Tapered bar element and a two-element bridge pier: 07-Str-B3, May 2017, Q1

Question 1: Tapered bar element and a two-element bridge pier (equal value)

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.

1.1 Stiffness matrix of a linearly tapered bar element

Given. A two-node axial element of length $L$ with one degree of freedom per node, a constant modulus $E$, and a cross-section that varies linearly from $A_0$ at node 1 to $A_1$ at node 2.

Find. The $2\times 2$ element stiffness matrix $[k]$ relating the end forces $\{f_1, f_2\}$ to the end displacements $\{u_1, u_2\}$.

Figure 1(a) — the tapered bar element, with the linear axial displacement field assumed between its two nodes.

Approach. Assume the standard two-node linear displacement field, form the strain–displacement matrix $[B]$, and integrate $[k]=\int_0^{L}[B]^{T}E\,A(x)\,[B]\,dx$ along the member — because $[B]$ is constant for a linear field, the taper survives only as the integral of the area.

Interpolate the axial displacement linearly between the nodes. With the origin at node 1, $$u(x)=N_1(x)\,u_1+N_2(x)\,u_2,\qquad N_1=1-\frac{x}{L},\quad N_2=\frac{x}{L}$$ where $u_1,u_2$ are the nodal axial displacements. This is the only field a two-node bar can represent, and it is the same field used for the prismatic element.
Differentiate to obtain the strain–displacement matrix. The axial strain is $\varepsilon = du/dx$, so $$\varepsilon = [B]\{d\},\qquad [B]=\left[\frac{dN_1}{dx}\ \ \frac{dN_2}{dx}\right]=\left[-\frac{1}{L}\ \ \frac{1}{L}\right]$$ Because the shape functions are linear, $[B]$ is constant along the element — the assumed strain is uniform even though the real strain in a tapered bar is not.
Write the area variation. A linear taper between the two end areas gives $$A(x)=A_0+\left(A_1-A_0\right)\frac{x}{L}$$ with $A(0)=A_0$ and $A(L)=A_1$ as required by Figure 1(a).
Integrate the stiffness definition. The element stiffness follows from the strain energy, $[k]=\int_{0}^{L}[B]^{T}E\,A(x)\,[B]\,dx$. Since $[B]$ and $E$ are constant they come outside the integral: $$[k]=\frac{E}{L^{2}}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\int_{0}^{L}A(x)\,dx$$ Only the area integral — that is, the volume of the element — carries the taper.
Evaluate the area integral. Integrating the linear variation, $$\int_{0}^{L}\!\!\left[A_0+\left(A_1-A_0\right)\frac{x}{L}\right]dx = A_0 L + \left(A_1-A_0\right)\frac{L}{2} = \frac{\left(A_0+A_1\right)L}{2}$$ which is simply the mean area times the length.
Assemble the result. Substituting the area integral back, $$\boxed{\;[k]=\frac{E\left(A_0+A_1\right)}{2L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}=\frac{E\bar{A}}{L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix},\qquad \bar{A}=\frac{A_0+A_1}{2}\;}$$ The tapered element is therefore identical to a prismatic element carrying the arithmetic-mean area. Setting $A_1=A_0$ recovers $EA/L$ exactly, as it must.
Check the matrix. Its determinant vanishes and its rows sum to zero, so $[k]\{1,1\}^{T}=\{0,0\}^{T}$: a rigid-body translation of both nodes generates no forces, which is the necessary singularity of an unsupported element stiffness matrix. It is also symmetric and positive semi-definite, as any stiffness derived from a strain-energy functional must be.

Final results — part 1.1
Quantity	Result
Shape functions	$N_1 = 1-x/L$, $N_2 = x/L$
Strain–displacement matrix	$[B]=[-1/L\ \ 1/L]$ (constant)
Effective area	$\bar{A}=(A_0+A_1)/2$ (arithmetic mean)
Element stiffness matrix	$[k]=\dfrac{E(A_0+A_1)}{2L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$
Rank / rigid-body check	$\det[k]=0$; $[k]\{1,1\}^{T}=\{0,0\}^{T}$

Topic. Direct derivation of a variable-section bar element from the principle of minimum potential energy.

Key relations. $u=[N]\{d\}$; $\varepsilon=[B]\{d\}$ with $[B]=d[N]/dx$; $[k]=\int_V [B]^{T}[D][B]\,dV = \int_0^L [B]^{T}E A(x)[B]\,dx$.

Why this works. The element stiffness is the second variation of the strain energy with respect to the nodal degrees of freedom, so it is completely determined once the displacement field is chosen. A two-node element admits only a linear field, which makes $[B]$ constant and pulls it out of the integral; every trace of the geometry then collapses into $\int A\,dx$. That is why a linearly tapered bar behaves exactly like a prismatic bar of the mean area — not because the taper is unimportant, but because the assumed uniform strain cannot see anything finer than the element volume. Logan §3.10 and Chandrupatla & Belegundu §3.6 derive the same matrix by this route.

Common pitfall. Writing $[k]=EA_{\text{avg}}/L$ by inspection and calling it exact. The exact stiffness of a tapered bar, obtained from the closed-form solution $u(x)=\int N\,dx/(EA)$, is $E(A_1-A_0)/\!\left[L\ln(A_1/A_0)\right]$ — the logarithmic mean area, which is always smaller than the arithmetic mean. The finite element is therefore always slightly too stiff, and refining the mesh is what closes that gap. A second pitfall is to integrate $A(x)$ but forget that $[B]^{T}[B]$ has already introduced the $1/L^{2}$, and so to divide by $L$ twice.

Source. Logan, A First Course in the Finite Element Method, 6th ed., §3.1–3.2 and §3.10; Cook et al., Concepts and Applications of FEA, 4th ed., §2.4; Przemieniecki, Theory of Matrix Structural Analysis, §5.3 (exact tapered-member stiffness).

1.2 Two-element analysis of the bridge pier

Given. The pier of Figure 1(b): a solid concrete member 2 m high, trapezoidal in front elevation from 0.5 m wide at the top to 1.5 m wide at the base, and of constant 0.5 m thickness out of plane. Deck plus traffic apply a uniform $20\ \text{kN/m}^{2}$ pressure over the pier head.

Given data
Quantity	Symbol	Value
Pier height	$H$	2.0 m
Head width × thickness	—	0.5 m × 0.5 m
Base width × thickness	—	1.5 m × 0.5 m
Deck + traffic pressure	$q$	20 kN/m$^{2}$
Unit weight of concrete	$\gamma$	25 kN/m$^{3}$
Modulus of elasticity	$E$	$28\times10^{6}$ kN/m$^{2}$
Mesh	—	2 tapered bar elements, 3 nodes at $x=0,1,2$ m

Find. The displacement field and the stress field of the pier from a two-element model, and the error of both against the exact (continuum) solution.

Figure 1(b) — the pier as drawn (front elevation and side view) and its two-element idealisation. The coordinate $x$ runs downward from the pier head; the base is fully restrained.

Approach. Build $A(x)$ from the two views, reduce the deck pressure and the self-weight to nodal forces (the self-weight through work-equivalent loads on each element), solve the $2\times2$ reduced system with the tapered-element stiffness of part 1.1, and compare against the closed-form axial solution obtained by integrating the equilibrium equation.

Establish the cross-sectional area as a function of depth. With $x$ measured downward from the head, the front-elevation width grows linearly from 0.5 m to 1.5 m over 2 m while the thickness stays at 0.5 m, so $$A(x)=0.5\left(0.5+\frac{1.5-0.5}{2}x\right)=0.25+0.25x \ \ \text{m}^{2}$$ giving $A(0)=0.25\ \text{m}^{2}$, $A(1)=0.50\ \text{m}^{2}$ and $A(2)=0.75\ \text{m}^{2}$. The area is linear in $x$, so the element of part 1.1 applies without approximation of the geometry.
Reduce the deck pressure to a concentrated head load. The pressure acts over the $0.5\ \text{m}\times0.5\ \text{m}$ head: $$P=q\,A(0)=20\times0.25=5.0\ \text{kN}$$
Compute the self-weight, which dominates here. The volume is $\int_0^2 A\,dx=\left[0.25x+0.125x^{2}\right]_0^2=1.00\ \text{m}^{3}$, so $$W=\gamma\!\int_0^{2}\!A\,dx = 25\times1.00 = 25.0\ \text{kN}$$ The base must therefore carry $\boxed{P+W=30.0\ \text{kN}}$ — five times the deck load. Any solution that omits the self-weight is wrong by a factor of six at the base.
Write the exact axial force and stress fields. Cutting at depth $x$ and summing the load above the cut, $$N(x)=P+\gamma\!\int_0^{x}\!A(\xi)\,d\xi = 5+25\left(0.25x+0.125x^{2}\right)=5+6.25x+3.125x^{2}\ \text{kN}$$ $$\sigma(x)=\frac{N(x)}{A(x)}=\frac{12.5x^{2}+25x+20}{x+1}\ \text{kPa}$$ which gives 20.00 kPa at the head, 28.75 kPa at mid-height and 40.00 kPa at the base. The head value must equal the applied pressure of 20 kPa, and it does — a useful check on the algebra.
Integrate to the exact displacement field. With the base fixed and $u$ measured downward, $du/dx=-\sigma/E$, so $u(x)=\frac{1}{E}\int_x^{2}\sigma(\xi)\,d\xi$. Polynomial division gives $\sigma=12.5x+12.5+7.5/(x+1)$, hence $$u(x)=\frac{1}{E}\Big[6.25\xi^{2}+12.5\xi+7.5\ln(\xi+1)\Big]_{x}^{2}$$ Evaluating, $u(0)=58.2396/E=2.0800\times10^{-6}$ m and $u(1)=34.2914/E=1.2247\times10^{-6}$ m. The logarithm is the fingerprint of the taper; a prismatic member would give a pure quadratic.
Form the two element stiffnesses. Each element is 1 m long, and part 1.1 says to use the mean area: $$\bar{A}_1=\tfrac{0.25+0.50}{2}=0.375\ \text{m}^{2},\qquad \bar{A}_2=\tfrac{0.50+0.75}{2}=0.625\ \text{m}^{2}$$ $$k_1=\frac{E\bar{A}_1}{L_e}=10.5\times10^{6}\ \text{kN/m},\qquad k_2=\frac{E\bar{A}_2}{L_e}=17.5\times10^{6}\ \text{kN/m}$$
Convert the self-weight to work-equivalent nodal loads. The body force per unit length is $w(x)=\gamma A(x)$, and the consistent load at node $i$ of an element is $f_i=\int N_i\,w\,dx$. For element 1, $w=6.25+6.25x$ on $0\le x\le1$: $$f_1^{(1)}=\int_0^{1}(1-x)\,w\,dx=4.1667\ \text{kN},\qquad f_2^{(1)}=\int_0^{1}x\,w\,dx=5.2083\ \text{kN}$$ and for element 2, with $w=12.5+6.25s$ on $0\le s\le1$, $$f_1^{(2)}=7.2917\ \text{kN},\qquad f_2^{(2)}=8.3333\ \text{kN}$$ Each pair sums to its own element weight (9.375 kN and 15.625 kN), so nothing has been lost. Notice that the split is not half-and-half: the heavier lower half of each element pushes more load onto its lower node.
Assemble the global load vector. Adding the deck load at node 1 and the shared element contributions at node 2, $$F_1=5+4.1667=9.1667\ \text{kN},\quad F_2=5.2083+7.2917=12.5000\ \text{kN},\quad F_3=8.3333\ \text{kN}$$ and $F_1+F_2+F_3=30.0$ kN, matching step 3.
Solve the reduced system. Node 3 is fixed, so the two free equations are $k_1(u_1-u_2)=F_1$ and $-k_1u_1+(k_1+k_2)u_2=F_2$. Adding them eliminates $u_1$ and gives $k_2u_2=F_1+F_2$, so $$u_2=\frac{21.6667}{17.5\times10^{6}}=1.2381\times10^{-6}\ \text{m},\qquad u_1=u_2+\frac{F_1}{k_1}=\boxed{2.1111\times10^{-6}\ \text{m}}$$ Physically, the second equation is just vertical equilibrium of everything above the mid-height node.
Recover the element stresses. A constant-strain element gives one stress per element, $\sigma_e=E\,(u_{j}-u_{i})/L_e$: $$\sigma_1=28\times10^{6}\times\frac{1.2381-2.1111}{1}\times10^{-6}=-24.44\ \text{kPa},\qquad \sigma_2=-34.67\ \text{kPa}$$ both compressive, as expected. The base reaction recovered from element 2 is $k_2u_2=21.67$ kN, which added to the 8.33 kN of element self-weight already sitting at node 3 returns the 30 kN of step 3.

Compare with the exact solution. The displacements are good and the stresses are good only where they are sampled correctly:

Two-element model against the exact solution
Quantity	Finite element	Exact	Error
$u$ at the head, $x=0$	$2.1111\times10^{-6}$ m	$2.0800\times10^{-6}$ m	+1.50 %
$u$ at mid-height, $x=1$ m	$1.2381\times10^{-6}$ m	$1.2247\times10^{-6}$ m	+1.09 %
$\sigma$ at element-1 mid-length, $x=0.5$ m	24.44 kPa	23.75 kPa	+2.92 %
$\sigma$ at element-2 mid-length, $x=1.5$ m	34.67 kPa	34.25 kPa	+1.22 %
$\sigma$ at the head, $x=0$	24.44 kPa	20.00 kPa	+22.2 %
$\sigma$ at the base, $x=2$ m	34.67 kPa	40.00 kPa	−13.3 %

The element stress is within about 3 % of the truth at the element mid-length and more than 20 % out at the ends — the classic signature of a constant-strain element, and the reason the peak stress at the base is under-predicted rather than over-predicted.

Exact fields against the two-element solution. The element stresses are step-constant and cross the exact curve near each element mid-length; the nodal displacements are accurate to about 1.5 % and interpolate linearly in between.

Final results — part 1.2
Quantity	Symbol	Value
Deck + traffic load on the head	$P$	5.00 kN
Self-weight of the pier	$W$	25.00 kN
Total base reaction	$R$	30.00 kN
Element stiffnesses	$k_1,\ k_2$	$10.5\times10^{6}$, $17.5\times10^{6}$ kN/m
Work-equivalent nodal loads	$F_1,F_2,F_3$	9.167, 12.500, 8.333 kN
Head settlement (FE)	$u_1$	$2.111\times10^{-6}$ m (exact $2.080\times10^{-6}$ m)
Mid-height settlement (FE)	$u_2$	$1.238\times10^{-6}$ m (exact $1.225\times10^{-6}$ m)
Element stresses (compression)	$\sigma_1,\ \sigma_2$	24.44 kPa, 34.67 kPa
Exact stress range	$\sigma(0)\ldots\sigma(H)$	20.00 kPa to 40.00 kPa
Worst displacement error	—	+1.5 % (head)

Topic. One-dimensional finite element analysis of a variable-section member under a body force, and where a constant-strain element may legitimately be believed.

Key relations. $A(x)=A_0+(A_1-A_0)x/L$; element stiffness $k_e=E\bar{A}_e/L_e$; work-equivalent body-force loads $f_i=\int N_i\,\gamma A\,dx$; element stress $\sigma_e=E(u_j-u_i)/L_e$; exact field $N(x)=P+\gamma\!\int_0^x\!A\,d\xi$, $u(x)=\frac{1}{E}\int_x^H \sigma\,d\xi$.

Why this works. A displacement-based element is exact in the energy norm over the space it can represent, so the nodal displacements converge fast — here two elements already give 1.5 % — while the stresses, which are one derivative down, converge one order more slowly. There is a compensating gift: for a linear-displacement bar the element stress equals the exact stress at the point where the assumed uniform strain crosses the true strain, and for a linearly tapered member sampled at the mean area that point is the element mid-length. Those mid-length points are the superconvergent (Barlow) stress points, and Cook et al. §9.9 shows that sampling there rather than at the nodes is worth roughly one mesh refinement for free. The work-equivalent load vector matters just as much: lumping the self-weight half-and-half at the nodes would put 4.69 kN rather than 4.17 kN at the head node and spoil the displacement to no purpose, since the consistent vector costs one extra integral.

Common pitfall. Reading the pier as a solid of revolution or applying the trapezoid to both views, which makes $A(x)$ quadratic and puts every number out; and reporting the FE stress at the base as the design stress. The FE base stress of 34.67 kPa is 13 % below the true 40.00 kPa, so a candidate who trusts it is unconservative exactly where the section is most heavily loaded. Extrapolate from the mid-length values, or refine the mesh near the base.

Source. Logan, 6th ed., §3.10 (tapered bar) and §3.11 (work-equivalent body-force loads); Cook, Malkus, Plesha & Witt, 4th ed., §9.9 (optimal stress-sampling points) and §9.4 (convergence rates); Chandrupatla & Belegundu, 4th ed., §3.6.

Question 1 of 3: Tapered bar element and a two-element bridge pier

Question 1: Tapered bar element and a two-element bridge pier (equal value)

1.1 Stiffness matrix of a linearly tapered bar element

1.2 Two-element analysis of the bridge pier