Chapter 4 shear strength of inclined section in flexure

symmetrical concentrated loads (CD segment called the 2 pure bending section; AC cut
curved section and DB section above)
τ2
cp
σ
2
>45o
σ tp σ cp

A'
3
C'
D'
B'
b) B' D A Vc Dc C B
Sa C B
VA
Ts
vd
c
a
Can solve?
MA
Beams without web reinforcement after diagonal crack appeared isolated body diagram
MB
z
A'
4.1.2 Beams without web reinforcement mechanism of shear: arch model
B'
A Vc
Dc
by the total cross section of shear resistance. Diagonal crack appears, by some section of the shear VA AA / resistance, shear pressure area is reduced ∴ AA / shear stress and compressive stress increases
σ tp
Point 3 : tensile stress makes tp increase and cp decrease, α < 45
。
τ3 σ
3
σ cp
3
<45o
σ tp
Oblique section failure reasons:
Since under the action of bending moment and shear tok parts: Ⅰ vault arch cross section, the
cross-section minimum, maximum stress; twoway compressive stress and shear stress together, concrete cutting pressure failure occurs. At this point, there are other parts of Ⅰ
VA
Sa C B Ts
vd
c
a
MB
图4-2 斜裂缝出现后的应力状态 Inclined cracks appear before the bar of any cross-section of longitudinal section of
tensile stress at the moment by the MB decision. Diagonal crack appears, the tensile stress in longitudinal reinforcement cross-section inclined cracks the top section from the bending moment at the MA decision. ∴ longitudinal reinforcement tensile stress increased significantly
––– said inclined section failure.
Shear curve section of the beam after diagonal cracks occur, the stress
state cross-section a qualitative change, or that the stress redistribution
Bending moment M and shear force under the action of Q:
Calculation of flexural strength - the main bar
As , As
Bending moment M and shear force under the action of Q:
respectively, in the produce section of the normal stress and shear stress, causing
the principal tensile stress and compressive stress, when the principal tensile stress tp> ft, that is generated oblique crack, the failure surface And the beam axis bias.
Point 2 : compressive stress makes tp decrease andcp increase, α > 45。
σ
τ2
2
σ cp
2
>45o
σ tp
Point 1 : neutral axis, 0, max, 45。
τ1
1
σ cp
1
45o
Oblique section strength calculation - web reinforcement
Asb , Asv
Web reinforcement: stirrups and bent (oblique) reinforced
Beams without web reinforcement: only the reinforced longitudinal beams without web reinforcement design
MA
z
Inclined cracks appear before the VA
A'
' the stress Taken along the diagonal crack isolation body on the inclined section A of
state. It can be seen, the resistance of inclined section are: (1)Shear pressure surface (AA) on the shear stress Dc and Vc; (2) vertical steel tension Ts; (3) friction, aggregate bite force Sa; (4) bolt longitudinal reinforcement of the sales force Vd. By the equilibrium conditions:
arch bearing potential.
Arch Ⅱ, Ⅲ: to pass the shear is very small, not fully functional.
Ⅲ
4.1.2 Beams without Web Reinforcement oblique section failure modes:
Beams state seen as a set of lever arch: the critical diagonal crack at the top of the basic arch and a group of small arch Ⅰ, Ⅱ, Ⅲ composition, the equivalent of longitudinal reinforcement rods. Ⅱ withstand the internal forces of small arches off bolt, the role of arch occlusal bearing Ⅰ Arch Ⅰ bear most of the shear


Oblique section of bending members the main factor for shear strength.
Flexural oblique shear strength formula for conditions. Such a high degree of web reinforcement beams of the preliminary design steps. Resisting moment mapping. Check all of bearing capacity and structural requirements. Construction oblique shear strength calculations.

In
My0 I0

VS 0 bI0

,

of working together, produce the principal tensile
stress and compressive stress
tp

2

2
4

2
P A C
P B D a
cp
Direction

2

2
4
2
a
1 2 arctg 2
P P
Pa
M (Ignore gravity)
V
F a A B
2 1 3
F a C B
σ τ
A'
C'
D'
B'
The picture shows a simply supported beam without web reinforcement, the role of two σ