T ORSION OF T HIN -W ALLED B EAMS
Chapter 16 Bending of open and closed,
20.4 Deflection of open and closed section beams
20.4 Deflection of open and closed section beams
Bending, shear and torsional deflections of thin-walled beams are readily obtained by application of the unit load method described in Section 5.5.
The displacement in a given direction due to torsion is given directly by the last of Eqs (5.21), thus
T =
L
T0T1
GJ dz (20.14)
where J, the torsion constant, depends on the type of beam under consideration. For an open section beam J is given by either of Eqs (18.11) whereas in the case of a closed section beam J=4A2/(
ds/t) (Eq. (18.4)) for a constant shear modulus.
Expressions for the bending and shear displacements of unsymmetrical thin-walled beams may also be determined by the unit load method. They are complex for the general case and are most easily derived from first principles by considering the com- plementary energy of the elastic body in terms of stresses and strains rather than loads and displacements. In Chapter 5 we observed that the theorem of the principle of the stationary value of the total complementary energy of an elastic system is equivalent to the application of the principle of virtual work where virtual forces act through real displacements.We may therefore specify that in our expression for total complementary energy the displacements are the actual displacements produced by the applied loads while the virtual force system is the unit load.
Considering deflections due to bending, we see, from Eq. (5.6), that the increment in total complementary energy due to the application of a virtual unit load is
−
L
A
σz,1εz,0dA
dz+1M
whereσz,1is the direct bending stress at any point in the beam cross-section correspond- ing to the unit load andεz,0 is the strain at the point produced by the actual loading system. Further,Mis the actual displacement due to bending at the point of applica- tion and in the direction of the unit load. Since the system is in equilibrium under the action of the unit load the above expression must equal zero (see Eq. (5.6)). Hence
M =
L
A
σz,1εz,0dA
dz (20.15)
From Eq. (16.18) and the third of Eqs (1.42) σz,1=
My,1Ixx−Mx,1Ixy IxxIyy−Ixy2
x+
Mx,1Iyy−My,1Ixy IxxIyy−Ixy2
y
εz,0= 1 E
My,0Ixx−Mx,0Ixy
IxxIyy−Ixy2
x+
Mx,0Iyy−My,0Ixy
IxxIyy−Ixy2
y
where the suffixes 1 and 0 refer to the unit and actual loading systems and x, y are the coordinates of any point in the cross-section referred to a centroidal system of
axes. Substituting forσz,1andεz,0in Eq. (20.15) and remembering that
Ax2dA=Iyy,
Ay2dA=Ixx, and
Axy dA=Ixy, we have
M = 1
E(IxxIyy−Ixy2)2
L{(My,1Ixx−Mx,1Ixy)(My,0Ixx−Mx,0Ixy)Iyy +(Mx,1Iyy−My,1Ixy)(Mx,0Iyy−My,0Ixy)Ixx
+[(My,1Ixx−Mx,1Ixy)(Mx,0Iyy−My,0Ixy)
+(Mx,1Iyy−My,1Ixy)(My,0Ixx−Mx,0Ixy)]Ixy}dz (20.16) For a section having either the x or y axis as an axis of symmetry, Ixy=0 and Eq. (20.16) reduces to
M = 1 E
L
My,1My,0
Iyy +Mx,1Mx,0 Ixx
dz (20.17)
The derivation of an expression for the shear deflection of thin-walled sections by the unit load method is achieved in a similar manner. By comparison with Eq. (20.15) we deduce that the deflectionS, due to shear of a thin-walled open or closed section beam of thickness t, is given by
S=
L
sect
τ1γ0t ds
dz (20.18)
where τ1 is the shear stress at an arbitrary point s around the section produced by a unit load applied at the point and in the directionS, andγ0is the shear strain at the arbitrary point corresponding to the actual loading system. The integral in parentheses is taken over all the walls of the beam. In fact, both the applied and unit shear loads must act through the shear centre of the cross-section, otherwise additional torsional displacements occur. Where shear loads act at other points these must be replaced by shear loads at the shear centre plus a torque. The thickness t is the actual skin thickness and may vary around the cross-section but is assumed to be constant along the length of the beam. Rewriting Eq. (20.18) in terms of shear flows q1and q0, we obtain
S =
L
sect
q0q1 Gt ds
dz (20.19)
where again the suffixes refer to the actual and unit loading systems. In the cases of both open and closed section beams the general expressions for shear flow are long and are best evaluated before substituting in Eq. (20.19). For an open section beam comprising booms and walls of direct stress carrying thickness tDwe have, from Eq. (20.6)
q0 = −
Sx,0Ixx−Sy,0Ixy IxxIyy−Ixy2
s 0
tDx ds+ n r=1
Brxr
−
Sy,0Iyy−Sx,0Ixy IxxIyy−Ixy2
s 0
tDy ds+ n r=1
Bryr
(20.20)
20.4 Deflection of open and closed section beams 575
and
q1= −
Sx,1Ixx−Sy,1Ixy IxxIyy−Ixy2
s 0
tDx ds+ n r=1
Brxr
−
Sy,1Iyy−Sx,1Ixy IxxIyy−Ixy2
s 0
tDy ds+ n r=1
Bryr
(20.21) Similar expressions are obtained for a closed section beam from Eq. (20.11).
Example 20.5
Calculate the deflection of the free end of a cantilever 2000 mm long having a channel section identical to that in Example 20.3 and supporting a vertical, upward load of 4.8 kN acting through the shear centre of the section. The effective direct stress carrying thickness of the skin is zero while its actual thickness is 1 mm. Young’s modulus E and the shear modulus G are 70 000 and 30 000 N/mm2, respectively.
The section is doubly symmetrical (i.e. the direct stress carrying area) and supports a vertical load producing a vertical deflection. Thus we apply a unit load through the shear centre of the section at the tip of the cantilever and in the same direction as the applied load. Since the load is applied through the shear centre there is no twisting of the section and the total deflection is given, from Eqs (20.17), (20.19), (20.20) and (20.21), by
= L
0
Mx,0Mx,1 EIxx dz+
L
0
sect
q0q1 Gt ds
dz (i)
where Mx,0= −4.8×103(2000−z), Mx,1= −1(2000−z) q0 = −4.8×103
Ixx n r=1
Bryr q1= − 1 Ixx
n r=1
Bryr
and z is measured from the built-in end of the cantilever. The actual shear flow dis- tribution has been calculated in Example 20.3. In this case the q1 shear flows may be deduced from the actual distribution shown in Fig. 20.8, i.e.
q1=q0/4.8×103 Evaluating the bending deflection, we have
M= 2000
0
4.8×103(2000−z)2dz
70 000×48×106 =3.81 mm The shear deflectionSis given by
S= 2000
0
1 30 000×1
1
4.8×103(62×200+122×400+62×200)
dz
=1.0 mm
The total deflectionis thenM+S=4.81 mm in a vertical upward direction.
Problems
P.20.1 Idealize the box section shown in Fig. P.20.1 into an arrangement of direct stress carrying booms positioned at the four corners and panels which are assumed to carry only shear stresses. Hence determine the distance of the shear centre from the left-hand web.
Ans. 225 mm.
10 mm
10 mm Angles
60 50 10 mm Angles
50 40 8 mm
10 mm 500 mm
8 mm
300 mm
Fig. P.20.1
P.20.2 The beam section shown in Fig. P.20.2 has been idealized into an arrange- ment of direct stress carrying booms and shear stress only carrying panels. If the beam section is subjected to a vertical shear load of 1495 N through its shear centre, booms 1, 4, 5 and 8 each have an area of 200 mm2and booms 2, 3, 6 and 7 each have an area of 250 mm2determine the shear flow distribution and the position of the shear centre.
Ans. Wall 12, 1.86 N/mm; 43, 1.49 N/mm; 32, 5.21 N/mm; 27, 10.79 N/mm;
remaining distribution follows from symmetry. 122 mm to the left of the web 27.
50 mm
50 mm 40 mm
80 mm
80 mm 40 mm
150 mm 200 mm 150 mm
8
7 6
5
4 3
2 1
Fig. P.20.2
Problems 577 P.20.3 Figure P.20.3 shows the cross-section of a single cell, thin-walled beam with a horizontal axis of symmetry. The direct stresses are carried by the booms B1to B4, while the walls are effective only in carrying shear stresses. Assuming that the basic theory of bending is applicable, calculate the position of the shear centre S. The shear modulus G is the same for all walls.
Cell area=135 000 mm2. Boom areas: B1=B4=450 mm2, B2=B3=550 mm2. Ans. 197.2 mm from vertical through booms 2 and 3.
Fig. P.20.3
Wall Length (mm) Thickness (mm)
12, 34 500 0.8
23 580 1.0
41 200 1.2
P.20.4 Find the position of the shear centre of the rectangular four boom beam section shown in Fig. P.20.4. The booms carry only direct stresses but the skin is fully effective in carrying both shear and direct stress. The area of each boom is 100 mm2.
Ans. 142.5 mm from side 23.
Fig. P.20.4
P.20.5 A uniform beam with the cross-section shown in Fig. P.20.5(a) is supported and loaded as shown in Fig. P.20.5(b). If the direct and shear stresses are given by the basic theory of bending, the direct stresses being carried by the booms and the shear stresses by the walls, calculate the vertical deflection at the ends of the beam when the loads act through the shear centres of the end cross-sections, allowing for the effect of shear strains.
Take E=69 000 N/mm2and G=26 700 N/mm2. Boom areas: 1, 3, 4, 6 = 650 mm2, 2, 5 =1300 mm2.
Ans. 3.4 mm.
Fig. P.20.5
P.20.6 A cantilever, length L, has a hollow cross-section in the form of a doubly symmetric wedge as shown in Fig. P.20.6. The chord line is of length c, wedge thickness is t, the length of a sloping side is a/2 and the wall thickness is constant and equal to t0. Uniform pressure distributions of magnitudes shown act on the faces of the wedge. Find the vertical deflection of point A due to this given load- ing. If G=0.4E, t/c=0.05 and L=2c show that this deflection is approximately 5600p0c2/Et0.
Problems 579
Fig. P.20.6
P.20.7 A rectangular section thin-walled beam of length L and breadth 3b, depth b and wall thickness t is built in at one end (Fig. P.20.7). The upper surface of the beam is subjected to a pressure which varies linearly across the breadth from a value p0 at edge AB to zero at edge CD. Thus, at any given value of x the pressure is constant in the z direction. Find the vertical deflection of point A.
Fig. P.20.7
Ans. p0L2(9L2/80Eb2+1609/2000G)/t.