3.36. Assume that the shear modulus G=27GPa and the allowable shear stress of τallow =187MPa.
cm t =0.3
Figure 3.36 Single thin-walled section
Solution:
Since the thickness of all walls are equal to t =0.3cm, we can obtain the allowable shear flow from allowable shear stress, that is
m / N 10 61 . 5 003 . 0 10 187 t
qallow=τallow = × 6× = × 5
Then we have the torque capability as
m N 22440 10
61 . 5 ) 2 . 0 1 . 0 ( 2 q
A 2
Tallow= allow = × × × 5 = −
--- ANS
3.7.1
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
3.8 A two-cell thin-walled member with the cross-section shown in Fig. 3.37 is subjected to a torque T. The resulting twist angle θ is . Find the shear flows of the applied torque, and the torsion constant. The material is aluminum alloy 2024-T3.
m / 3o
Figure 3.37 Two-cell section
Solution:
(a) Assume the material is linearly elastic under the twist angle θ. For aluminum alloy 2024-T3, we have the shear modulus
GPa ) 27
33 . 0 1 ( 2
72 )
1 ( 2
G E =
= +
= +
υ
(b) We denote the shear flow on the left cell , and the shear flow on the right cell . The shear flow in the vertical web is
q1
q2 q12 =q1 −q2, are the positive directions as shown in the figure above.
Also, we have the torque for two-cell sections
2
1 2
1 2
2Aq A q
T = + (3.8.1)
where 2
2 2
2
1 0.098m
8 ) 5 . 0 ( 8 A d
A = =π =π =
, The twist angle of the left cell is
)) (
( 2
1 2
1
2 1 12
12 1 1
1 1 1
1
1 q q
t q s t s A t G
qds A
G cell = + −
=
∫
θ (3.8.2)
where 0.785m 2
s1 =πd =
is the length of the left side wall, and is the length of the vertical web.
m s12 =0.5
The twist angle of the right cell is
)) (
( 2
1 2
1
2 1 12 12 2 2 2 2 2
2
2 q q
t q s t s A t G
qds A
G cell = − −
=
∫
θ (3.8.3)
Again, we have 0.785m 2
s2 =πd =
, the length of the right side wall.
Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition
m and by substituting all the known quantities,
q1 q2
Back substituting into (3.8.2) and (3.8.4), we have
9 1
Subsequently from (3.8.5) we obtain q2 =1.687q1 =765000N/m
--- ANS (c) The applied torque
From (3.8.1), we compute the applied torque
m (d) The torsion constant J
From the fundamental relationship of torque and twist angle, we have T =GJθ So the torsion constant can be derived as
4
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
3.9 For the bar of Fig. 3.37, find the maximum torque if the allowable shear stress is
allow =187MPa
τ . What is the corresponding maximum twist angle θ ?
Figure 3.37 Two-cell section
Solution:
(a) Assume the material is linearly elastic under the twist angle θ. For aluminum alloy 2024-T3, we have the shear modulus
GPa ) 27
33 . 0 1 ( 2
72 )
1 ( 2
G E =
= +
= +
υ
(b) We denote the shear flow on the left cell as and that on the right cell as . The shear flow in the vertical web is
q1 q2
2 1
12 q q
q = − . The positive directions for the shear flows are shown in the figure above.
The torque for two-cell section is
2
1 2
1 2
2Aq A q
T = + (3.9.1)
where 2
2 2
2
1 0.098m
8 ) 5 . 0 ( 8 A d
A = =π =π =
, The twist angle of the left cell is
)) (
( 2
1 2
1
2 1 12
12 1 1
1 1 1
1
1 q q
t q s t s A t G
qds A
G cell = + −
=
∫
θ (3.9.2)
where 0.785m 2
s1=πd =
is the length of the left side wall, and is the length of the vertical web.
m s12 =0.5
Also we have the twist angle of the right cell as
)) (
( 2
1 2
1
2 1 12 2 2 2 2 2
2
2 q q
t q s t s A t G
qds A
G cell = − −
=
∫
θ (3.9.3)
where 0.785m 2
s2 =πd =
is the length of the right side wall.
(c) Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition
θ θ
θ1 = 2 = (3.9.4)
From (3.9.2) to (3.9.4) and note that A1 = A2, we derive the relation between and by substituting all the known quantities,
q1
Back substituting (3.9.5) into (3.9.2) and (3.9.4), we have
9 1
From the above quantities of shear flow, we can then compute the shear stress in each wall by
(e) From the above stresses (3.9.9) to (3.9.11), because the negative value just denote the negative direction, the maximum absolute magnitude of shear stress is
6 allow
9
1 =8.66×10 θ ≤τ =187×10 τ
Therefore the maximum twist angle is m (f) The maximum torque can be solved by using (3.9.1), (3.9.6), (3.9.7) and the
maximum twist angle, that is m
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
3.10 Find the shear flow and twist angle in the two-cell three-stringer thin-walled bar with the cross-section shown in Fig. 3.38. The material is Al2024-T3. The applied torque is 2×105N⋅m.
Figure 3.38 Two-cell three-stringer thin-walled section
Solution:
(a) Assume the material is linearly elastic under the applied torque. For aluminum alloy 2024-T3, we have the shear modulus
E GPa
G 27
) 33 . 0 1 ( 2
72 )
1 (
2 =
= +
= +
υ
(b) Denote the shear flow on the left cell as , and the shear flow on the right cell as
; both are considered positive if counterclockwise. The shear flow in the vertical web is , which is positive if it is in the same direction as .
q1
q2
2 1
12 q q
q = − q1
We have the torque for the two-cell section as
2
1 2
1 2
2Aq A q
T = + (3.10.1)
where 2
2 2
1 0.565
8 ) 2 . 1 (
8 m
A =πd =π =
,
and 2 1.2 2
2 ) 2 . 1 ( 2
2 m
A = bh = =
The twist angle of the left. cell is
)) (
( 2
1 2
1
2 1 1 12 1 1 1 1 1
1
1 q q
t q s t s A t G
qds A
G cell = + −
=
∫
θ (3.10.2)
where d m
s 1.88
1 =π2 =
is the length of the left half circular wall, and is the length of the vertical web.
m s12 =1.2
The twist angle of the right. cell is
)) the right cell.
Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition
θ θ
θ1 = 2 = (3.10.4)
Solving the two equations, (3.10.2) and (3.10.4), we obtain 005 )
(c) To find the shear flow , we back substitute (3.10.5) into (3.10.1) and have
1 (d) For the twist angle, we can utilize the shear flows and equations (3.10.2) and
(3.10.4) to get,
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
3.11 What is the maximum torque for the structure of Fig. 3.38 if the allowable twist angle θ is 2o/m?
Figure 3.38 Two-cell three-stringer thin-walled section
Solution:
(a) Assume the material used is still Aluminum alloy 2024-T3. For aluminum alloy 2024-T3, we have the shear modulus
GPa ) 27
33 . 0 1 ( 2
72 )
1 ( 2
G E =
= +
= +
υ
(b) Denote the shear flow on the left cell as , and the shear flow on the right cell as . Both are assumed positive in the counterclockwise direction. The shear flow in the vertical web is
q1
q2
2 1
12 q q
q = − , from bottom to top.
The torque for two-cell section is
2
1 2
1 2
2Aq A q
T = + (3.11.1)
where 2
2 2
1 0.56m
8 ) 2 . 1 ( 8
A =πd =π =
,
and 2 1.2 2
2 ) 2 . 1 ( 2
2 m
A = bh = =
The twist angle of the left cell is
)) (
( 2
1 2
1
2 1 1 12 1 1 1 1 1
1
1 q q
t q s t s A t G
qds A
G cell = + −
=
∫
θ (3.11.2)
where 1.88m 2
s1 =πd =
is the length of the left half circular wall, and is the length of the vertical web.
m s12 =1.2
The twist angle of the right cell is
)) (
( 2
1 2
1
2 1 1 12 2 3 3 2 2
2 2 2
2
2 q q
t q s t q s t s A t G
qds A
G cell = + − −
=
∫
θ (3.11.3)
wheres2 =2m is the length of the lower straight wall of thickness t2, and
Since the entire thin-wall section must rotate as a rigid body in the plane, we require the compatibility condition
θ θ
θ1 = 2 = (3.11.4)
From (3.10.2) to (3.10.4), we can derive the relation between and by substituting all the known quantities,
q1 q2
allowable 0
2
(d) To find the maximum torque, we can use equation (3.11.1)
m
Therefore the maximum torque is m It should be noted that under this torque the shear stress has already exceeded the yield condition of Al 2024-T3. Consequently, this solution may not be of practical significance if allowable stress condition is to be satisfied too.
3.11.2
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
3.12 The two shafts of thin-walled cross-sections shown in Fig. 3.39a and b, respectively. Contain the same amount of aluminum alloy. Compare the torsional rigidities of the two shafts without end constraints.
(a) (b)
Figure 3.39 Cross-sections of two shafts
Solution:
(a) Fig. 3.39a is a cross-section of an open thin-wall, its torsional rigidity is GJa
4 3
i 3 i i
a )(200)(3) G 5400Gmm
3 (1 3 t 3b G 1
GJ =
∑
= =--- ANS (b) Fig. 3.39b is a cross-section of a closed thin-wall, its torsional rigidity is GJb
=
∫
t ds G A GJb
4 2
, where 4 3b2 A= ,
4 6 3
2 4
10 4 6
) 3 ( 4
3
4 b tG Gmm
b t G b
t ds G A
GJb = = = = ×
∫
--- ANS (c) The ratio of the torsional rigidities is
G 1111 5400
G 10 6 GJ
GJ 6
a
b = × =
--- ANS
3.13 Find the distributions of the primary warping displacement on the cross-sections shown in Fig. 3.39b. Due to symmetry, the center of twist coincides with the centroid of the section, and warp at the midpoint of each flat sheet section is zero.
Sketch the warping displacement along the wall.
(b)
Figure 3.39 Cross-sections of two shafts
Solution:
(a) Observation.
Because of the symmetry, the center of twist coincides with the centroid of the section, and warp at the midpoint of each flat sheet section is zero.
So from the figure above we set w=0 at the midpoint of each flat sheet. First we assume the warp at point A is positive of z-direction. While going from A to B, we pass the midpoint and then the warp goes from positive into negative part, then end at point B with the maximum negative warping. Using the same concept on sheet BC will result in a maximum positive warping at point C. Now we consider the sheet CA by using the same conclusion, we will surprisingly find the warping at A is negative of z-direction. Hence it contradicts our assumption of A being
3.13.1
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
positive. By applying the assumption of A is negative direction we will conclude in another contradiction. Therefore, we can confidently assure that there is no warping in this equilateral triangular thin-walled cross-section. In the following we will approve it by further derivatives.
(b) For the closed thin-walled section, we have ρθ
γ +
∂
= ∂
∂ +∂
∂
= ∂
s w z u s
w s
sz , (3.13.1)
where ρ is the distance from the center of twist to the tangent line of point P of interest, w is the warping that we are seeking. Also, we have
Gt q G
s sz
sz =τ =
γ (3.13.2)
where is the shear flow along s-direction, t is the thickness of the wall and G is the shear modulus.
qs
Again, recall from the relationship between applied torque and shear flow, we have
A qs T
2
= (3.13.3)
Combining (3.13.1) to (3.13.3) results in Gt
A T s
w
2
=
∂ +
∂ ρθ , or = −ρθ
∂
∂
Gt A
T s w
2
=> s sρθ s ds Asθ
Gt A ds T
ds Gt A w T
s
w 2
2 2
) 0 ( )
( − =
∫
0 −∫
0 =∫
0 − (3.13.4)Also the twist angle can be derived from
∫
= t
ds A G 2
θ 1 (3.13.5)
(c) Assume the applied torque is uniformly applied to the cross-section. Also, the material is isotropic so that the shear modulus is constant.
For the equilateral triangular section, we have 4
3b2
A= (3.13.6)
And since the section is symmetric, we can just take the sheet CA into consideration and applied to all other sheets. Assume the origin of s is on the midpoint of sheet CA, so w(0)=0, then we have
12
As = 3bs (3.13.7)
From (3.13.4) to (3.13.7), we obtain
) ) 16 / 3 (
) 8 / 3 ( 3 ( 2 )
( 4
) 3 12 )(
( 3 2 2 2
) 2
( 4
2 2 2
0 b
b Gt b
Ts t
A G
T bs b t
A G A Ts
Gtds A s T
w =
∫
s − sθ = − = −=> w(s)=0
This approves our observation in part (a).
--- ANS
3.13.3
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN