Transcript 4-Allowable stresses

4-Chapter
Allowable stresses
contents
• Introduction
• 2.6.1(p8) Compression element, Axial or
bending
• Axial tension
• Allowable shear stress qall
• Axial compression
• Bending stress
• Allowable crippling stress in web qall
• Combined stresses
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Introduction

The actual stresses in any part of steel
bridge must not exceed the elastic limit of
the
material
otherwise
permanent
deformation would occur. All structural
calculations are approximate even if all
loads are carefully considered. In trusses
neglect the secondary stresses due to the
rigidity of joints.
The forces in members are determined under the
assumption that the connections are hinge and the
forces along the members are axial. Only the primary
stresses can be calculated. In some cases the
secondary stresses may reach 30 – 60 % of the
primary stresses. The analyses neglect also the
torsion in the main girders due to the deflections of
the X-girders. The unequal distribution of stresses
over the cross section due to bolts holes hasn’t taken
into consideration
•The allowable stresses (maximum stresses used in the
calculation) must therefore be lower than the elastic limit.
The more accurate calculations of steel bridge and the
better shop work, the higher allowable stresses may be
taken. Also, in the calculation if all possible forces are
taken into account the allowable stresses can be taken
higher than in case that only D.L., L.L, and Impact are
considered.
The permissible stresses for standard grade
structural steel determined according to the grade of
steel. Structural sections shall be classified,
depending on dw/tw for web and c/tf for flanges
under compression, axial bending, to compact,
noncompact, and slender sections as shown Fig(4-1)
Figure-4.1
Fy and Fu (t/cm2) depend on the thickness
t(1.4-p2).
t40mm
100mm t >400mm
Grade of
steel
Fy
Fu
Fy
Fu
ST37
2.40
3.60
2.15
3.40
ST44
2.80
4.40
2.55
4.10
ST52
3.60
5.20
3.35
4.90
2.2.1(p6) Primary + additional stresses (wind load
or
earthquake loads, lateral shock, etc.)
2.2.3(p7) Additional stresses
Additional stresses (allowable) = Primary
stress 1.20
2.3(p7) Secondary stresses in truss members
1. Chord member's depth > 1/10 of their length.
Diagonal member's depth > 1/15 of their length.
2. Truss with sub-panel.
Reduce 20 % of the allowable stress
back
2.6.1(p8) Compression element, Axial or
bending
 Compact
 Noncompact
 Selender





 dw
Cf

 t and t
f
 w
 Factor


Fy

Factor depends on:
1. Support of element ((One side (unstiffened
element) or two sides (stiffened element))
and shape of the cross section, I, C, , L, etc.
2. Load on element [(N) or (M) or (M+N)]
(p9,10,11-Table 2.1.a,b&c)
2.6.2(p13) Axial tension
Ft  0 . 58 F y 
t  40 mm  F y


40 mm  t  100 mm  F y 
 F rom clause 1.4 , get F y
Hence for,
 Ft  1 . 40 t / cm 2

2
t  40 mm   Ft  1 . 60 t / cm

2
F

2
.
10
t
/
cm
 t
 Ft  1 . 30 t / cm 2

2
40 mm  t  100 mm   Ft  1 . 50 t / cm

2
F

2
.
00
t
/
cm
 t
ST 37
ST 4 4
ST 5 2
ST 37
ST 4 4
ST 5 2
back
2.6.3(p13) Allowable shear stress
qall
q all  0 . 35 F y

t  40 mm  F y


40 mm  t  100 mm  F y 
 F rom clause 1.4, get Fy
Hence for,
q
all  0 . 84 t / cm


t  40 mm   q all  0 . 98 t / cm

q
 1 . 26 t / cm

 all
2
ST 37
2
ST 4 4
2
ST 5 2
2
q

0
.
75
t
/
cm
all


2
40 mm  t  100 mm   q all  0 . 89 t / cm

2
q

1
.
17
t
/
cm

 all
ST 37
ST 4 4
ST 5 2
2.6.3.1(p13) Effective web area
Rolled section = Total height tw
Built up section = Web height tw
2.6.3.2(p14) Shear buckling of web
dw
tw

105
 
Fy
d1
d
Stiffened web
d
 1 Kq  4
5 . 34

2
d1
  1  K q  5 . 34 
4

2
Unstiffened web
 =   Kq = 5.34
dw
If
Kq
 45
tw
,
Fy
 q  0 . 80
 no web buckling occur
dw
If,
 45
tw
 qp = 0.35 Fy
Kq
Fy
 Check web buckling
dw
q 
tw
57
Fy
  0 . 80   q
0 . 80    1 . 20 
q
b
q

q
(no web buckling occur)
Kq
 0 . 35 F y
 q b  1 . 50  0 . 625  q   0 . 35 F y
 1 . 20   q b 
0 . 90
q
 0 . 35 F y


back
2.6.4(p15) Axial compression
 
k  l
r
Fc  0 . 58 F y 
( 0 . 58 F y  0 . 75 )
10
4
t  40 mm
  t  40 mm
 
 get F y from 1.4
  t  40 mm
2
40 < t < 100 mm
Grade
of steel
Fc (t/cm2)
Fc (t/cm2)
2
Fc  1 . 30  0 . 000055 
2
2
Fc  1 . 50  0 . 000075 
2
ST37
Fc  1 . 40  0 . 000065 
ST44
Fc  1 . 60  0 . 000085 
ST52
Fc  2 . 1  0 . 000135 
2
Fc  2 . 0  0 . 000125 
2
For compact and Non-compact
sections use full area(Table2.1-p9-11).
•-
•For slender sections use effective
area(Tables 2.3&2.4-p23&24).
•For one angle reduce Fc by 40
%(p15).
back
2.6.5(p16) Bending stress
1For compact sections and the laterally
unsupported length (Lu )of the compression flange is
limited by:
(Lu is the smaller of)
-
Box sections
Lu 
84
Fy
 bf
Or
Lu

M1  bf

  137  84 
 F
M
2 
y

I-shape sections
Lu 
20 b f
Fy
Or
Lu 
1380 b f
d  Fy
 Cb
Cb
-
From Table2.2
Then
· Fb = 0.64 Fy
(Mx ) Boxand I-shapes
· Fb = 0.72 Fy
(My)
· Fb = 0.64 Fy
(My)
I-shapes
Box shapes
Clause 2.6.5.1
Clause 2.6.5.2
Clause 2.6.5.3
1- 1- For Non-compact sections:
· Fb = 0.58 Fy (Mx & My)
Box shapes
Clause 2.6.5.4
22- For slender (Box and I-shapes) and Non-compact (Ishapes) sections:
-
Tension
Clause 2.6.5.5
·
Fbt = 0.58 Fy
-
Compression
1·
2·
Clause 2.6.5.5
Lu  Lall
Fbc = 0.58 Fy
Lu > Lall
i – ( Shallow thick flanged section Luxtf / bfxd >10 (P18))
For any value of
·
F ltb 1 
·
800
Lu d / Af
 C b  0 . 58 F y
(eq 2.23)
ii - ( Deep thin flanged section Luxtf / bfxd <0.4 (P18))
Cb
84

Lu
Fy
rt
(eq2.24)
F ltb 2  0 . 58 F y
84
Cb

Fy
 188
rt
Cb
Fy
2

( L u / rT )  F y
  0 . 64 
5

1
.
176

10
Cb

F ltb 2
Lu
Lu
 188
Fltb 
(eq 2.25)
Cb
rt
F ltb 2 

  F  0 . 58 F
y
y


Fy
·
12000
( L u / rT )
2
2
 C b  0 . 58 F y
2
Fltb 1  Fltb 2  0 . 58 F y
(eq2.26)
(eq2.27)
II - For Channels( p21) Fltb;
F ltb 
800
Lu d / Af
 C b  0 . 58 F y ( M x )
(eq2.29)
III - For slender sections use effective width (be)
and the stress for non-compact(p21).
·
Effective width be for slender sections(Table
2.3& 2.4 – p23&24);
 
f2
f1
K 
· For
1   
16
2
 0 . 112 1   

2 0 .5
 1   
( Table 2.3 )
any value of  get K  from tables 2.3, and 2.4 for
stiffened and unstiffened elements respectively.
Calculate
 
Calculate
be =
 b
b /t
Fy
44
K

(plate slenderness)
     0 . 15  0 . 05 

2
   1 .0
Summary Table for Lateral Torsional Buckling
(Lu > Lall)
F ltb 1 
800
Lu d / Af
Cb
84

Fy
84
Cb

Fy
Lu
rt
For all
 188
 Cb
Lu
Fltb 2  0 . 58 F y
rt
Lu
 188
rt
Cb
Fy
Cb
Fy
F ltb  0 . 58 F y
F ltb 2
2

( L u / rT )  F y
  0 . 64 
5

1
.
176

10
Cb

F ltb 2 
Fltb 
12000
( L u / rT )
2
2

F
y


 Cb
2
Fltb 1  Fltb 2
back
2.6.6 (p22)Allowable crippling stress in web qall
Fcrp  0 . 75 Fy

t  40 mm
t  40 mm
 F rom clause 1.4, get Fy
In tension members we get smaller cross
sections by using high tensile stresses St. 52.
While in compression members we get smaller
section if l/i is less than 100 but if l/i is more than
100 we get same section for all kinds of steel.
2.6.7 Combined stresses
In a continuous beam we have a state of
combined shear and bending
back
F1 , 2 
f

2
 f 
2

q
 f pt  f pc
 
 2 
This stress may be greater than the bending stress
in the outside fibers.
The modern theory of equivalent structure is given
by;
Fe 
 f 2
 3q
2
 1 . 10  f pt
2.6.7.1- Axial Compression And Bending
f ca
f bcx

Fc
F bcx
1 
f ca
When
f bcy
F bcy
 2  1 .0

 0 . 15
 1   2  1.0
Fc
1 
C mx

f ca
1 

F EX

F EX 
7500

2
Cmx , Cmy




,
2 
, FEY 
7500

from code
2
C my

f ca
1 

F EY





2.6.7.2- Axial Tension And Bending
f ta
Ft

f btx
Fbtx

f bty
Fbty
 1 .0