清华大学土木工程系结构力学(英)Chapter3_399306542

32
Graphical Construction (1)
Ya 168kN
Find Reactions
Yb 392kN
Find shear forces at key points
Va 168kN
Vb,left 232kN Vb,right 160kN
33
Graphical Construction (2)
8
Sign Convention (1)
Over the years, structural engineers have adopted almost universally a sign convention for internal forces in beams to facilitate graphical construction.
wL/2
wL/2
V ( x) wL / 2 wx
x wx( L x) M ( x) ( wL / 2) x ( wx) 2 2
15
Internal Force Diagram (1)
Positive shear drawn above the axis.
Moment drawn on the tension side.
F V S ( x ) Const
— Shear diagram consists of horizontal lines
q(x)
F M Linear S ( x ) Const
F V S ( x) 0
P
F V S ( x) 0
V
M
30
Interpretations (3)
② q( x) Const
13
Example (2) Determine V&M at
A 3m YA=2kN
1 1 5m
5kN 2 2
B
sections 1-1 and 2-2
Sol.: (1)Shear 1-1Sec.： 2-2Sec.： (2) M
YB =3kN
（Left） F RA V S 1 Y A 2kN 1
24
25
Differential Equations
q(x)
26
Integral Equations (1)
q(x) q(x)
27
Integral Equations (2)
28
Interpretations (1)
q(x)
q (x )
q(x)
29
Interpretations (2) ① q ( x) 0
21
Free-Body of Small Beam Piece (1)
q(x)
q(x) c
M(x) V(x)
M(x)+dM(x)
V(x)+dV(x)
dx
Y 0 :
F VS ( x ) dF V VS ( x ) [F S ( x )] q ( x )dx 0
dF V S ( x) q( x ) dx
M C ,left 5kNm M C ,right 2.5kNm
5
M (kNm)
38
Superposition
q=5kN/m C A 7.5kNm 1m B 2.5m
7.5 2.5
5
M (kNm)
A
7.5kNm 1m B
C
7.5
2.5m
M1 (kNm)
15 2.5
A
q=5kN/m C 1m B 2.5m
q(x)
F Linear V S ( x ) Const
— Shear diagram consists of inclined straight lines
F M Parabola S ( x ) Const
( q(x)q
q(x)+
)
V
M
31
Procedures for Graphical Construction (1)
12
Section Method for Internal Forces
q0a l Vc
Mc YA
1 qa Vc YA 0 a 2 l
1 q0 a a M c YA a a 2 l 3
Shear——Algebraic sum of vertical force components acting to the left/right of the considered section (Left-Upward; Right-Downward) Moment——Algebraic sum of couples about the considered section produced by loads acting to the left/right （concave upward-positive, convex upward-negative）
Chapter 3 Analysis of Statically Determinate Beam and Frame Structures
1
Beam Structures

Beams are usually long slender members which resist large lateral load relative to their owe weight.
q0a l Vc
Mc YA Mc a Vc l
q0
YB
Y 0
Vc Y A
1 q0 a a 0 2 l
2
q 0 l 3a Vc 6l
2

M
c
0
1 q0 a a M c YA a a 0 2 l 3
q 0 al 2 3a 2 Mc 6l
16
Internal Force Diagram (2)
17
Concentrated Loads (1)
For x a : Pa V ( x) L Pb M ( x) x P( x a) L
18
Concentrated Loads (2)
“Jump” at each concent. load
RC: 3~5 times larger Steel: 8~10 times larger

The cross-section dimensions must be << than the length for the beam theory idealization to apply h < L/10
V (kN)
37
Example (3)
Construct V&M diagrams for the structure Solution:
(1) Reactions:
q=5kN/m C A 7.5kNm 1m B 2.5m
5
+
VA 5kN
M A 7.5kNm
7.5
V (kN)
2.5
(2) Moments at key points:
M2 (kNm)
39
Principle of Superposition
2
Beam Model
Simply-supported
Continuous
3
Simply-Supported Beam
4
Continuous Beam
5
Cantilever Beam
6
Beam Type
7
Internal Forces in Beams


Internal forces must be known to design correctly the cross section. The beam is a “shear-bending” member,
q0 l 2 (Resultant) q0
A a YA c l B
YB
q0l 2l M A 0 : YB l 2 3 0 q0l l M B 0 : YA l 2 3 0
q0l YB 3 q0l YA 6
11
Example (1)
(2)FBD: Segment AC will be considered since it yields the simplest solution
Construct V&M diagrams for the structure Solution:
(1) Reactions:
q=5kN/m C A 7.5kNm 1m B 2.5m
VA 5kN
M A 7.5kNm
5 +
(2) Shear forces at key points:
VC 5kN
Va 168kN
Vb,left 232kN
Vb,right 160kN
Ya 168kN
168
Yb 392kN
160
V(kN)
4.2m 232
34
Graphical Construction (3)
d
Hale Waihona Puke Ya 168kNYb 392kN
Find bending moments at key points
Ma 0
M b 320kNm
M c 0 M d 352.8kNm
35
Graphical Construction (3)
Ma 0
d
M b 320kNm
Yb 392kN