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

Question

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.

Accepted Answer

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

[latex]\Delta_M=\frac{1}{E} \int_L\left(\frac{M_{y, 1} M_{y, 0}}{I_{y y}}+\frac{M_{x, 1} M_{x, 0}}{I_{x x}} ight) \mathrm{d} z[/latex] (20.17)

[latex]\Delta_S=\int_L\left(\int_{ ext {sect }} au_1 \gamma_0 t \mathrm{~d} s ight) \mathrm{d} z[/latex] (20.18)

[latex]\begin{aligned}q_0=&-\left(\frac{S_{x, 0} I_{x x}-S_{y, 0} I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_{y, 0} I_{y y}-S_{x, 0} I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)\end{aligned}[/latex] (20.20)

[latex]\begin{aligned}q_1=&-\left(\frac{S_{x, 1} I_{x x}-S_{y, 1} I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_{y, 1} I_{y y}-S_{x, 1} I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)\end{aligned}[/latex] (20.21)

[latex]\Delta=\int_0^L \frac{M_{x, 0} M_{x, 1}}{E I_{x x}} \mathrm{~d} z+\int_0^L\left(\int_{\operatorname{sect}} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex] (i)

where [latex]M_{x, 0}=-4.8 imes 10^3(2000-z), M_{x, 1}=-1(2000-z)[/latex]

[latex]q_0=-\frac{4.8 imes 10^3}{I_{x x}} \sum_{r=1}^n B_r y_r \quad q_1=-\frac{1}{I_{x x}} \sum_{r=1}^n B_r y_r[/latex]

and z is measured from the built-in end of the cantilever. The actual shear flow distribution has been calculated in Example 20.3. In this case the [latex]q_1[/latex] shear flows may be deduced from the actual distribution shown in Fig. 20.8, i.e.

[latex]q_1=q_0 / 4.8 imes 10^3[/latex]

Evaluating the bending deflection, we have

[latex]\Delta_{\mathrm{M}}=\int_0^{2000} \frac{4.8 imes 10^3(2000-z)^2 \mathrm{~d} z}{70000 imes 48 imes 10^6}=3.81 \mathrm{~mm}[/latex]

The shear deflection [latex]\Delta_{\mathrm{S}}[/latex] is given by

[latex]\begin{aligned}\Delta_{\mathrm{S}} &=\int_0^{2000} \frac{1}{30000 imes 1}\left[\frac{1}{4.8 imes 10^3}\left(6^2 imes 200+12^2 imes 400+6^2 imes 200 ight) ight] \mathrm{d} z \&=1.0 \mathrm{~mm}\end{aligned}[/latex]

The total deflection Δ is then [latex]\Delta_{\mathrm{M}}+\Delta_{\mathrm{S}}=4.81[/latex] mm in a vertical upward direction.

Question 20.5: Calculate the deflection of the free end of a cantilever 200...

Related Answered Questions

Verified Answer:

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. ...

Verified Answer:

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 B1 to B4, while the walls are effective only in carrying shear stresses. Assuming that the basic theory of bending is applicable, ...

Verified Answer:

The beam section shown in Fig. P.20.2 has been idealized into an arrangement 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 mm² ...

Verified Answer:

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. ...

Verified Answer:

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 ...

Verified Answer:

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 ...

Verified Answer:

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. ...

Verified Answer:

The thin-walled single cell beam shown in Fig. 20.11 has been idealized into a combination of direct stress carrying booms and shear stress only carrying walls. If the section supports a vertical shear load of 10 kN acting in a vertical plane through booms 3 and 6, calculate the distribution ...

Verified Answer:

Calculate the shear flow distribution in the channel section shown in Fig. 20.7 produced by a vertical shear load of 4.8 kN acting through its shear centre. Assume that the walls of the section are only effective in resisting shear stresses while the booms, each of area 300 mm², ...

Verified Answer: