# BEAMS: SHEAR AND MOMENT DIAGRAMS

## Transcription

BEAMS: SHEAR AND MOMENT DIAGRAMS
```LECTURE
Third Edition
BEAMS: SHEAR AND MOMENT
DIAGRAMS (GRAPHICAL)
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
13
Chapter
5.3
by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park
Slide No. 1
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Example 8
The beam is loaded and supported as
shown in the figure. Write equations for
the shear V and bending moment M for
any section of the beam in the interval AB.
6 kN/m
y
A
x
B
9m
1
Slide No. 2
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Example 8 (cont’d)
A free-body diagram for the beam is shown
Fig. 17. The reactions shown on the
diagram are determined from equilibrium
equations as follows:
 6 × 9  1 
= 0; RA (9) − 
 9 ×  = 0
 2  3 
∴ RA = 9 kN
+
∑M
B
+ ↑ ∑ Fy − 0; RB + 9 −
6×9
=0
2
∴ RB = 18 kN
Slide No. 3
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Example 8 (cont’d)
6 kN/m
Figure 17
y
w 6
2
= ⇒w= x
x 9
3
6 kN/m
x
A
x
B
RA = 9 kN
x
3
y
A
9 kN
x
x
S
Rb = 18 kN
2
x kN/m
3
M
V
9m
2  1
+ ↑ ∑ Fy = 0; − V + 9 −  x (x ) = 0
3  2
2
x
∴V = 9 −
for 0 < x < 9
3
2  x  x 
+ ∑ M S = 0; − M + 9 x − x   = 0
3  2  3 
3
x
∴ M = 9x −
for 0 < x < 9
9
2
Slide No. 4
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Example 9
A timber beam is loaded as shown in Fig.
18a. The beam has the cross section
shown in Fig.18.b. On a transverse cross
section 1 ft from the left end, determine
a) The flexural stress at point A of the cross
section
b) The flexural stress at point B of the cross
section.
Slide No. 5
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Example 9 (cont’d)
y
8′′
600 lb
·
B
300 lb/ft
5600 ft-lb
A
3500 ft-lb
2′′
x
6 ft
6′′
A
·
10 ft
900 lb
Figure 18a
5 ft
2′′
8′′
1500 lb
Figure 18b
3
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Example 9 (cont’d)
First, we have to determine the moment of
inertia Ix. From symmetry, the neutral axis
is located at a distance y = 5 in. either from
the bottom or the upper edge. Therefore,
8′′
·
5′′
Slide No. 6
2′′
 8(5)3 6(3)3 
4
−
I x = 2
 = 558.7 in
3 
 3
6′′
2′′
8′′
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 7
Example 9 (cont’d)
600 lb
y
300 lb/ft
5600 ft-lb
A
3500 ft-lb
x
1500 lb
1 ft
5600 ft-lb
A
900 lb
M
+ ↑ ∑ Fy = 0; − V + 1( 200) = 0
∴V = 200 lb
+
V
∑M
S
for 0 < x < 6
= 0; − M − 5600 + (1)(200)(0.5) = 0
∴ M = −5600 + 100 = −5500 ft - lb at x = 1 ft
4
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams
Example 9 (cont’d)

a) Stress at A
8′′
5′′
2′′
·
Slide No. 8
σA = −
6′′
2′′
8′′
(− 5500 ×12)(− 3) = −354.4 psi
Mr y
=−
Ix
558.7
= 354.4 psi Compression (C)
b) Stress at B
σB = −
(− 5500 ×12)(5) = +591 psi
Mr y
=−
558.7
Ix
= 591 psi Tension (T)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 9
Moment Relationships
– In cases where a beam is subjected to
several concentrated forces, couples, and
discussed previously can be tedious
because it would then require several cuts
and several free-body diagrams.
– In this section, a simpler method for
constructing shear and moment diagrams
are discussed.
5
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 10
Moment Relationships
L
b
a
w2
P
w1
O
x1
x2
x3
Figure 19
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 11
Moment Relationships
– The beam shown in the Figure 20 is
w = w(x) and a series of concentrated
forces and couple moments.
– We will consider the distributed load w to
shown in Figure 20.
6
Slide No. 12
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Moment Relationships
P
P
ML
∆x
w
w
M R = M L + ∆M
VL
C
C
x
x
Figure 20a
∆x
2
∆x
2
Figure 20b
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

VR = VL + ∆V
Slide No. 13
Moment Relationships
– In reference to Fig. 20a at some location x,
the beam is acted upon by a distributed
a concentrated couple C.
– A free-body diagram segment of the beam
centered at the location x is shown in
Figure 20b.
7
Slide No. 14
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Moment Relationships
The element must be in equilibrium,
therefore,
+ ↑ ∑ Fy = 0; VL + wavg ∆x + P − (VL + ∆V ) = 0
From which
∆V = P + wavg ∆x
(28)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 15
Moment Relationships
In Eq. 28, if the concentrated force P and
distributed force w are both zero in some
region of the beam, then
∆V = 0
or
VL = VR
(29)
This implies that the shear force is
constant in any segment of the beam
8
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 16
Moment Relationships
If the concentrated load P is not zero, then
in the limit as ∆x → 0,
(30)
∆V = P
or
VR = VL + P
That is, across any concentrated load P,
the shear force graph (shear force versus
x) jumps by the amount of the
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 17
Moment Relationships
Furthermore, moving from left to right
along the beam, the shear force graph
jumps in the direction of the concentrated
If the concentrated load is zero, then in the
limit as ∆x → 0, we have
(31)
∆V = w ∆x → 0
avg
9
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 18
Moment Relationships
And the shear force is a continuous
function at x. Dividing through by ∆x in Eq.
31, gives
∆V dV
=
=w
lim
(32)
dx
∆x →0 ∆X
That is, the slope of the shear force
graph at any section x in the beam is
section of the beam.
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 19
Moment Relationships
Moving from left to right along the beam, if
the distributed force is upward, the slope of
the shear force graph (dV/dx = w) is
positive and the shear force graph is
increasing (moving upward).
If the distributed force is zero, then the
slope of the shear graph (dV/dx = 0) and
the shear force is constant.
10
Slide No. 20
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Moment Relationships
In any region of the beam in which Eq. 32
is valid (any region in which there are no
concentrated loads), the equation can be
integrated between definite limits to obtain
V2
x2
V1
x1
∆V = V2 − V1 = ∫ dV = ∫ w dx
Slide No. 21
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

dV
= w( x)
(33)
dx
Slope of
Distributed
=
Shear Diagram
11
Slide No. 22
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

V2
x2
V1
x1
∆V2−1 = V2 − V1 = ∫ dV = ∫ w( x) dx
Change in
Shear
=
(34)
Curve between x1 and x2
Slide No. 23
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Moment Relationships
Similarly, applying moment equilibrium to
the free-body diagram of Fig. 20b we
obtain
+
∑M
c
= 0; - (M L + ∆M ) − M L − VL
∆x
∆x
− (VL + ∆V ) + a (wavg ∆x ) = 0
2
2
From which
∆x
∆M = C + VL ∆x + ∆V
− a wavg ∆x
2
(
)
(35)
12
Slide No. 24
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Moment Relationships
P
P
ML
∆x
w
w
M R = M L + ∆M
VL
C
C
x
x
Figure 20a
∆x
2
∆x
2
Figure 20b
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

VR = VL + ∆V
Slide No. 25
Moment Relationships
In which -∆x/2 < a < ∆x/2 and in the limit as
∆x → 0, a → 0 and wavg → w.
Three important relationships are clear from
Eq. 35. First, if the concentrated couple is
not zero, then in the limit as ∆x → 0,
∆M = C
or
M R = M L + C (36)
13
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 26
Moment Relationships
That is, across any concentrated couple C,
the bending moment graph (bending
moment versus x) jumps by the amount of
the concentrated couple.
Furthermore, moving from left to right
along the beam, the bending moment
graph jumps upward for a clockwise
concentrated couple and jumps downward
for a concentrated counterclockwise C.
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 27
Moment Relationships
Second, if the concentrated couple C and
concentrated force P are both zero, then in
the limit as ∆x → 0, we have
∆M = VL ∆x + ∆V
∆x
− a (wavg ∆x ) → 0
2
and dividing by ∆x, gives
lim
∆x →0
∆M dM
=
=V
∆x
dx
(37)
(38)
14
Slide No. 28
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Moment Relationships
That is, the slope of the bending
moment graph at any location x in the
beam is equal to the value of the shear
force at that section of the beam.
Moving from left to right along the beam, if
V is positive, then dM/dx = V is positive,
and the bending moment graph is
increasing
Slide No. 29
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Moment Relationships
– In the region of the beam I which Eq. 38 is
valid (any region in which there are no
equation can be integrated between
definite limits to get
∆M = M 2 − M 1 =
M2
x2
∫ dM = ∫ V dx
M1
(39)
x1
15
Slide No. 30
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Shear Force and Bending Moment
Relationships
dM
=V
dx
Slope of
Moment Diagram
(40)
= Shear
Slide No. 31
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Shear Force and Bending Moment
Relationships
∆M 2−1 = M 2 − M 1 =
Change in
Moment
=
M2
x2
M1
x1
∫ dM = ∫ V dx
Area under Shear
(41)
diagram between x1 and x2
16
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 32
Shear and Moment Diagrams
– Procedure for Analysis
• Support Reactions
– Determine the support reactions and resolve the
forces acting on the beam into components which
are perpendicular and parallel to the beam’s axis
• Shear Diagram
– Establish the V and x axes and plot the values of the
shear at two ends of the beam. Since dV/dx = w, the
slope of the shear diagram at any point is equal to
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 33
Shear and Moment Diagrams
– If a numerical value of the shear is to be determined
at the point, one can find this value either by using
the methods of establishing equations (formulas)for
each section under study or by using Eq. 34, which
states that the change in the shear force is equal to
Since w(x) is integrated to obtain V, if w(x) is a curve
of degree n, then V(x) will be a curve of degree n + 1.
For example, if w(x) is uniform, V(x) will be linear.
• Moment Diagram
– Establish the M and x axes and plot the values of the
moment at the ends of the beam.
17
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 34
Shear and Moment Diagrams
– Since dM/dx = V, the slope of the moment diagram at
any point is equal to the intensity of the shear at the
point. In particular, note that at the point where the
shear is zero, that is dM/dx = 0, and therefore this
may be a point of maximum or minimum moment. If
the numerical value of the moment is to be
determined at a point, one can find this value either
by using the method of establishing equations
(formulas) for each section under study or by using
Eq. 41, which states that the change in the moment
is equal to the area under the shear diagram. Since
w(x) is integrated to obtain V, if w(x) is a curve of
degree n, then V(x) will be a curve of degree n + 1.
For example, if w(x) is uniform, V(x) will be linear.
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Slide No. 35
Shear and Moment Diagrams
Shear Diagram,
dV
=w
dx
dM
Moment Diagram, dx
=V
18
Slide No. 36
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear Forces and Bending
Moments in Beams

Shear and Moment Diagrams
Shear Diagram,
dV
=w
dx
dM
Moment Diagram, dx
Slide No. 37
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

=V
Example 10
Draw the shear and bending moment
diagrams for the beam shown in Figure
21a.
600 lb
40 lb/ft
A
12 ft
20 ft
1000 lb · in
Figure 21a
19
Slide No. 38
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 10 (cont’d)
– Support Reactions
• The reactions at the fixed support can be
calculated as follows:
+ ↑ ∑ Fy = 0; RA − 40(12 ) − 600 = 0 → RA = 1080 lb
+ ∑ M A = 0; − M + 40(12)(6 ) + 600(20) + 1000 = 0
Figure 21b
∴ M = 15,880 lb ⋅ in
600 lb
40 lb/ft
A
12 ft
M = 15,880 lb · ft
1000 lb · ft
20 ft
RA = 1080 lb
Slide No. 39
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 10 (cont’d)
– Shear Diagram
• Using the established sign convention, the
shear at the ends of the beam is plotted first.
For example, when x = 0, V = 1080; and when
x = 20, V = 600
40 lb/ft
V ( x ) = 1080 − 40 x
for 0 < x < 12
M
A
15,880 lb · ft
x
1080 lb
2
V M = 1080 x − 40 x − 15,880 for 0 < x < 12
2
20
Slide No. 40
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams
Example 10 (cont’d)

40 lb/ft
A
15,880 lb · in
M
12 ft
x V
1080 lb
V ( x ) = 600
for 12 < x < 20
M = −15,880 − 40(12 )(x - 6 ) + 1080 x
for 12 < x < 20
Slide No. 41
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 10 (cont’d)
40 lb/ft
Shear Digram
12 ft
600 lb
20 ft
1000 lb · in
1000
V (lb)
800
600
400
V ( x ) = 1080 − 40 x
for 0 < x < 12
10
15
200
0
0
5
20
x (ft)
21
Slide No. 42
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 10 (cont’d)
40 lb/ft
600 lb
Bending Moment Diagram
12 ft
0
-2000 0
5
10
15
20 ft
1000 lb · in
20
M (ft . lb)
-4000
-6000
-8000
-10000
-12000
-14000
-16000
x (ft)
Slide No. 43
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 10 (cont’d)
600 lb
40 lb/ft
12 ft
M = 15,880 lb · ft
RA = 1080 lb
20 ft
1000 lb · ft
1080
V (lb)
(+)
600
(-)
-1000
M (ft·lb)
x
x
22
Slide No. 44
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 11
Draw complete shear and bending moment
diagrams for the beam shown in Fig. 22
8000 lb
y
2000 lb/ft
A
B
C
x
12 ft
4 ft
D
Figure 22a
8 ft
Slide No. 45
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 11 (cont’d)
– The support reactions were computed from
equilibrium as shown in Fig. 22.b.
y
8000 lb
2000 lb/ft
A
12 ft
RA = 11,000 lb
B
C
x
4 ft
D
Figure 22a
8 ft
RC = 21,000 lb
23
Slide No. 46
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Example 11 (cont’d)
2000 lb/ft
A
8000 lb
B
C
x
12 ft
11,000 lb
D
4 ft
8 ft
21,000 lb
11,000
V (lb)
(+)
8,000 lb
(+)
5.5 ft
(-)
13,000 lb
30,250
M (ft -lb)
(-)
(-)
12,000
64,000
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3)
Shear and Moment Diagrams

Slide No. 47
Example 11 (cont’d)
12
11,000
x
x
12 − x
12
=
⇒x=
= 5.5
11,000 13,000
2.18
13,000
24
```