Deflections and Slope of Statically Determinate and Indeterminate Structures

Strain energy expressions

For a linear elastic member (energy stored equals work done by gradually applied loads):

For slender beams the bending term dominates; the others are usually neglected.

Castigliano's second theorem

Statement: For a linearly elastic structure in equilibrium, the partial derivative of the total strain energy with respect to a load $P_i$ equals the displacement $\delta_i$ of the point of application of that load, in the direction of the load.

$$\delta_i = \frac{\partial U}{\partial P_i},\qquad \theta_i = \frac{\partial U}{\partial M_i}$$

Proof: Let loads $P_1,\dots,P_n$ act and produce energy $U$. Increase $P_i$ by $dP_i$; the energy becomes $U + \dfrac{\partial U}{\partial P_i}dP_i$.

Now apply the loads in the reverse order: first $dP_i$ alone, then $P_1\ldots P_n$. The energy is

$$\tfrac12\,dP_i\,d\delta_i + U + dP_i\,\delta_i$$

where the last term is the work of $dP_i$ moving through the displacement $\delta_i$ already caused by the other loads (it acts at full value). The final state and so the energy are independent of loading order. Equating, and neglecting the second-order term $dP_i\,d\delta_i$:

$$\frac{\partial U}{\partial P_i}\,dP_i = \delta_i\,dP_i \quad\Rightarrow\quad \delta_i = \frac{\partial U}{\partial P_i}$$

If no load exists at the point of interest, apply a fictitious load $Q$ (or moment $M_0$), differentiate, then set $Q = 0$.

For beams, differentiating under the integral sign:

$$\delta = \int \frac{M}{EI}\frac{\partial M}{\partial P}\,dx$$

Unit (dummy) load method

Apply a unit load (or unit moment for slope) at the point and in the direction of the required deflection, and let $m$ be the bending moment due to it. By the principle of virtual work (external work $1\cdot\delta$ = internal work):

$$1\cdot\delta = \int \frac{M\,m}{EI}\,dx$$

For trusses: $\delta = \sum \dfrac{F\,f\,L}{AE}$, where $F$ is the member force from the real load and $f$ from the unit load. Because the derivation uses only virtual work, the method can also be used for non-linear material by replacing $M/EI$ by the actual curvature.

Use

• Deflections of beams, bent frames, hooks, curved bars, trusses where double integration is lengthy.
• Deflection at a point where no load acts, or when the loading is discontinuous.
• Solution of statically indeterminate structures: choose the redundant $R$ and impose a compatibility condition, e.g. $\partial U/\partial R = 0$ when the support does not move.

Data

$L = 2$ m, $w = 10$ kN/m, $P = 8$ kN, $EI = 200\times10^{9}\times 8\times10^{-5} = 16\times10^{6}$ N m$^2$ $= 16000$ kN m$^2$.

Bending moment

Take $x$ from the free end. Add a fictitious couple $M_0$ at the free end for the slope (to be set to zero after differentiation). Let $P$ be treated as a variable load for the deflection.

$$M(x) = -\left(Px + \frac{w x^2}{2} + M_0\right)$$

(hogging; the sign cancels in $M\,\partial M/\partial P$.)

  P = 8 kN   w = 10 kN/m (all along)
   |  v v v v v v v v v v v v
   v
   o==============================# fixed end
   x = 0  <---------- 2 m ------>

Deflection at the free end

$$\frac{\partial M}{\partial P} = -x$$ $$\begin{aligned} \delta &= \frac{\partial U}{\partial P} = \frac{1}{EI}\int_0^L M\frac{\partial M}{\partial P}\,dx = \frac{1}{EI}\int_0^L \left(Px + \frac{wx^2}{2}\right)x\,dx \\ &= \frac{1}{EI}\left[\frac{PL^3}{3} + \frac{wL^4}{8}\right] \end{aligned}$$ $$\delta = \frac{1}{16000}\left[\frac{8\times2^3}{3} + \frac{10\times 2^4}{8}\right] = \frac{21.333 + 20}{16000} = 2.583\times10^{-3}\ \text{m}$$

Here $PL^3/3EI = 1.333$ mm (point load) and $wL^4/8EI = 1.250$ mm (UDL).

Slope at the free end

$$\frac{\partial M}{\partial M_0} = -1$$ $$\theta = \frac{\partial U}{\partial M_0}\Big|_{M_0=0} = \frac{1}{EI}\int_0^L\left(Px + \frac{wx^2}{2}\right)dx = \frac{1}{EI}\left[\frac{PL^2}{2} + \frac{wL^3}{6}\right]$$ $$\theta = \frac{16 + 13.333}{16000} = 0.00183\ \text{rad}$$

The positive signs mean that the free end moves in the direction of the load (downward) and rotates in the sense of the applied moment (clockwise).

Answer: Deflection at the free end $= 2.58$ mm downward; slope $= 0.00183$ rad ($0.105^\circ$).

Method

The beam is statically indeterminate to the first degree. Take the prop reaction $R_B$ as the redundant. Since the prop does not allow vertical movement at B:

$$\delta_B = \frac{\partial U}{\partial R_B} = 0$$

Bending moment

Take $x$ from B toward the fixed end A ($L = 4$ m, $w = 12$ kN/m):

$$M(x) = R_B\,x - \frac{w x^2}{2},\qquad \frac{\partial M}{\partial R_B} = x$$

 A (fixed)                       B
 |#==============================o   <- prop R_B
 |#  v v v v v v v v v v v v v
 |#  w = 12 kN/m,  L = 4 m

Compatibility

$$\begin{aligned} \frac{1}{EI}\int_0^L \left(R_B x - \frac{wx^2}{2}\right)x\,dx &= 0 \\ \frac{R_B L^3}{3} - \frac{wL^4}{8} &= 0 \\ R_B &= \frac{3wL}{8} = \frac{3\times12\times4}{8} = 18.0\ \text{kN} \end{aligned}$$

Reactions at A

$$\begin{aligned} R_A &= wL - R_B = 48 - 18 = 30.0\ \text{kN} \\ M_A &= \frac{wL^2}{2} - R_B L = \frac{12\times16}{2} - 18\times4 = 96 - 72 = 24.0\ \text{kN m (hogging)} \end{aligned}$$

Check: $M_A = wL^2/8 = 24$ kN m. $\checkmark$

Contraflexure

$M(x) = 18x - 6x^2 = 0 \Rightarrow x = 3$ m from B (i.e. 1 m from the fixed end).

Maximum sagging moment

Shear force $= R_B - wx = 0 \Rightarrow x = 18/12 = 1.5$ m from B:

$$M_{max} = 18(1.5) - 6(1.5)^2 = 27 - 13.5 = 13.5\ \text{kN m (sagging)}$$

Bending moment diagram (shape)

 -24 at A                  0 at B
  \                          .
   \____ 0 at x=3 m _____ +13.5 at x=1.5 m

Answer: $R_B = 18$ kN, $R_A = 30$ kN, $M_A = 24$ kN m (hogging), contraflexure 1 m from the fixed end, maximum sagging moment $13.5$ kN m at 1.5 m from the prop.

Data

$P = 5$ kN, $a = BC = 1.0$ m, $b = AB = 1.5$ m, $EI = 8000$ kN m$^2$.

  B ------------- C
  |      a        | P = 5 kN
  |               v
  | b
  |
 ###  A (fixed)

Real bending moments

• In BC (x measured from C, $0 \le x \le a$): $M = -Px$ (hogging).
• In AB (y measured downward from B, $0 \le y \le b$): the load acts at a lever arm $a$, so $M = -Pa$, constant.

Vertical deflection of C

Unit vertical load at C downward: $m_{BC} = -x$, $m_{AB} = -a$.

$$\begin{aligned} \delta_v &= \frac{1}{EI}\left[\int_0^a (Px)(x)\,dx + \int_0^b (Pa)(a)\,dy\right] \\ &= \frac{P}{EI}\left[\frac{a^3}{3} + a^2 b\right] \end{aligned}$$ $$\delta_v = \frac{5}{8000}\left[0.3333 + 1.5000\right] = \frac{5\times 1.8333}{8000} = 1.146\times10^{-3}\ \text{m}$$

Horizontal deflection of C

Unit horizontal load at C acting to the right. In BC the unit load passes along the member, so $m_{BC} = 0$. In AB it gives $m_{AB} = -y$ (arm $y$ measured from the level of C, which is the level of B plus $y$ at a section $y$ below B).

$$\delta_h = \frac{1}{EI}\int_0^b (Pa)(y)\,dy = \frac{Pa\,b^2}{2EI}$$

With $b^2/2 = 1.125$:

$$\delta_h = \frac{5\times1.0\times1.125}{8000} = 0.703\times10^{-3}\ \text{m}$$

The positive sign shows C moves in the direction of the unit load (to the right). The arm BC rotates with the top of the column, B tilts to the right, and C moves right as well.

Answer: Vertical deflection of C $= 1.15$ mm downward; horizontal deflection $= 0.70$ mm to the right.