Bending - fast10.vsb.cz
Transcript of Bending - fast10.vsb.cz
Bending
β’ Calculation of normal bending stress.β’ Neutral axis, cross-sectional modulus.β’ Bending of beams of asymmetrical cross-section.β’ Design and assessment of beams with elastic behavior
of the material according to the ultimate limit state.
1. Bending stress = normal stress from My
Load qz , Fz , My β bending moment My β normal stress
Οπππ₯ =ππ¦
πΌπ¦β π =
ππ¦
ππ¦
Stress course along the height of the cross section :
linear, changing with increasing distance from the
member axis, maximum in the extreme fibers
1. Stress β normal stress Οπ₯
πΌπ¦ [π4]
Effect of bending moment πy
Οπ₯ =ππ¦
πΌπ¦β π§
- moment of inertia
(The shear force Vz causes shear stresses in the cross section β
following topic)
ππ¦ =πΌπ¦
π
- section modulus Wy
derived from Iy for edge
fibers
ππ¦ [π3]
+
M
e
e
neutrΓ‘lnΓ‘ osa
Deformation: deflection w [m] and angle of rotation Ο [rad]
Calculation:
2. Deformation
πΈ. πΌπ¦ . π = βΰΆ±ππ¦. dπ₯ + πΆ1
πΈ. πΌπ¦. π€ = βΰΆ± ΰΆ±ππ¦. dπ₯ . dπ₯ + πΆ1. π₯ + πΆ2
In following topic
-
+
Ο = 0
Tensile fibers
Pressed fibersMy My
x
z
y
1. Ultimate limit state:
MRd carrying capacity in design value
2. Serviceability limit state :
Normal stress Ο [MPa]
ππΈπ = ππππ₯ maximal bending moment in design value
ππ π β₯ ππΈπ
The limiting deflection (entered) must be less than the actual deflection from the given load calculated in characteristic values.
Design and assessment of bending stress elements(symmetrical cross sections)
π€πππ β₯ π€ππππ
ππ π = ππ¦π β ππ¦
fyd = ππ¦,π
πΎπ= Οallow
In following topicsy
z
Οmax,2
Οmax,1
ee
Ο = 0
Cross section
Οπππ₯ =ππ¦
πΌπ¦β π =
ππ¦
ππ¦
+
M
ππΈπ = ππππ₯
- Section modulus, we use for the design and
assessment of the cross-section, as it applies
to the extreme fibers of the cross-section,
where the stress is maximum
ππ¦ [π3]
- yield strength, material strength, maximum normal stress that the cross section transmits in the elastic state, for steel the same in tension and pressure
ππ¦ [ππ]
We calculate the stress in absolute value and determine the sign according to the sign of bending moments, for positive bending moments the lower fibers are tensile (+), for negative moments the upper fibers are tensile (+).
Οπππ₯ =ππ¦
ππ¦+
-
Cross-sectional characteristics in bending :
β’ simple shapes and rolled profiles β tables
β’ compound shapes - calculation
1. Symmetrical cross sections:
2. Asymmetric cross sections :
Οπππ₯, 1 = Οπππ₯, 2 =π
ππ¦
π1 = π2 = W
ππ¦ =πΌπ¦π
Ο1 β Ο 2
π1 β π2e2
e1
x
)(tlakW
M
hor
hor
)(tahW
M
doldol
y
z
Οπππ₯ = Ο1 =π
ππ¦, 1
The maximum stress arises in the fibers farther from the cross-sectional axis, eg at the T-section the lower fibers are (Wy, min):
πΌ =π β π4
64
π =π β π3
32
πΌ =1
12πβ3
π =1
6πβ2 π =
1
6π3
πΌ =1
12π4
ππ¦ =πΌπ¦
π[m3]
β’ Moment of inertia Iy [m4]
β’ Section modulus Wy [m3]
π1 β π 2
π1 = π2 = e
2
1
Οπππ₯ = Ο1 =π
ππ¦, 1
Ο2 =π
ππ¦,2
π =β
2π =
π
2π =
π
2
Determine the value of maximum of normal stress in the upper and thelower fibres of the simple beam of the welded T profile.
1. Reactions, distribution of V (shear forces) and M (bending moments)
flange(160 x 16)
web (10 x 220)
T
q = 11 kN/m
l = 6 m
V [kN]
M [kNm]
Mmax = 49,5 kNm
2Β°
1Β°
Example 1
33
-33
1) placing the centre of gravity from y
2) Distance of the centres of gravity of parts from
the centre of gravity of the whole section
3) central moment of inertia
4) Section modulus for the edge fibres
2. Cross-sectional characteristics
22
01
6
T1
T
10
c 1c 2
e 1e 2
y
z
zT = 62,54 mm
c1 = 63,46 mm
c2 = 54,54 mm
Iy = 2,54027 . 107 mm4 = 2,54027 . 10-5 m4
ππ¦,1 =πΌπ¦π1=2,54027 β 107
173,46= 1,4644 β 105 mm3
ππ¦,2 =πΌπ¦π2
=2,54027 β 107
62,54= 4,0618 β 105 mm3
e2 = zT =
e1 = (220+16) - zT =
Wy,min = Wy1 Maximal stress is in the lower fibers
Example 1
T2
y
160
z T
The signs of the stress are determined from the moment distribution (tensile / pressed fibers)
neutr. axesβ Ο = 0
-108,6 MPa
301,1 MPa
45 m1054027,2 yI
π2 = 62,54mm
π1 = 173,46mm
π2 =ππππ₯
ππ¦,2= 121,88MPa (pressure)
π1 = ππππ₯ =ππππ₯
ππ¦,1= 339,04MPa(tensile fibers)
Ο in place of Mmax
3. Normal stress in the upper, lower fibers of the cross-section and in the place below the flange
ππ¦,1 = 1,4644 β 105 mm3
ππ¦,2 = 4,0618 β 105 mm3
22
01
6
T2
T1
T
10
160c 1
c 2
e 1e 2
y
-80,65 MPa
πππππππ =ππππ₯
πΌπ¦. (π2 β 0,016) =
49,5.103
2,54027.10β5. 0,0465 = 90,61MPa
Οπ₯ =ππ¦
πΌπ¦β π§
π1 = ππππ₯ =ππππ₯
πΌπ¦β e1= 339,04 MPa (tensile fibers)
π2 =ππππ₯
πΌπ¦β e2= 121,88 MPa (pressed fibers)
π§
π§ = π2 β 16
Alternatively:
+
M
-
+
Tensile fibers
Pressed fibersMy My
z(pressure)
Example 1
Example 2 - design and assessment of a circular cross-section according to ULS
Design and assess a steel beam of circular cross-sectionaccording to ULS, qk = 1,25 kNm-1, Ξ³Q =1,5, Ξ³M =1,00.Fe430 / S275.
q
3 0,51a b
2. Reactions and internal forces
1. ππ = ππ β Ξ³Q = 1,25 β 1,5 = 1,875 kN/m
ππ¦,π =ππ¦,ππΎπ
=275
1,0= 275 MPa
πΉπ,π§ = 0:β4,923 β 3,516 + 1,875 β 4,5 = 0
+
ππ,π = 0:
ππ,π = 0:
π ππ§,π = 4,923 ππ
π ππ§,π = 3,516 ππ
ππ = 1,875 kN/m
ππ,1 β 0,5 β ππ,2 β 1,5 β ππ,3 β 3,25 + π ππ§,π β 3 = 0
ππ,1 β 3,5 + ππ,2 β 1,5 β ππ,3 β 0,25 β π ππ§,π β 3 = 0
ΰ·¨ππ,1 = ππ β 1
ΰ·¨ππ,2 = ππ β 3
ΰ·¨ππ,3 = ππ β 0,5
ππ
3 0,51a b
ΰ·¨ππ,1 ΰ·¨ππ,2 ΰ·¨ππ,3
π ππ§,π π ππ§,π
Example 2
q
3 0,51
a b
-2,577-1,875
3,048
n
0,939
a bc
V
[kN]
xnP= 1,374xn
L=1,626
d
2Β°
2Β°
2Β°n
a bc d
1,53
-0,94-0,23
M
[kNm]
1Β° 1Β°
π₯ππΏ =
πππππ
=3,048
1,875= 1,626 m
π₯ππ =
πππππ
=2,577
1,875= 1,374 m
1Β°
ππ = ππ =0
πππΏ = βππ β
π₯2
2= β1,875 β
1,02
2= β0,94 kNm
πππΏ = +π ππ§,π β π₯π
πΏ β ππ βπ₯ππΏ + 1 2
2= 1,53 kNm
c d
Critical point:
πππ = βππ β
π₯2
2= β1,875 β
0,52
2= β0,23 kNm
πππ = +π ππ§,π β π₯π
π β ππ βπ₯ππ+0,5
2
2= 1,53 kNm
Extremes of moments on the beam:
ππ,π = 1,53 kNm
ππ,π = β0,94 kNm
ππ,π = β0,23 kNmπ΄πππ = π΄π¬π = π, ππ π€ππ¦
You can also calculate bending moments with the help of substitute forces
π ππ§,π = 4,92 ππ π ππ§,π = 3,52ππ
3. Calculation of the minimum cross-section of the beam according to ULS :
πy,πππ=ππΈπ
ππ¦π
πππππ3
32= 5,564 . 10-6
ππππ= 0,038 β 10β3π
ππΈπ = 1,53 kNmππ¦π β₯ Οπππ₯ =ππΈπ
ππ¦
Example 2
ππ¦π β₯ππΈπ
ππ¦
Wy,min = 5,564 . 10-6 m3
5. Assessment:
4. Design
ππ π = ππ¦π β πy
We round the value of the diameter d upwards :
π = 0,04 mDesign:
ππ π β₯ ππΈπ
πy =π β π3
32==
π β (0,04)3
32= 6,28 . 10 β6m3Section modulus:
MRd= 1,73 kN > MRd= 1,53kN
d
π = 0,04 m
Answer:The proposed cross-section with a diameter of d = 0.04 m complies with the ultimate limit state.0
In the case of the design of a square
cross β section,we substitute for ππ¦, πππ=1
6π3
π =π β π3
32
Example 3 Crosssetional characteristics1. The IPN pair is next to each other2. The IPN profiles are on top of each other
Tables: IPN 160: Wy = 1,17.105 mm3 = 1,17. 10-4 m3
1. The IPN pair is next to
each other
2. The IPN profiles are on
top of each other
Wy,1 = 1,17 β 10-4 = m31.
2.
Wy= 2 x 1,17 β 10-4 = 2,34 β 10-4 m3
y
z
2 x IPN160
2 x IPN160
80
y
z
Iy = 2 β 9,35 β 10-6 = 18,7 β 10-6 m4
Iy = 9,35. 106 mm4 = 9,35. 10-6 m4
ππ¦ =πΌπ¦π=
18,7 β 10β6
0,08= 2,34 β 10β4 m3
Iy = 2 β 9,35 β 10-6 + 2 β 2,28 β 10-3 β 0,082 = 4,79 β 10-5 m4
ππ¦ =πΌπ¦π=
4,79 β 10β5
0,16= 2,99 β 10β4 m3
e=160
e=80Iy calculate using Steiner's theorem, Wy compute
Wy can be taken directly as a double from the tables, it can be verified by calculating from Iy
A = 2,28. 103 mm2 = 2,28. 10-3 m2Tables IPN 160:
Example 4 - section from a pair of profiles
Design and assess a Fe360 / S235 steel beam loaded witha variable load qk= 15 kNm-1, gQ= 1,5, gM= 1,00.Cross section: UPN
l = 6 m
qd
U
U2. Reactions
1.
ππ = ππ β Ξ³Q = 15 β 1,5 = 22,5 kN/m
ππ¦,π =ππ¦,ππΎπ
=235
1,0= 235 MPa
ΟπΉπ,π§ = 0:β22,5 β 45 +1
2β qd β 6 = 0
+
ππ,π = 0:
ππ,π = 0:
π π,π = 22,5 kN
π π,π = 45 kN
ππ = 22,5 kN/m
β1
2β ππ β 6 β 4 + π π,π β 6 = 0
1
2β ππ β 6 β 2 β π π,π β 6 = 0π π,π
ΰ·¨π =1
2β qd β 6
l = 6 m
q
a b
a b
π π,π
ππ = ππ =0
ππππ₯πΏ = +π π,π β π₯π β
ππ β π₯π3
6 β π= 51,96 kNm
Critical point:
l = 6 m
qd =22,5kN/m
45
+
-
22,5
na b
V
[kN]
Xn= 3,464 2Β°
-45
a bM
[kNm]+
xMmax = 51,96
3Β°
ΰ·¨π =1
2β qd β 6
a b
22,5
π₯ππΏ =
2 β ππβ π
ππ=
2 β 22,5 β 6
22,5= 3,464 m
3. Design according to ULS:
πy,πππ=ππΈπ
ππ¦π=
51,96
275β103= 2,211.10-4 m3
ππΈπ = 51,96 kNm
ππ¦π β₯ππΈπ
πy
Wy,min,1 = 2,211β10β4
2=1,105 β 10-4 m3 =1,105 β 105 mm3
UPN160: Wy,1 = 1,16 β 105 mm3 = 1,16 β 10-4 m3
For 2U:
For 1U:
Tables:1U profil
2xUPN160: Wy = 2 β 1,16.10-4 = 2,32 β 10-4 m3Design:
Assessment:
MRd = ππ¦π β ππ¦ = 235 β 103 β 2,32 . 10β4 = 54,52kNm
MRd= 54,52kNm > MEd = 51,96 kNm
Example 3
Answer:The proposed cross-section with a diameter of d = 0.04 m complies with the ultimate limit state.
Example 5 - homework Design and assess a steel beam from a rolled IPN profileaccording to ULS. Fe 360/S235, Ξ³M=1,00, Ξ³Q=1,5
qk = 7,5 kNm-1
6 1,5
1) Reactions Ra, Rb
2) V, M
3) Mmax= MEd
4) Determine Wy,min :
5) Design section:
6) Carrying capacity MRd
7) Assessment:
= 44,50 kNm
2 extrΓ©mes, chose Mmax
ππ¦π β₯ ππππ₯
ππ¦π β₯ππΈπ
ππ¦
ππ¦, πππ=ππΈπ
ππ¦π= 1,8936.10-4m3
=2,14.10-4m3IPN200: Wy
β satisfyingEdRd MM
ππ π = ππ¦π β ππ¦
MRd = 50,29 kNm