB.8m = 0
8 tm – 8 VB = 0
HA
VA = 3 t
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A
B
w = 1t/m . 4m = 4 t
4m
VB = 1 t
ΣMB=0
- w.6m + VA.8m = 0
-4t.6m +VA.8m = 0
24 tm – 8 VA = 0
B
4m
VA
ΣH=0
HA = 0
HA = 0
ΣV=0
VA + VB = w
3t+1t=4t
4t=4t
VB
OK !!!
Problems (1)
Step 2 :
q =1t/m
Section the part
of structure to its
axis at each
distance “x “ and
make the freebody diagram
HA
A
B
w = 1t/m . 4m = 4 t
4m
VA
q =1t/m
HA
4m
VB
B
A
w = 1t/m . “4m = 4 t
VB
VA
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Problems (1)
Step 2 :
q =1t/m
Section the part
of structure to its
axis at each
distance “x “ and
make the freebody diagram
HA
A
B
w = 1t/m . 4m = 4 t
4m
VA
VB
x2
x1
q =1t/m
HA
4m
B
A
w = 1t/m . “4m = 4 t
VB
VA
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Problems (1)
Step 3 :
x1
On the free-body
diagram , should be
shown their
Mx1
q =1t/m
HA
Nx1
A
Dx1
possitive
direction ( see
sign convention )
w = 1t/m . “4m = 4 t
VA
Dx2
x2
Nx2
Mx2
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B
VB
Problems (1)
Step 4 .a:
N is obtained from ΣFx = 0
D is obtained from ΣFY = 0
M is obtained by summing
moments about the point of
cutting location ΣMi = 0
ΣFx=0
HA + NX1 = 0
0 + NX1 = 0
If….
x1 = 0
X1 = 4
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Dx1 = 3 t
Dx1 = -1 t
Mx1
q =1t/m
HA
Nx1
A
Dx1
w = 1t/m . “4m = 4 t
VA
Nx1 = 0
ΣFy=0
VA – w – Dx1 = 0
3 t – ( 1t/m.x ) – Dx1 = 0
Dx1 = 3 – x
Dx1 = 3 – x
x1
ΣMx=0
w.( ½ .x) –VA.x + Mx1 = 0
( 1t/m.x ) .( ½ .x) – 3.x + Mx1 = 0
Mx1 = 3x – ½ x2
Mx1 = 3x – ½ x2
If….
x1 = 0
X1 = 1
x1 = 2
x1 = 3
x1 = 4
Mx1 = 0
Mx1 = 2 ½ tm
Mx1 = 4 tm
Mx1 = 4 ½ tm
Mx1 = 4 tm
Check !!!
dDx2/dx = w(x)
1=1
Check !!!
dMx2/dx = D(x2)
3 -x=3- x
OK !!!
Problems (1)
Step 4 .b:
Dx2
N is obtained from ΣFx = 0
D is obtained from ΣFY = 0
M is obtained by summing
moments about the point of
0 < x2 < 4
x2
Nx2
B
Mx2
cutting location ΣMi = 0
ΣFx=0
NX2 = 0
Nx2 = 0
VB
ΣFy=0
VB + Dx2 = 0
1 t + Dx2 = 0
Dx2 = 1
Dx2 = 1
ΣMx=0
- VB.x + Mx2 = 0
- 1t.x + Mx2 = 0
Mx2 = x
Mx1 = x
If….
x2 = 0
X2 = 4
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Check !!!
dDx2/dx = w(x)
0 =0
Check !!!
dMx2/dx = D(x2)
1 =1
Mx2 = 0
Mx2 = 4 tm
OK !!!
Step 5 a. :
Problems (1)
q =1t/m
Plot to diagrams
A
B
Normal Force
4m
0 < x1 < 4
4m
Nx1 = 0
0 < x2 < 4
Nx2 = 0
N
0
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0
Step 5 b. :
Problems (1)
q =1t/m
Plot to diagrams
A
B
Shear Force
4m
4m
0 < x1 < 4
Dx1 = 3 – x
3
If….
x1 = 0
X1 = 4
Dx1 = 3 t
Dx1 = -1 t
0 < x2 < 4
D
+
0
-1
Dx2 = 1
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+
1
0
Step 5 c. :
Problems (1)
Plot to diagrams
q =1t/m
A
Moment
0 < x1 < 4
B
If….
x1 = 0
X1 = 1
x1 = 2
x1 = 3
x1 = 4
Mx1 = 0
Mx1 = 2 ½ tm
Mx1 = 4 tm
Mx1 = 4 ½ tm
Mx1 = 4 tm
M
0
0
0 < x2 < 4
Mx1 = x
If….
x2 = 0
X2 = 4
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Mx2 = 0
Mx2 = 4 tm
4m
4m
Mx1 = 3x – ½ x2
+
2½
4
4
4½
q =1t/m
A
B
4m
4m
N
0
03
Solution
D
M
+
+
-
0
-1
+
4
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4
4½
0
0
0
2½
1
Problems (2)
Determine and draw the Shear Force and
Moment Diagrams for the structure below :
P =2 t
4m
A
B
4m
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4m
Problems (2)
P =2 t
Step 1 a :
Idealized
Structured
4m
P =2 t
A
B
4m
4m
A
B
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4m
4m
4m
Problems (2) P =2 t
Step 1 b:
Determine all
reaction forces
4m
ΣMA=0
P.4m - VB.8m = 0
2t.4m - VB.8m = 0
8 tm – 8 VB = 0
HA
A
B
4m
VA
VB = 1 t
ΣMB=0
- P.4m + VA.8m = 0
-2t.4m +VA.8m = 0
8 tm – 8 VA = 0
VA = 1 t
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4m
ΣH=0
HA = 0
HA = 0
ΣV=0
VA + VB = P
1t+1t=2t
2t=2t
VB
OK !!!
Problems (2)
Step 2a :
Section the part
of structure to its
axis at each
distance “x “ and
make the freebody diagram
P =2 t
4m
A
B
4m
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4m
Problems (2)
Step 2 b:
Section the part
of structure to its
axis at each
distance “x “ and
make the freebody diagram
0 < x1 < 4
HA
A
x2
x1
B
VA
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0 < x2 < 4
VB
Problems (2)
Step 2 c:
0 < x3 < 4
Section the part
of structure to its
axis at each
distance “x “ and
make the freebody diagram
x3
HA
A
B
VA
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x3
VB
Problems (2)
Step 2 c:
0 < x4 < 4
Section the part
of structure to its
axis at each
distance “x “ and
make the freebody diagram
HA
P =2 t
x4
x4
P =2 t
A
B
VA
VB
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Problems (2)
Step 3 a:
0 < x1 < 4
Nx1
On the free-body
diagram , should be
shown their
Dx1
Mx1
possitive
direction ( see
sign convention )
HA
A
x1
VA
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Problems (2)
Step 3 b:
Mx2
Dx2
possitive
direction ( see
sign convention )
0 < x2 < 4
Nx2
On the free-body
diagram , should be
shown their
x2
B
VB
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Problems (2)
Step 3 c:
0 < x3 < 4
On the free-body
diagram , should be
shown their
x3
possitive
direction ( see
sign convention )
HA
Nx3
Dx3
A
VA
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Mx3
Problems (2)
Step 3 d:
On the free-body
diagram , should be
shown their
possitive
direction ( see
sign convention )
0 < x3 < 4
Dx3
x3
Nx3
Mx3
B
VB
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Problems (2)
Step 3 e:
0 < x4 < 4
On the free-body
diagram , should be
shown their
P =2 t
Dx4
A
VA
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Mx4
Nx4
possitive
direction ( see
sign convention )
HA
x4
Problems (2)
Step 3 f
0 < x4 < 4
On the free-body
diagram , should be
shown their
possitive
direction ( see
sign convention )
Dx3
x4
P =2 t
Nx3
Mx3
B
VB
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Problems (2)
……..To continue the solving problem , you may
try as your exercise & after that we will discuss
it in the class
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