BE Civil Engineering (IOE, TU) Theory of Structures II (IOE, CE 602) Question Paper 2078 Nepal

Setup — choose redundant

The propped cantilever is statically indeterminate to the first degree. Release the prop reaction $R_B$ (vertical) as the redundant. The primary structure is a cantilever fixed at $A$, free at $B$.

Compatibility condition: the vertical deflection at $B$ in the released structure must be zero (because the prop prevents deflection):

$$\delta_{B0} + R_B \, \delta_{BB} = 0$$

Step 1 — Deflection at free end due to UDL ($\delta_{B0}$, downward positive)

For a cantilever with UDL $w$ over span $L$:

$$\delta_{B0} = \frac{wL^4}{8EI}$$

Step 2 — Deflection at free end due to unit upward load at $B$ ($\delta_{BB}$)

For a cantilever with point load at the free end:

$$\delta_{BB} = \frac{L^3}{3EI}$$

Step 3 — Apply compatibility (downward = positive)

Net deflection at $B = 0$:

$$\frac{wL^4}{8EI} - R_B \frac{L^3}{3EI} = 0$$ $$R_B = \frac{wL^4/8}{L^3/3} = \frac{3wL}{8}$$

Substitute $w = 15\,\text{kN/m}$, $L = 8\,\text{m}$:

$$R_B = \frac{3 \times 15 \times 8}{8} = \frac{360}{8} = 45\,\text{kN (upward)}$$

(a) Prop reaction $R_B = \mathbf{45\,kN\ (upward)}$

Step 4 — Reaction at $A$ (vertical equilibrium)

Total load $= wL = 15 \times 8 = 120\,\text{kN}$.

$$R_A = 120 - 45 = 75\,\text{kN (upward)}$$

Step 5 — Fixing moment at $A$

Standard result $M_A = \dfrac{wL^2}{8}$ (hogging). Verify by taking moments of all forces about $A$:

$$M_A = wL\cdot\frac{L}{2} - R_B\cdot L = 120 \times 4 - 45 \times 8 = 480 - 360 = 120\,\text{kN·m (hogging)}$$

Check: $\dfrac{wL^2}{8} = \dfrac{15 \times 64}{8} = 120\,\text{kN·m}$. ✓

(b) Fixing moment at $A = \mathbf{120\,kN\cdot m\ (hogging)}$

Step 6 — Bending moment diagram and maximum sagging moment

Measure $x$ from prop $B$. Bending moment (sagging positive):

$$M(x) = R_B x - \frac{w x^2}{2} = 45x - 7.5x^2$$

Maximum-moment location: $\dfrac{dM}{dx} = 45 - 15x = 0 \Rightarrow x = 3\,\text{m}$ from $B$.

$$M_{max} = 45(3) - 7.5(3)^2 = 135 - 67.5 = 67.5\,\text{kN·m (sagging)}$$

Point of contraflexure (M = 0): $45x - 7.5x^2 = 0 \Rightarrow x(45 - 7.5x) = 0 \Rightarrow x = 6\,\text{m}$ from $B$ (i.e. 2 m from A), plus at $B$ itself.

BMD (schematic):

 A (-120 kN·m hogging)
  |\
  | \______ point of contraflexure at 6 m from B (2 m from A)
  |        \___
  |   sagging    \___
  |  +67.5 max         \___ B (0)
  +------------------------+
   x=3 m from B (peak sag)

(c) Hogging $120\,\text{kN·m}$ at $A$, zero at $B$, maximum sagging $\mathbf{67.5\,kN\cdot m}$ at $\mathbf{3\,m}$ from $B$, point of contraflexure $2\,\text{m}$ from $A$.

Joints/unknowns: $A$ is fixed $\Rightarrow \theta_A = 0$. $C$ is a simple end support. Unknown rotations: $\theta_B$ and $\theta_C$. No sway. $EI$ constant. (Clockwise member-end moment positive.)

Step 1 — Fixed-end moments (FEM)

Span $AB$ (UDL, $w=20$, $L=6$):

$$\text{FEM}_{AB} = -\frac{wL^2}{12} = -\frac{20 \times 36}{12} = -60\,\text{kN·m}, \quad \text{FEM}_{BA} = +60\,\text{kN·m}$$

Span $BC$ (central point load, $P=40$, $L=4$):

$$\text{FEM}_{BC} = -\frac{PL}{8} = -\frac{40 \times 4}{8} = -20\,\text{kN·m}, \quad \text{FEM}_{CB} = +20\,\text{kN·m}$$

Step 2 — Slope-deflection equations ($M = \text{FEM} + \frac{2EI}{L}(2\theta_{near} + \theta_{far})$, no sway)

$$M_{AB} = -60 + \frac{2EI}{6}(2\theta_A + \theta_B) = -60 + \tfrac{EI}{3}\theta_B$$ $$M_{BA} = +60 + \frac{2EI}{6}(2\theta_B + \theta_A) = 60 + \tfrac{2EI}{3}\theta_B$$ $$M_{BC} = -20 + \frac{2EI}{4}(2\theta_B + \theta_C) = -20 + EI\theta_B + \tfrac{EI}{2}\theta_C$$ $$M_{CB} = +20 + \frac{2EI}{4}(2\theta_C + \theta_B) = 20 + EI\theta_C + \tfrac{EI}{2}\theta_B$$

Step 3 — Equilibrium / boundary conditions

At joint $B$: $M_{BA} + M_{BC} = 0$. At end $C$ (simple support, no applied moment): $M_{CB} = 0$.

From $M_{CB} = 0$:

$$20 + EI\theta_C + \tfrac{EI}{2}\theta_B = 0 \Rightarrow EI\theta_C = -20 - \tfrac{EI}{2}\theta_B \quad (i)$$

Joint $B$:

$$\left(60 + \tfrac{2EI}{3}\theta_B\right) + \left(-20 + EI\theta_B + \tfrac{EI}{2}\theta_C\right) = 0$$ $$40 + \tfrac{5EI}{3}\theta_B + \tfrac{1}{2}(EI\theta_C) = 0$$

Substitute $(i)$:

$$40 + \tfrac{5EI}{3}\theta_B + \tfrac{1}{2}\left(-20 - \tfrac{EI}{2}\theta_B\right) = 0$$ $$40 + \tfrac{5EI}{3}\theta_B - 10 - \tfrac{EI}{4}\theta_B = 0$$ $$30 + EI\theta_B\left(\tfrac{5}{3} - \tfrac14\right) = 0, \quad \tfrac{5}{3} - \tfrac14 = \tfrac{20 - 3}{12} = \tfrac{17}{12}$$ $$30 + \tfrac{17}{12}EI\theta_B = 0 \Rightarrow EI\theta_B = -\frac{360}{17} = -21.176\,\text{kN·m}^2$$

From $(i)$: $EI\theta_C = -20 - \tfrac12(-21.176) = -20 + 10.588 = -9.412\,\text{kN·m}^2$

(b) $EI\theta_B = -21.18\,\text{kN·m}^2$, $EI\theta_C = -9.41\,\text{kN·m}^2$.

Step 4 — Member-end moments

$$M_{AB} = -60 + \tfrac{1}{3}(-21.176) = -60 - 7.059 = \mathbf{-67.06\,kN\cdot m}$$ $$M_{BA} = 60 + \tfrac{2}{3}(-21.176) = 60 - 14.118 = \mathbf{+45.88\,kN\cdot m}$$ $$M_{BC} = -20 + (-21.176) + \tfrac12(-9.412) = -20 - 21.176 - 4.706 = \mathbf{-45.88\,kN\cdot m}$$ $$M_{CB} = 20 + (-9.412) + \tfrac12(-21.176) = 20 - 9.412 - 10.588 = \mathbf{0\,kN\cdot m} \;✓$$

Check joint B: $M_{BA} + M_{BC} = 45.88 - 45.88 = 0$ ✓

(c) Final member-end moments:

Step 1 — Fixed-end moments

Span $AB$ (central point load): $\text{FEM}_{AB} = -\dfrac{PL}{8} = -\dfrac{60 \times 5}{8} = -37.5\,\text{kN·m}$, $\text{FEM}_{BA} = +37.5\,\text{kN·m}$.

Span $BC$ (UDL): $\text{FEM}_{BC} = -\dfrac{wL^2}{12} = -\dfrac{18 \times 16}{12} = -24\,\text{kN·m}$, $\text{FEM}_{CB} = +24\,\text{kN·m}$.

Step 2 — Stiffness and distribution factors at B

Span $BA$: far end $A$ fixed $\Rightarrow$ stiffness $= \dfrac{4EI}{5} = 0.80EI$.

Span $BC$: far end $C$ simply supported $\Rightarrow$ modified stiffness $\dfrac{3EI}{4} = 0.75EI$.

Total $= 1.55EI$.

$$DF_{BA} = \frac{0.80}{1.55} = 0.516, \qquad DF_{BC} = \frac{0.75}{1.55} = 0.484$$

Step 3 — Release end C first. Because the modified stiffness assumes a pin at $C$, first release $C$: its unbalanced moment $+24$ is removed, carrying over half to $B$. Carry-over to $B$: $\tfrac12 \times (-24) = -12\,\text{kN·m}$ (we add the balancing of $C$, which carries $\tfrac12$ of the released moment). New $\text{FEM}_{BC} = -24 - 12 = -36\,\text{kN·m}$, and $M_{CB} \to 0$.

Step 4 — Moment distribution table

Step 5 — Check joint B equilibrium

$$M_{BA} + M_{BC} = 36.73 - 36.73 = 0 \;✓$$

(c) Final end moments

$$M_{AB} = \mathbf{-37.89\,kN\cdot m}, \quad M_{BA} = \mathbf{+36.73\,kN\cdot m}$$ $$M_{BC} = \mathbf{-36.73\,kN\cdot m}, \quad M_{CB} = \mathbf{0\,kN\cdot m}$$

Step 1 — Geometry (lengths & direction cosines)

Member 1, $A(0,0)\to C(0,3)$: $L_1 = \sqrt{0^2 + 3^2} = 3\,\text{m}$. Direction cosines (A→C): $c_1 = 0/3 = 0$, $s_1 = 3/3 = 1$ (vertical member).

Member 2, $B(4,0)\to C(0,3)$: $L_2 = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{16+9} = 5\,\text{m}$. Direction cosines (B→C): $c_2 = (0-4)/5 = -0.8$, $s_2 = (3-0)/5 = 0.6$.

Step 2 — Member stiffness contributions at joint $C$

$$k = \frac{AE}{L}\begin{bmatrix} c^2 & cs \\ cs & s^2 \end{bmatrix}$$

Member 1: $\dfrac{AE}{L_1} = \dfrac{60000}{3} = 20000\,\text{kN/m}$.

$$k_1 = 20000\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 20000 \end{bmatrix}$$

Member 2: $\dfrac{AE}{L_2} = \dfrac{60000}{5} = 12000\,\text{kN/m}$, with $c^2 = 0.64$, $s^2 = 0.36$, $cs = (-0.8)(0.6) = -0.48$:

$$k_2 = 12000\begin{bmatrix} 0.64 & -0.48 \\ -0.48 & 0.36 \end{bmatrix} = \begin{bmatrix} 7680 & -5760 \\ -5760 & 4320 \end{bmatrix}$$

Step 3 — Reduced global stiffness matrix (free DOFs $u_C, v_C$)

$$K = k_1 + k_2 = \begin{bmatrix} 7680 & -5760 \\ -5760 & 24320 \end{bmatrix}\,\text{kN/m}$$

Step 4 — Load vector and solve

$F = \begin{bmatrix} 0 \\ -90 \end{bmatrix}\,\text{kN}$. Solve $K d = F$.

$\det K = 7680\times24320 - (-5760)^2 = 186{,}777{,}600 - 33{,}177{,}600 = 153{,}600{,}000$.

$$u_C = \frac{24320(0) - (-5760)(-90)}{\det K} = \frac{-518400}{153600000} = -0.003375\,\text{m}$$ $$v_C = \frac{-(-5760)(0) + 7680(-90)}{\det K} = \frac{-691200}{153600000} = -0.004500\,\text{m}$$

(c) Displacements: $u_C = \mathbf{-3.375\,mm}$, $v_C = \mathbf{-4.500\,mm}$.

Step 5 — Axial forces $\;N = \dfrac{AE}{L}(c\,u_C + s\,v_C)$ (supports fixed).

Member 1 ($c=0, s=1$): $N_1 = 20000(0 + 1(-0.0045)) = -90\,\text{kN}$ → 90 kN compression.

Member 2 ($c=-0.8, s=0.6$): $N_2 = 12000(-0.8(-0.003375) + 0.6(-0.0045)) = 12000(0.0027 - 0.0027) = 0\,\text{kN}$.

Check joint $C$ vertical equilibrium: member 1 vertical carries $90$ kN (compression pushing up on $C$? no — member 1 from $A$ below carries $90$ kN that balances the $90$ kN downward load); member 2 carries $0$. Equilibrium satisfied. ✓

Axial forces: $N_1 = \mathbf{90\,kN\ (compression)}$, $N_2 = \mathbf{0\,kN}$.

Setup. Fixed-base portal, uniform $M_p$. Possible plastic hinge locations: bases $A$, $D$; corners $B$, $C$; beam mid-span $E$.

(a) Mechanisms: (1) Beam, (2) Sway, (3) Combined.

Mechanism 1 — Beam mechanism

Hinges at $B$, $E$, $C$. Beam rotation $\theta$ at ends; mid-span deflection $\delta = \tfrac{L}{2}\theta = 3\theta$; hinge at $E$ rotates $2\theta$. Internal work $= M_p(\theta) + M_p(2\theta) + M_p(\theta) = 4M_p\theta$. External work $= W\delta = 80(3\theta) = 240\theta$.

$$4M_p\theta = 240\theta \Rightarrow M_p = 60\,\text{kN·m}$$

Mechanism 2 — Sway mechanism

Hinges at $A$, $B$, $C$, $D$; sway $\Delta = h\theta = 4\theta$. Internal work $= 4M_p\theta$. External work $= H\Delta = 40(4\theta) = 160\theta$.

$$4M_p\theta = 160\theta \Rightarrow M_p = 40\,\text{kN·m}$$

Mechanism 3 — Combined mechanism

Add beam + sway; the hinge at $B$ cancels (equal and opposite rotations). Remaining hinges: $A(\theta)$, $D(\theta)$, $C(2\theta)$, $E(2\theta)$. External work $= H(h\theta) + W(\tfrac{L}{2}\theta) = 160\theta + 240\theta = 400\theta$. Internal work $= M_p(\theta + \theta + 2\theta + 2\theta) = 6M_p\theta$. (Equivalently $4M_p\theta + 4M_p\theta - 2M_p\theta = 6M_p\theta$.)

$$6M_p\theta = 400\theta \Rightarrow M_p = \frac{400}{6} = 66.67\,\text{kN·m}$$

(c) Governing mechanism

Required $M_p$: beam $60$, sway $40$, combined $66.67\,\text{kN·m}$. For the given loads the section must satisfy the largest required $M_p$, so the combined mechanism governs.

$$\boxed{M_p = 66.67\,\text{kN·m, governed by the COMBINED mechanism}}$$

(Equivalently, for a fixed $M_p$ the combined mechanism gives the lowest collapse load factor and hence governs collapse.)

Definitions

• Static indeterminacy ($D_s$): the number of unknown forces/reactions in excess of the available equations of static equilibrium; the number of redundant force quantities that must be released to render the structure statically determinate.
• Kinematic indeterminacy ($D_k$): the number of independent unknown joint displacement components (degrees of freedom).

Static indeterminacy

For a plane rigid frame: $D_s = 3m + r - 3j$ (m members, r reaction components, j joints). Here $m = 3$, $j = 4$, $r = 3 + 3 = 6$ (two fixed supports).

$$D_s = 3(3) + 6 - 3(4) = 9 + 6 - 12 = 3$$

A single-bay single-storey fixed-base portal is a closed frame, confirming $\mathbf{D_s = 3}$.

Kinematic indeterminacy (axial deformation neglected)

Free joints: $B$, $C$ (bases fixed). Unknowns:

• Rotations $\theta_B$, $\theta_C$ → 2.
• Translations: inextensible columns prevent vertical movement of $B$, $C$; the inextensible beam forces $B$ and $C$ to share the same horizontal sway → 1 independent sway.

$$D_k = 2 + 1 = 3$$ $$\boxed{D_s = 3, \quad D_k = 3 \text{ (axial deformation neglected)}}$$

(If axial deformation is included, $D_k = 3 \times 2 = 6$.)

Theorem of three moments (Clapeyron), constant $EI$, no settlement

For three consecutive supports $A$, $B$, $C$ on spans $L_1 = AB$, $L_2 = BC$:

$$M_A L_1 + 2M_B(L_1 + L_2) + M_C L_2 = -6\left(\frac{A_1 \bar{x}_1}{L_1} + \frac{A_2 \bar{x}_2}{L_2}\right)$$

where $A\bar{x}$ are moment-area terms of the free (simply-supported) BMD of each span. For a span with UDL $w$ over length $L$, $\dfrac{6A\bar{x}}{L} = \dfrac{wL^3}{4}$.

Application

End supports simply supported $\Rightarrow M_A = M_C = 0$. Equal spans $L = 5\,\text{m}$, $w = 12\,\text{kN/m}$.

$$2M_B(L + L) = -\left(\frac{wL^3}{4} + \frac{wL^3}{4}\right)$$ $$4 M_B L = -\frac{wL^3}{2} \Rightarrow M_B = -\frac{wL^2}{8}$$ $$M_B = -\frac{12 \times 25}{8} = -\frac{300}{8} = -37.5\,\text{kN·m}$$ $$\boxed{M_B = -37.5\,\text{kN·m (hogging)}}$$

Portal method — concept and assumptions

An approximate method for laterally-loaded building frames. Key assumptions:

1. A point of contraflexure occurs at the mid-height of every column.
2. A point of contraflexure occurs at the mid-span of every beam.
3. Storey shear is shared so that each interior column carries twice the shear of an exterior column (shear proportional to bays supported).

Best suited to low-to-medium-rise frames where shear behaviour dominates.

Application — single bay (two exterior columns)

Total storey shear $V = 30\,\text{kN}$, resisted by 2 exterior columns; with only exterior columns the shear splits equally:

$$V_{col} = \frac{30}{2} = 15\,\text{kN per column}$$

Inflection point at mid-height ($h/2 = 2\,\text{m}$). Base moment of each column:

$$M_{base} = V_{col} \times \frac{h}{2} = 15 \times 2 = 30\,\text{kN·m}$$ $$\boxed{\text{Shear per column} = 15\,\text{kN}, \quad \text{Base moment per column} = 30\,\text{kN·m}}$$

The top-of-column moment is equal in magnitude ($30\,\text{kN·m}$) and opposite in sense, consistent with the mid-height inflection point.

Concept. For a fixed beam, a relative support settlement $\Delta$ induces equal-magnitude end moments. From slope-deflection with $\theta_A = \theta_B = 0$ and chord rotation $\psi = \Delta/L$:

$$M_{AB} = \frac{2EI}{L}\left(2\theta_A + \theta_B - 3\psi\right) = -\frac{6EI\Delta}{L^2}, \qquad M_{BA} = -\frac{6EI\Delta}{L^2}$$

so $|M_{AB}| = |M_{BA}| = \dfrac{6EI\Delta}{L^2}$.

Numerical evaluation

$\Delta = 10\,\text{mm} = 0.010\,\text{m}$, $L = 6\,\text{m}$, $EI = 24000\,\text{kN·m}^2$:

$$|M| = \frac{6 \times 24000 \times 0.010}{6^2} = \frac{1440}{36} = 40\,\text{kN·m}$$ $$\boxed{M_{AB} = M_{BA} = 40\,\text{kN·m (magnitude)}}$$

With $B$ settling, the moment at $A$ is hogging and at $B$ is sagging; the beam takes an S-shape with a point of contraflexure at mid-span ($x = 3\,\text{m}$).

Definitions

• Shape factor ($S$): $S = M_p/M_y = Z_p/Z_e$ — the ratio of plastic moment to first-yield moment (equivalently plastic to elastic section modulus); a measure of reserve strength beyond first yield.
• Plastic moment capacity ($M_p$): the moment at which the whole section has yielded, $M_p = f_y Z_p$.

Shape factor of a rectangle

Elastic modulus $Z_e = \dfrac{bd^2}{6}$; plastic modulus $Z_p = \dfrac{bd^2}{4}$ (two half-areas $\tfrac{bd}{2}$ each at lever arm $\tfrac{d}{4}$: $2\cdot\tfrac{bd}{2}\cdot\tfrac{d}{4} = \tfrac{bd^2}{4}$).

$$S = \frac{bd^2/4}{bd^2/6} = \frac{6}{4} = 1.5$$

Plastic moment of 200 × 300 mm section

$$Z_p = \frac{bd^2}{4} = \frac{200 \times 300^2}{4} = \frac{200 \times 90000}{4} = 4.5\times10^6\,\text{mm}^3$$

$f_y = 250\,\text{N/mm}^2$:

$$M_p = f_y Z_p = 250 \times 4.5\times10^6 = 1.125\times10^9\,\text{N·mm} = 1125\,\text{kN·m}$$ $$\boxed{S = 1.5, \quad M_p = 1125\,\text{kN·m}}$$

Check: $M_y = f_y Z_e = 250 \times 3\times10^6 = 750\,\text{kN·m}$, and $M_p/M_y = 1125/750 = 1.5$ ✓

Beam element stiffness matrix (Euler–Bernoulli, 2 nodes × 2 DOF)

DOF order $\{v_1, \theta_1, v_2, \theta_2\}$:

$$[k] = \frac{EI}{L^3}\begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^2 & -6L & 2L^2 \\ -12 & -6L & 12 & -6L \\ 6L & 2L^2 & -6L & 4L^2 \end{bmatrix}$$

It is symmetric and singular (rigid-body translation and rotation give zero force).

Leading coefficient $k_{11}$

$$k_{11} = \frac{12EI}{L^3}$$

(the shear force at end 1 for a unit transverse displacement there, other DOFs zero).

Numerical value ($EI = 32000\,\text{kN·m}^2$, $L = 4\,\text{m}$, $L^3 = 64\,\text{m}^3$):

$$k_{11} = \frac{12 \times 32000}{64} = \frac{384000}{64} = 6000\,\text{kN/m}$$ $$\boxed{k_{11} = 6000\,\text{kN/m}}$$

(For reference $k_{12} = \tfrac{6EI}{L^2} = 12000\,\text{kN}$, $k_{22} = \tfrac{4EI}{L} = 32000\,\text{kN·m/rad}$.)