Question 7 of 8: Deflection of a tie-rod-supported beam by virtual work

Question 7: Deflection of a tie-rod-supported beam by virtual work (22 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A horizontal beam pinned to a wall at point 1 $(0,\,0)$ and running 9 m to a free end at point 4. A 48 kN downward load acts at point 2, 3 m from the pin; a tie rod runs from point 3, 6 m from the pin, up to a second wall pin at point 5, 8 m above point 1. The beam has $EI = 36\,000$ kN·m$^2$ and is inextensible; the tie rod has $EA = 25\,000$ kN.

Find. (a) The vertical deflection at point 4. (b) The vertical deflection at point 2 when the same 48 kN load is moved to point 4.

Approach. The structure is determinate — two pin reactions and the tie force against three equations — so find the real bending moments and tie force by statics, repeat with a unit load at point 4, and combine the bending and axial contributions.

For a structure containing both flexural and axial members the unit-load statement is

where $M$ and $N$ come from the real loading and $m$ and $n$ from a unit load placed at and along the required displacement. Because the beam is stated to be inextensible, its axial force does no work and only the tie rod appears in the second sum.

1. Real tie force. The tie runs from $(6,0)$ to $(0,8)$, so its length is 10 m and its direction cosines are $-0.6$ and $+0.8$. Taking moments about the pin at point 1, $$ 0.8\,T(6) = 48(3) \;\Rightarrow\; \boxed{T = 30.0\ \text{kN (tension)}} $$
2. Real reactions. Resolving, $H_1 = 0.6(30) = \boxed{18.0\ \text{kN}}$ and $V_1 = 48 - 0.8(30) = \boxed{24.0\ \text{kN}\ (\uparrow)}$.
3. Real bending moment. From the pin, $M = 24x$ up to the load, so $M(3) = 72$ kN·m; from the load to the tie, $M = 144 - 24x$, which vanishes at $x = 6$ m; and beyond the tie the beam is unloaded, so $M \equiv 0$ from point 3 to the free end.
4. Virtual system. Remove the 48 kN load and apply 1 kN downward at point 4. Moments about the pin give $0.8\,t(6) = 1(9)$, so $t = 1.875$, and the virtual vertical reaction is $1 - 0.8(1.875) = -0.5$ kN, that is 0.5 kN downward. Hence $m = -0.5x$ from 0 to 6 m and $m = x - 9$ from 6 m to the free end, which is zero at point 4 as it must be.
5. Flexural contribution. Only the length 0 to 6 m carries real moment, so $$ \int_0^3 (24x)(-0.5x)\,\mathrm{d}x + \int_3^6 (144-24x)(-0.5x)\,\mathrm{d}x = -108 - 216 = -324\ \text{kN}^2\!\cdot\!\text{m}^3 $$ and dividing by $EI$, $$ \Delta_{\text{bending}} = \frac{-324}{36\,000} = \boxed{-9.0\ \text{mm}} $$ the negative sign meaning that bending alone lifts point 4, which is right: the span 1–3 sags, so its tangent at the prop rotates and the unloaded cantilever beyond swings upward.
6. Axial contribution of the tie. $$ \Delta_{\text{tie}} = \frac{N n L}{EA} = \frac{30(1.875)(10)}{25\,000} = \boxed{+22.5\ \text{mm}} $$
7. Combine. $$ \Delta_4 = -9.0 + 22.5 = \boxed{13.5\ \text{mm downward}} $$
8. Independent check of the tie term. The tie stretches by $TL/(EA) = 30(10)/25\,000 = 12$ mm. Since the beam is inextensible and pinned at point 1, point 3 must move so that its component along the tie equals that stretch; the tie makes an angle whose vertical cosine is 0.8, so point 3 drops $12/0.8 = 15$ mm. Point 4, being on the same rigid line 9 m from the pin, drops $15 \times 9/6 = 22.5$ mm, exactly the axial term computed above.
9. (b) The reciprocal case. Maxwell's reciprocal theorem states that for a linearly elastic structure the displacement at $A$ due to a load at $B$ equals the displacement at $B$ due to the same load at $A$. Hence the vertical deflection at point 2 caused by 48 kN at point 4 is the same as the deflection at point 4 caused by 48 kN at point 2: $$ \boxed{\Delta_2 = 13.5\ \text{mm downward}} $$ No further analysis is required, and the marks for part (b) are for saying exactly that.