Simple torsion in beams at circle crosssection and pipes
Axial loading
Torsion
Rule of the right hand, similarity to N and V forces
l3
F3
F2
l2
l1
F1
x
l3
N
N1 = − F1
N 2 = − F1 − F2
-Rx,a
Mx3
Mx,a
N 3 = − F1 − F2 − F3
Mx2
Mx1
l2
l1
N σ – normal stress
σx =
A
T1 = − M x 1
T2 = − M x 1 − M x 2
T3 = − M x 1 − M x 2 − M x 3
τ =
T
r
Ip
τ – shear stress
σ = const.
N i li
∆l = ∑
Ei Ai
G=
- deformation [m]
x
T
-Mx,a
Rx,a
E
2(1 + ν )
Ti ⋅ li
ϕL = ∑
Gi ⋅ I pi
τ max on the
circumference
-Deformation
angle in torsion
[rad ]
180
π
[deg ]
Simple torsion in beams of the circle
cross-section and pipe
τmax
τmax
Cross-sectional characteristic of
circle and pipe in torsion is
polar moment of inertia: IP
r
R
(IP = Iy + Iz = π r4 / 4 + π r4 / 4 )
Circle
It =
πd 4
32
Pipe
=
πr 4
2
It =
(
D
32
π
4
−d4
)
Alternative expression:
τ max =
Assessment:
τ allow ≥ τ max
Tmax
=
r
It
T max
Wt
Wt ... Section modullus
[m3]
Example 1 – shear stress, angel of twist, carrying capacity
Steel bar of the circle cross-section loaded by the torsion moment according to the picture.
Determine:
maximal shear stress τmax and angle of twist ϕl.
draw distribution of shear stress at the end of the bar.
Determine carrying capacity in torsion of the bar
Mx = 0,75 kNm
Mx,a
d = 32 mm, Fe360 / S235
G = 81 000 MPa
γM=1.0, γG=1.35
steel : τ all =
f yd
3
Ip =
Polar moment of inertia:
Maximal shear stress:
τ max
π
r4
2
Tmax
=
r
Ip
Tl
l = 2,5 m
T
+
Angle of twist:
φl = G Ip [rad]
Carrying capacity in torsion :
TRd = τ all
Ip
r
=
f yd I p
3 r
= τ all
2I p
d
Mxd=1,0125kNm=TEd , Ip=1,02944.10-7m4, τall=135,677MPa, τmax=157,368MPa, φ=0,225rad,
TRd=0,873kNm ….not satisfying
Example 2 – design and assessment of the circle bar
Steel bar of the circle cross-section loaded by the torsion moment according to the picture.
Determine:
1.
reaction, diagram of T forces ( Tk, Td )
2.
4.
5.
6.
7.
make the design of the circle rod – r, d
make the assessment (TRd > TEd)- from designed values
determine φL at the end of the whole bar – in radians and degrees (from characteristic values)
determine maximum of shear stress τmax
Draw the distribution of the stress in sections at the free end of the bar
Mxk= 1,6 kNm
l=6m
Tk
+
Required radius rreq:
τ all =
TEd
T
⋅ rnut = Ed4 ⋅ rreq ⇒ rreq = 3 ...
π rreq
Ip
2
Fe430 / S275
G = 81 000 MPa
γQ=1,5
γM=1,0
steel : τ all =
f yd
diagram of T forces
TEd
= 2,4kNm
rreq, dreq = 0,0213m; 0,0426m
rreal , dreal = 0,022m; 0,044m;
τallow
= 158,77MPa
τmax
= 144,118MPa
Ip
= 3,6797.10-7m4
TRd ≥ TEd = 2,656 kNm (satisfying)
φl
= 0,324rad
3
Example 3
The circle bar is loaded by torsion moments as you see the picture, d1=0,042m, d2=0,036m, Fe360/S235,
γM = 1,00, γG = 1,35. G = 81 000 MPa
1.
2.
3.
4.
5.
reaction, diagram of T forces ( Tk, Td )
determine maximum of shear stress tmax on the both parts of the bar
determine φL of the whole bar – in radians and degrees (from characteristic values)
Make the assessment in both parts (TRd > TEd)- from designed values
Draw the distribution of the stress in sections in both parts
πd 4
Ip =
Mxk,1 = 0,5 kNm
1
Mxk,2 = 0,2 kNm
2
32
steel : τ all =
τ max =
1,0
1,2
-
Tk
Tk2 =0,2
Tk1 =- 0,3
1
2
τmax,1
τmax,2
ϕ
=
πr 4
2
f yd
3
Tmax
r
It
ϕL = ∑
Ti ⋅ li
Gi ⋅ I pi
Ip1 = 3,0549.10-7m4
diagram of T forces Ip2 =1,6489.10-7m4
τmax,1 = -20,62 MPa
Tk1, Tk2
τmax,2 = 21,83 MPa
Td1, Td2
TRd,1 = 1,97kNm > TEd,1 = 0,675kNm – satisf.
TRd,2 = 1,24kNm > TEd,2 = 0,27kNm – satisf.
φL = -0,0145+0,0149=4,74*10-4 rad
Example 4
Steel bar of the circle cross-section loaded by the torsion moment see the picture.
Determine:
1.
Diagram of internal forces
2.
Maximum of the shear stress on the both parts of the bar
3.
Assessment of the both parts of the bar
4.
Angle of the torsion of the free end of the bar
Modullus in the shear is G = 81 GPa, fyk = 235MPa, γG = 1,35, γM = 1,0
l1 = 2,0m, l2 = 1,0m, d1 = 0,04m, d2 = 0,03m, Mx = 0,3kNm,
Mx = 0,30 kNm
2
1
l1
l2
ϕ
Ip1 = 7,952.10-8m4
Ip2 = 2,5133.10-7m4
τmax,1 = 76,396 MPa
τmax,2 = 32,243 MPa
TRd,1 = 0,7193kNm > TEd,1 = 0,405kNm – satisf.
TRd,2 = 1,705kNm > TEd,2 = 0,405kNm – satisf.
φ = 0,076rad = 4,36°