工程力学英文版课件11 Shear Forces and Bending Moments in Beams
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• The shear V is obtained by summing forces perpendicular to the beam’s axis. • The moment M is obtained by summing moments about the sectioned end of the segment.
P P/2 P/2
18
Shear and Moment Functions. The beam is sectioned at an arbitrary distance x from the support A, extending within region AB, and the free-body diagram of the left segment is shown in figure. The unknowns V and M are indicated acting in the positive sense on the right-hand face of the segment according to the established sign convention.
13
Fig. (c)
qL a x2 Fig. (c)
Y qL Q2 q( x2 a ) 0
B
M2
Q2
Q2 q( x2 a L)
mB (Fi ) 0 , 1 qLx2 M 2 q( x2 a)2 0 2
1 2 M 2 q( x2 a) qLx2 2
4
Often beams are classified as to how they are supported. For example, a simply supported beam is pinned at one end and roller-supported at the other, whereas a cantilevered beam is fixed at one end and free at the other. Beams with overhanging end (ends) are overhanging beam. The actual design of a beam requires a detailed knowledge of the variation of the internal shear force V and bending moment M acting at each point along the axis of the beam.
14
§13-2 Shear-Force and Bending-Moment Diagrams
PROCEDURE FOR ANALYSIS The shear and bending-moment diagrams for a beam can be constructed using the following procedure. Support Reactions • Determine all the reactive forces and couple moments acting on the beam and resolve all the forces into components acting perpendicular and parallel to the beam’s axis. 15
[Example 3] Draw the shear and moment diagrams for the beam shown in figure.
L/2
A P L/2 C
B
Solution: Support Reactions. The support reactions have been determined.
V M
V
M
8
[Example 1] Determine the resultant internal loadings acting on the cross section at C of the beam shown in figure.
270 N/m
A 3m
C 6m
B
9
Solution: The beam is sectioned at point C, and the free-body diagram of the right segment is shown in figure. The unknowns V and M are indicated acting in the positive sense on the left-hand face of the segment according to the established sign convention.
11
[Example 2] Determine the resultant internal loadings acting on 1—1, 2—2 section as shown in figure (a).
qL 1 2 q
1 a qL
A x1
2
b Fig. (a)
M1
Fig. (b)
Solution：Determine internal forces by the method of section. Free body diagram of the left portion of 1—1 section is shown in figure (b).
The variations of V and M as functions of the position x along the beams axis can be obtained by using the method of sections.
6
Sign Convention Before presenting a method for determining the shear and bending moment as functions of x and later plotting these functions (shear and bendingmoment diagrams), it is first necessary to establish a sign convention so as to define a ―positive‖ and ―negative‖ shear force and bending moment acting in the beam. [This is analogous to assigning coordinate directions x positive to the right and y positive upward when plotting a function y = f (x).]
5
Before an analysis of the internal forces can be made, the beam reactions must be calculated. After this force and bending-moment analysis is complete, one can then use the theory of mechanics of materials and an appropriate engineering design code to determine the beam’s required cross-sectional area.
2
§13-1 Shear Forces and Bending Moments in Beams
Introduction Beam is a member acted on by loads that produce bending.
P1 q P2
M
The plane of symmetry
3
Beams are structural member which are designed to support loadings applied perpendicular to their axes. In general, beams are long, straight bars having a constant cross-sectional area. The internal reactions in the beam are shear forces and bending moments and lesser extent axial forces.
7
Although the choice of a sign convention is arbitrary, here we will choose the one used for the majority of engineering applications. Here the positive directions are denoted by an internal shear force that causes clockwise rotation of the member on which it acts, and by an internal moment that causes compression on the upper part of the member.
1
Shear Forces and Bending Moments in Beams
§13–1 Shear Forces and Bending Moments in Beams §13–2 Shear-Force and Bending-Moment Diagrams §13–3 Relations among Loads, Shear Forces, and Bending Moments
12
Q1
qL A x1 Q1 Fig. (b) qL a x2 B M2 Q2 M1
Y qL Q
Q1 qL
1
0
m
A
( Fi ) qLx1 M 1 0
M 1 qLx1
Free body diagram of the left portion of 2—2 section is shown in figure (c).
180 N/m
M V
C 6m
B
10
180 N/m
M V
C 6m
B
Applying the equilibrium equations yields
1 Fy 0 V 2 180 6 0 V 540N 1 M C 0 M 2 180 6 2 0 M 1080 N m
Shear and Moment Diagrams • Plot the shear diagram and the moment diagram. If computed values of the functions describing V and M are positive, the values are plotted above the x axis, whereas negative values are plotted below the x axis. 17
Shear and Moment Functions • Specify separate coordinates x having an origin at the beam’s left end and extending to regions of the beam between concentrated forces and/or couple moments, or where there is no discontinuity of distributed loading. • Section the beam perpendicular to its axis at each distance x and draw the free-body diagram of one of the segments. Be sure V and M are shown acting in their positive sense, in accordance with the sign convention given in previous section.
• The shear V is obtained by summing forces perpendicular to the beam’s axis. • The moment M is obtained by summing moments about the sectioned end of the segment.
P P/2 P/2
18
Shear and Moment Functions. The beam is sectioned at an arbitrary distance x from the support A, extending within region AB, and the free-body diagram of the left segment is shown in figure. The unknowns V and M are indicated acting in the positive sense on the right-hand face of the segment according to the established sign convention.
13
Fig. (c)
qL a x2 Fig. (c)
Y qL Q2 q( x2 a ) 0
B
M2
Q2
Q2 q( x2 a L)
mB (Fi ) 0 , 1 qLx2 M 2 q( x2 a)2 0 2
1 2 M 2 q( x2 a) qLx2 2
4
Often beams are classified as to how they are supported. For example, a simply supported beam is pinned at one end and roller-supported at the other, whereas a cantilevered beam is fixed at one end and free at the other. Beams with overhanging end (ends) are overhanging beam. The actual design of a beam requires a detailed knowledge of the variation of the internal shear force V and bending moment M acting at each point along the axis of the beam.
14
§13-2 Shear-Force and Bending-Moment Diagrams
PROCEDURE FOR ANALYSIS The shear and bending-moment diagrams for a beam can be constructed using the following procedure. Support Reactions • Determine all the reactive forces and couple moments acting on the beam and resolve all the forces into components acting perpendicular and parallel to the beam’s axis. 15
[Example 3] Draw the shear and moment diagrams for the beam shown in figure.
L/2
A P L/2 C
B
Solution: Support Reactions. The support reactions have been determined.
V M
V
M
8
[Example 1] Determine the resultant internal loadings acting on the cross section at C of the beam shown in figure.
270 N/m
A 3m
C 6m
B
9
Solution: The beam is sectioned at point C, and the free-body diagram of the right segment is shown in figure. The unknowns V and M are indicated acting in the positive sense on the left-hand face of the segment according to the established sign convention.
11
[Example 2] Determine the resultant internal loadings acting on 1—1, 2—2 section as shown in figure (a).
qL 1 2 q
1 a qL
A x1
2
b Fig. (a)
M1
Fig. (b)
Solution：Determine internal forces by the method of section. Free body diagram of the left portion of 1—1 section is shown in figure (b).
The variations of V and M as functions of the position x along the beams axis can be obtained by using the method of sections.
6
Sign Convention Before presenting a method for determining the shear and bending moment as functions of x and later plotting these functions (shear and bendingmoment diagrams), it is first necessary to establish a sign convention so as to define a ―positive‖ and ―negative‖ shear force and bending moment acting in the beam. [This is analogous to assigning coordinate directions x positive to the right and y positive upward when plotting a function y = f (x).]
5
Before an analysis of the internal forces can be made, the beam reactions must be calculated. After this force and bending-moment analysis is complete, one can then use the theory of mechanics of materials and an appropriate engineering design code to determine the beam’s required cross-sectional area.
2
§13-1 Shear Forces and Bending Moments in Beams
Introduction Beam is a member acted on by loads that produce bending.
P1 q P2
M
The plane of symmetry
3
Beams are structural member which are designed to support loadings applied perpendicular to their axes. In general, beams are long, straight bars having a constant cross-sectional area. The internal reactions in the beam are shear forces and bending moments and lesser extent axial forces.
7
Although the choice of a sign convention is arbitrary, here we will choose the one used for the majority of engineering applications. Here the positive directions are denoted by an internal shear force that causes clockwise rotation of the member on which it acts, and by an internal moment that causes compression on the upper part of the member.
1
Shear Forces and Bending Moments in Beams
§13–1 Shear Forces and Bending Moments in Beams §13–2 Shear-Force and Bending-Moment Diagrams §13–3 Relations among Loads, Shear Forces, and Bending Moments
12
Q1
qL A x1 Q1 Fig. (b) qL a x2 B M2 Q2 M1
Y qL Q
Q1 qL
1
0
m
A
( Fi ) qLx1 M 1 0
M 1 qLx1
Free body diagram of the left portion of 2—2 section is shown in figure (c).
180 N/m
M V
C 6m
B
10
180 N/m
M V
C 6m
B
Applying the equilibrium equations yields
1 Fy 0 V 2 180 6 0 V 540N 1 M C 0 M 2 180 6 2 0 M 1080 N m
Shear and Moment Diagrams • Plot the shear diagram and the moment diagram. If computed values of the functions describing V and M are positive, the values are plotted above the x axis, whereas negative values are plotted below the x axis. 17
Shear and Moment Functions • Specify separate coordinates x having an origin at the beam’s left end and extending to regions of the beam between concentrated forces and/or couple moments, or where there is no discontinuity of distributed loading. • Section the beam perpendicular to its axis at each distance x and draw the free-body diagram of one of the segments. Be sure V and M are shown acting in their positive sense, in accordance with the sign convention given in previous section.