shear and moment diagrams (graphical) - assakkaf
TRANSCRIPT
1
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
Third EditionLECTURE
135.3
Chapter
BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL)
byDr. Ibrahim A. Assakkaf
SPRING 2003ENES 220 – Mechanics of Materials
Department of Civil and Environmental EngineeringUniversity of Maryland, College Park
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 1ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Example 8The 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.
A x
y
B
9 m
6 kN/m
2
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 2ENES 220 ©Assakkaf
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:
kN 18
02
969 ;0
kN 9
0319
296)9( ;0
=∴
=×
−+−↑+
=∴
=
×
×
−=+
∑
∑
B
By
A
AB
R
RF
R
RM
Shear Forces and Bending Moments in Beams
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 3ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Example 8 (cont’d)
A x
y
B
6 kN/m
RA = 9 kN Rb = 18 kN
A x
y
9 kN
M
Vx S
( )
90for 9
9
0323
29 ;0
90for 3
9
021
329 ;0
3
2
<<−=∴
=
−+−=+
<<−=∴
=
−+−=↑+
∑
∑
xxxM
xxxxMM
xxV
xxVF
S
y
6 kN/m
x9 m
xwxw
32
96
=⇒=
kN/m 32 x3
x
Figure 17
3
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 4ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Example 9A 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, determinea) The flexural stress at point A of the cross
sectionb) The flexural stress at point B of the cross
section.
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 5ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Example 9 (cont’d)
A x
y
6 ft 10 ft 5 ft
5600 ft-lb 300 lb/ft
600 lb
3500 ft-lb
900 lb 1500 lb
6 ′′
8 ′′
2 ′′
2 ′′
8 ′′
Figure 18aFigure 18b
·
·
A
B
4
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 6ENES 220 ©Assakkaf
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,
6 ′′
8 ′′
2 ′′
2 ′′
8 ′′
·5 ′′
( ) ( ) 433
in 7.558336
3582 =
−=xI
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 7ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Example 9 (cont’d)
A x
5600 ft-lb 300 lb/ft600 lb
3500 ft-lb
900 lb
1500 lb
A
5600 ft-lb
1 ft
M
V ft 1at lb-ft 55001005600
0)5.0)(200)(1(5600 ;0 60for lb 200
0)200(1 ;0
=−=+−=∴
=+−−=+
<<=∴
=+−=↑+
∑
∑
xMMM
xVVF
S
y
y
5
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 8ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Example 9 (cont’d)a) Stress at A
b) Stress at B
( )( )
(C)n Compressio psi 4.354
psi 4.3547.558
3125500
=
−=−×−
−=−=x
rA I
yMσ
( )( )
(T)Tension psi 591
psi 5917.558
5125500
=
+=×−
−=−=x
rB I
yMσ
6 ′′
8′′
2′′
2′′
8′′
·5′′
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 9ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships– In cases where a beam is subjected to
several concentrated forces, couples, and distributed loads, the equilibrium approach 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.
6
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 10ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
Pw1
Lb
a
x1x2
x3
O
w2
Figure 19
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 11ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships– The beam shown in the Figure 20 is
subjected to an arbitrary distributed loading w = w(x) and a series of concentrated forces and couple moments.
– We will consider the distributed load w to positive when the loading acts upward as shown in Figure 20.
7
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 12ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
wP
∆x
x
C
P w
2x∆
2x∆
C
x
MMM LR ∆+=
VVV LR ∆+=
LMLV
Figure 20a Figure 20b
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 13ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships– In reference to Fig. 20a at some location x,
the beam is acted upon by a distributed load w(x), a concentrated load force P, and a concentrated couple C.
– A free-body diagram segment of the beam centered at the location x is shown in Figure 20b.
8
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 14ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
The element must be in equilibrium, therefore,
From which
( )∑ =∆+−+∆+=↑+ 0 ;0 avg VVPxwVF LLy
xwPV ∆+=∆ avg(28)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 15ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
In Eq. 28, if the concentrated force P and distributed force w are both zero in some region of the beam, then
This implies that the shear force is constant in any segment of the beam where there are no loads.
RL VVV ==∆ or 0 (29)
9
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 16ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
If the concentrated load P is not zero, then in the limit as ∆x → 0,
That is, across any concentrated load P, the shear force graph (shear force versus x) jumps by the amount of the concentrated load.
PVVPV LR +==∆ or (30)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 17ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
Furthermore, moving from left to right along the beam, the shear force graph jumps in the direction of the concentrated load.If the concentrated load is zero, then in the limit as ∆x → 0, we have
0avg →∆=∆ xwV (31)
10
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 18ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
And the shear force is a continuous function at x. Dividing through by ∆x in Eq. 31, gives
That is, the slope of the shear force graph at any section x in the beam is equal to the intensity of loading at that section of the beam.
wdxdV
XV
x==
∆∆
→∆lim
0(32)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 19ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending 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.
11
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 20ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending 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
∫∫ ==−=∆2
1
2
1
12
x
x
V
V
dxwdVVVV
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 21ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load and Shear Force Relationships
Intensity LoaddDistribute
DiagramShear of Slope
)(
=
= xwdxdV
(33)
12
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 22ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load and Shear Force Relationships
21
1212
and between CurveLoadingunder Area
Shear in Change
)( 2
1
2
1
xx
dxxwdVVVVx
x
V
V
=
==−=∆ ∫∫−
(34)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 23ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
Similarly, applying moment equilibrium to the free-body diagram of Fig. 20b we obtain
From which
( ) ( ) ( )∑ =∆+∆
∆+−∆
−−∆+=+ 022
- ;0 avg xwaxVVxVMMMM LLLLc
( )xwaxVxVCM L ∆−∆
∆+∆+=∆ avg2(35)
13
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 24ENES 220 ©Assakkaf
Load, Shear Force, and Bending Moment Relationships
wP
∆x
x
C
P w
2x∆
2x∆
C
x
MMM LR ∆+=
VVV LR ∆+=
LMLV
Figure 20a Figure 20b
Shear Forces and Bending Moments in Beams
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 25ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending 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,
CMMCM LR +==∆ or (36)
14
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 26ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending 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) Slide No. 27ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships
Second, if the concentrated couple C and concentrated force P are both zero, then in the limit as ∆x → 0, we have
and dividing by ∆x, gives( ) 0
2 avg →∆−∆
∆+∆=∆ xwaxVxVM L (37)
Vdx
dMx
Mx
==∆∆
→∆lim
0(38)
15
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 28ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending 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
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 29ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Load, Shear Force, and Bending Moment Relationships– In the region of the beam I which Eq. 38 is
valid (any region in which there are no concentrated loads or couples), the equation can be integrated between definite limits to get
∫∫ ==−=∆2
1
2
1
12
x
x
M
M
dxVdMMMM (39)
16
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 30ENES 220 ©Assakkaf
Shear Force and Bending Moment Relationships
ShearDiagramMoment
of Slope
=
=Vdx
dM (40)
Shear Forces and Bending Moments in Beams
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 31ENES 220 ©Assakkaf
Shear Force and Bending Moment Relationships
21
1212
and between diagramShearunder Area
Moment in Change
2
1
2
1
xx
dxVdMMMMx
x
M
M
=
==−=∆ ∫∫−
(41)
Shear Forces and Bending Moments in Beams
17
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 32ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
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 the intensity of the distributed loading at the point.
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 33ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
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 the area under the distributed loading 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.
• Moment Diagram– Establish the M and x axes and plot the values of the
moment at the ends of the beam.
18
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 34ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
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) Slide No. 35ENES 220 ©Assakkaf
Shear Forces and Bending Moments in Beams
Shear and Moment Diagrams
Moment Diagram, Shear Diagram, Loading wdxdV
= Vdx
dM=
19
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 36ENES 220 ©Assakkaf
Shear and Moment DiagramsMoment Diagram, Shear Diagram, Loading w
dxdV
= Vdx
dM=
Shear Forces and Bending Moments in Beams
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 37ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 10Draw the shear and bending moment diagrams for the beam shown in Figure 21a.
40 lb/ft 600 lb
1000 lb · in
Figure 21a
12 ft
20 ft
A
20
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 38ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 10 (cont’d)– Support Reactions
• The reactions at the fixed support can be calculated as follows:
( )( ) ( )
inlb 880,15
01000206006)12(40 ;0
lb 108006001240 ;0
⋅=∴
=+++−=+
=→=−−=↑+
∑∑
MMM
RRF
A
AAy
40 lb/ft 600 lb
1000 lb · ftRA = 1080 lb
M = 15,880 lb · ft
Figure 21b
12 ft
20 ft
A
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 39ENES 220 ©Assakkaf
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
1080 lb
15,880 lb · ft x V
M40 lb/ft
( ) 120for 401080 <<−= xxxV
120for 880,152
4010802
<<−−= xxxM
A
21
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 40ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 10 (cont’d)
1080 lb
15,880 lb · in
40 lb/ft
x V
M
12 ft
( ) 2012for 600 <<= xxV
( )( ) 2012for 10806-1240880,15 <<+−−= xxxM
A
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 41ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 10 (cont’d)
Shear Digram
0
200
400
600
800
1000
0 5 10 15 20x (ft)
V (l
b)
( ) 120for 401080 <<−= xxxV
40 lb/ft 600 lb
1000 lb · in
12 ft20 ft
22
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 42ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 10 (cont’d)40 lb/ft 600 lb
1000 lb · in
12 ft20 ft
Bending Moment Diagram
-16000
-14000
-12000
-10000
-8000
-6000
-4000
-2000
00 5 10 15 20
x (ft)
M (f
t . lb
)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 43ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 10 (cont’d)40 lb/ft 600 lb
1000 lb · ftRA = 1080 lb
M = 15,880 lb · ft12 ft
20 ft
6001080
V (lb)
M (ft·lb)
(+)
(-) -1000
x
x
23
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 44ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 11Draw complete shear and bending moment diagrams for the beam shown in Fig. 22
AB C D
12 ft 4 ft 8 ft
8000 lb
2000 lb/ft
y
x Figure 22a
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 45ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 11 (cont’d)– The support reactions were computed from
equilibrium as shown in Fig. 22.b.
AB C D
12 ft 4 ft 8 ft
8000 lb
2000 lb/ft
y
x Figure 22a
RA = 11,000 lb RC = 21,000 lb
24
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 46ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 11 (cont’d)
AB C D
12 ft 4 ft 8 ft
8000 lb
2000 lb/ft
x11,000 lb
21,000 lb11,000
8,000 lb
13,000 lb
V (lb)5.5 ft
M (ft -lb)
30,250
12,00064,000
(+)
(-)
(+)(-)
(-)
LECTURE 13. BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) (5.3) Slide No. 47ENES 220 ©Assakkaf
Shear and Moment Diagrams
Example 11 (cont’d)
11,000
13,000
x
12
5.518.2
12 000,13
12000,11
==⇒−
= xxx