BE Civil Engineering (IOE, TU) Theory of Structures II (IOE, CE 602) Question Paper 2080 Nepal

Step 1 — Choose redundant and form primary structure.

The propped cantilever is indeterminate to the first degree ($D_s = 1$). Remove the prop reaction $R_B$ (vertical) and treat it as the redundant. The primary (released) structure is a cantilever fixed at $A$, free at $B$, carrying the UDL $w$.

Step 2 — Deflection at $B$ due to the UDL on the cantilever ($\delta_{B0}$).

For a cantilever of length $L$ with UDL $w$, the free-end deflection (downward) is

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

Step 3 — Deflection at $B$ due to a unit upward load at $B$ ($\delta_{BB}$).

For a cantilever with a point load at the free end, the free-end deflection per unit load is

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

Step 4 — Compatibility. Net vertical deflection at the real prop is zero:

$$\delta_{B0} - R_B\,\delta_{BB} = 0$$ $$R_B = \frac{\delta_{B0}}{\delta_{BB}} = \frac{wL^4/8EI}{L^3/3EI} = \frac{3wL}{8}$$

Numerically:

$$R_B = \frac{3 \times 20 \times 6}{8} = \frac{360}{8} = 45\text{ kN (upward)}$$

Step 5 — Remaining reactions (statics).

Total load $W = wL = 20 \times 6 = 120\text{ kN}$.

Vertical equilibrium: $R_A = W - R_B = 120 - 45 = 75\text{ kN (upward)}$.

Fixing moment at $A$ (take moments of all forces about $A$, sagging +):

$$M_A = R_B\,L - \frac{wL^2}{2} = 45 \times 6 - \frac{20 \times 6^2}{2} = 270 - 360 = -90\text{ kN·m}$$

The negative sign means hogging. By the standard formula $M_A = \dfrac{wL^2}{8} = \dfrac{20 \times 36}{8} = 90\text{ kN·m}$ (hogging) — consistent.

Step 6 — Point of maximum sagging moment.

Shear is zero where $R_B - w x' = 0$ measuring $x'$ from $B$: $x' = R_B/w = 45/20 = 2.25\text{ m}$ from $B$. Max sagging moment:

$$M_{max} = R_B x' - \frac{w x'^2}{2} = 45 \times 2.25 - \frac{20 \times 2.25^2}{2} = 101.25 - 50.625 = 50.625\text{ kN·m}$$

Results (bold):

• $R_B = 45\text{ kN}$ (up)
• $R_A = 75\text{ kN}$ (up)
• $M_A = 90\text{ kN·m (hogging)}$
• Max sagging $= 50.625\text{ kN·m}$ at 2.25 m from $B$

BMD (schematic):

   A                              B
   |------------------------------|
  -90 kN·m (hogging at A)
        \___        ___/
            \__  __/   point of contraflexure
              \/
        +50.6 kN·m (sag, 2.25 m from B)

Point of contraflexure: where $M=0$, i.e. $45x' - 10x'^2 = 0 \Rightarrow x' = 4.5\text{ m}$ from $B$.

Fixed-end moments (FEM).

Span $AB$ (central point load $P = 60$ kN, $L = 4$ m):

$$\text{FEM}_{AB} = -\frac{PL}{8} = -\frac{60 \times 4}{8} = -30\text{ kN·m},\quad \text{FEM}_{BA} = +30\text{ kN·m}$$

Span $BC$ (UDL $w = 15$ kN/m, $L = 6$ m):

$$\text{FEM}_{BC} = -\frac{wL^2}{12} = -\frac{15 \times 36}{12} = -45\text{ kN·m},\quad \text{FEM}_{CB} = +45\text{ kN·m}$$

(Sign convention: clockwise moment on member end positive; sagging FEM at left end negative.)

Slope-deflection equations (no support settlement, $\psi = 0$):

$$M_{AB} = \frac{2EI}{4}(2\theta_A + \theta_B) - 30$$ $$M_{BA} = \frac{2EI}{4}(2\theta_B + \theta_A) + 30$$ $$M_{BC} = \frac{2EI}{6}(2\theta_B + \theta_C) - 45$$ $$M_{CB} = \frac{2EI}{6}(2\theta_C + \theta_B) + 45$$

Boundary conditions. $A$ and $C$ are simple ends: $M_{AB} = 0$ and $M_{CB} = 0$.

Let $k = EI$. From $M_{AB}=0$: $0.5k(2\theta_A + \theta_B) = 30 \Rightarrow 2\theta_A + \theta_B = 60/k$ …(i) From $M_{CB}=0$: $\tfrac{k}{3}(2\theta_C + \theta_B) = -45 \Rightarrow 2\theta_C + \theta_B = -135/k$ …(ii)

Joint equilibrium at $B$: $M_{BA} + M_{BC} = 0$.

$$0.5k(2\theta_B + \theta_A) + 30 + \tfrac{k}{3}(2\theta_B + \theta_C) - 45 = 0$$ $$k\theta_B + 0.5k\theta_A + \tfrac{2k}{3}\theta_B + \tfrac{k}{3}\theta_C - 15 = 0$$ $$\tfrac{5k}{3}\theta_B + 0.5k\theta_A + \tfrac{k}{3}\theta_C = 15$$

…(iii)

From (i): $\theta_A = (60/k - \theta_B)/2 = 30/k - 0.5\theta_B$. From (ii): $\theta_C = (-135/k - \theta_B)/2 = -67.5/k - 0.5\theta_B$.

Substitute into (iii):

$$\tfrac{5k}{3}\theta_B + 0.5k(30/k - 0.5\theta_B) + \tfrac{k}{3}(-67.5/k - 0.5\theta_B) = 15$$ $$\tfrac{5k}{3}\theta_B + 15 - 0.25k\theta_B - 22.5 - \tfrac{k}{6}\theta_B = 15$$

Collect $\theta_B$ terms: $\left(\tfrac{5}{3} - 0.25 - \tfrac{1}{6}\right)k\theta_B$. In sixths: $\tfrac{10}{6} - \tfrac{1.5}{6} - \tfrac{1}{6} = \tfrac{7.5}{6} = 1.25$, so $1.25 k\theta_B$. Constants: $15 - 22.5 = -7.5$; move to RHS: $1.25k\theta_B = 15 + 7.5 = 22.5$.

$$\theta_B = \frac{22.5}{1.25k} = \frac{18}{k}$$

Then $\theta_A = 30/k - 0.5(18/k) = 30/k - 9/k = 21/k$. $\theta_C = -67.5/k - 0.5(18/k) = -67.5/k - 9/k = -76.5/k$.

Support moments.

$$M_{BA} = 0.5k(2(18/k) + 21/k) + 30 = 0.5(36 + 21) + 30 = 28.5 + 30 = 58.5\text{ kN·m}$$ $$M_{BC} = \tfrac{k}{3}(2(18/k) + (-76.5/k)) - 45 = \tfrac{1}{3}(36 - 76.5) - 45 = \tfrac{-40.5}{3} - 45 = -13.5 - 45 = -58.5\text{ kN·m}$$

Check: $M_{BA} + M_{BC} = 58.5 - 58.5 = 0$ ✓. $M_{AB}=M_{CB}=0$ ✓.

So $M_B = 58.5\text{ kN·m}$ (hogging over the support).

Reactions (free-body of each span; $M_B = 58.5$ kN·m hogging).

Span $AB$ (length 4 m, central 60 kN). Moments about $B$:

$$R_A \times 4 - 60 \times 2 + 58.5 = 0 \Rightarrow R_A = \frac{120 - 58.5}{4} = \frac{61.5}{4} = 15.375\text{ kN}$$

Shear at $B$ from span $AB$: $V_{BA} = 60 - 15.375 = 44.625\text{ kN}$.

Span $BC$ (length 6 m, UDL 15 kN/m → total 90 kN). Moments about $C$:

$$R'_{B} \times 6 - 90 \times 3 - 58.5 = 0 \Rightarrow R'_B = \frac{270 + 58.5}{6} = \frac{328.5}{6} = 54.75\text{ kN}$$

Then $R_C = 90 - 54.75 = 35.25\text{ kN}$.

$R_B = V_{BA} + R'_B = 44.625 + 54.75 = 99.375\text{ kN}$.

Final results (bold):

• $R_A = 15.375\text{ kN}$
• $R_B = 99.375\text{ kN}$
• $R_C = 35.25\text{ kN}$
• $M_B = 58.5\text{ kN·m (hogging)}$

Check total: $15.375 + 99.375 + 35.25 = 150\text{ kN} = 60 + 90$ ✓.

Step 1 — Fixed-end moments.

Span $AB$ (UDL $w=24$, $L=5$):

$$\text{FEM}_{AB} = -\frac{wL^2}{12} = -\frac{24 \times 25}{12} = -50\text{ kN·m},\quad \text{FEM}_{BA} = +50\text{ kN·m}$$

Span $BC$ (central point load $P=80$, $L=4$):

$$\text{FEM}_{BC} = -\frac{PL}{8} = -\frac{80 \times 4}{8} = -40\text{ kN·m},\quad \text{FEM}_{CB} = +40\text{ kN·m}$$

Step 2 — Stiffness and distribution factors.

$A$ is fixed; $C$ is a simple (pinned) end → use modified stiffness $\tfrac{3}{4}\cdot\tfrac{4EI}{L} = \tfrac{3EI}{L}$ for member $BC$ and release the pin at $C$.

Stiffness of $BA$ (far end fixed): $k_{BA} = \dfrac{4EI}{5} = 0.8EI$. Stiffness of $BC$ (far end pinned, modified): $k_{BC} = \dfrac{3EI}{4} = 0.75EI$.

Distribution factors at $B$:

$$DF_{BA} = \frac{0.8}{0.8+0.75} = \frac{0.8}{1.55} = 0.516$$ $$DF_{BC} = \frac{0.75}{1.55} = 0.484$$

Step 3 — Release the pin at $C$ first. $\text{FEM}_{CB} = +40$ is released by applying $-40$ at $C$; carry-over of $-20$ to the $BC$ end at $B$. After release $M_{CB}=0$. Updated $M_{BC}$ before balancing $B$: $-40 + (-20) = -60\text{ kN·m}$.

Step 4 — Balance joint $B$.

Unbalanced moment at $B = 50 + (-60) = -10$; balancing moment $= +10$, distributed:

• to $BA$: $+10 \times 0.516 = +5.16$
• to $BC$: $+10 \times 0.484 = +4.84$ Carry-over to $A$ (factor ½): $+5.16/2 = +2.58$. No carry-over to $C$ (already released).

Final moments:

$$M_{AB} = -50 + 2.58 = -47.42\text{ kN·m}$$ $$M_{BA} = +50 + 5.16 = +55.16\text{ kN·m}$$ $$M_{BC} = -60 + 4.84 = -55.16\text{ kN·m}$$ $$M_{CB} = 0$$

Check joint $B$: $M_{BA} + M_{BC} = 55.16 - 55.16 = 0$ ✓.

Final moments (bold):

• $M_A = -47.42\text{ kN·m}$ (hogging at fixed end)
• $M_B = 55.16\text{ kN·m}$ (hogging over support)
• $M_C = 0$ (simple support)

Step 1 — Geometry and direction cosines.

Member 1 ($A\to C$): $\Delta x = 3,\ \Delta y = 4$, length $L_1 = \sqrt{3^2+4^2} = 5\text{ m}$. $c_1 = 3/5 = 0.6,\quad s_1 = 4/5 = 0.8$.

Member 2 ($B\to C$): $\Delta x = 3-6 = -3,\ \Delta y = 4$, length $L_2 = \sqrt{(-3)^2+4^2} = 5\text{ m}$. $c_2 = -3/5 = -0.6,\quad s_2 = 4/5 = 0.8$.

Step 2 — Member axial stiffness. $\dfrac{AE}{L} = \dfrac{200000}{5} = 40000\text{ kN/m}$ for each member.

Step 3 — Assemble global stiffness for the free DOFs at $C$ ($u_C, v_C$).

Each member contributes $\dfrac{AE}{L}\begin{bmatrix} c^2 & cs \\ cs & s^2 \end{bmatrix}$.

Member 1: $c^2 = 0.36,\ s^2 = 0.64,\ cs = 0.48$. Times 40000:

$$K_1 = \begin{bmatrix} 14400 & 19200 \\ 19200 & 25600 \end{bmatrix}$$

Member 2: $c^2 = 0.36,\ s^2 = 0.64,\ cs = (-0.6)(0.8) = -0.48$. Times 40000:

$$K_2 = \begin{bmatrix} 14400 & -19200 \\ -19200 & 25600 \end{bmatrix}$$

Assembled at $C$:

$$K_{CC} = K_1 + K_2 = \begin{bmatrix} 28800 & 0 \\ 0 & 51200 \end{bmatrix}\text{ kN/m}$$

Step 4 — Solve $K_{CC}\,\{d\} = \{F\}$ with $F = \{50, 0\}^T$ kN.

Off-diagonal terms cancel, so:

$$u_C = \frac{50}{28800} = 1.7361\times10^{-3}\text{ m} = 1.736\text{ mm}$$ $$v_C = \frac{0}{51200} = 0$$

Displacements (bold): $u_C = 1.736\text{ mm}\ (+x),\quad v_C = 0$.

Step 5 — Member axial forces.

$N = \dfrac{AE}{L}\,(c\,u_C + s\,v_C)$.

Member 1: $N_1 = 40000\,(0.6 \times 1.7361\times10^{-3} + 0.8 \times 0) = 40000 \times 1.0417\times10^{-3} = 41.67\text{ kN (tension)}$.

Member 2: $N_2 = 40000\,((-0.6) \times 1.7361\times10^{-3} + 0.8 \times 0) = 40000 \times (-1.0417\times10^{-3}) = -41.67\text{ kN (compression)}$.

Member forces (bold): $N_1 = +41.67\text{ kN (T)},\quad N_2 = -41.67\text{ kN (C)}$.

Statics check at $C$: Horizontal $= N_1 c_1 + N_2 c_2 = 41.67(0.6) + (-41.67)(-0.6) = 25 + 25 = 50\text{ kN}$ ✓. Vertical $= 41.67(0.8) + (-41.67)(0.8) = 0$ ✓.

Step 1 — Collapse mechanism.

A fixed-ended beam under UDL forms three plastic hinges at collapse: one at each fixed end ($A$ and $B$) and one at mid-span ($C$). Degree of redundancy = 2, so hinges needed = $2 + 1 = 3$ → complete beam mechanism.

Step 2 — Virtual work (kinematic method).

Give the mid-span hinge a virtual deflection $\delta$. Each half-span rotates by $\theta = \delta/(L/2) = 2\delta/L$. The central hinge rotation is $2\theta$.

Internal work ($M_p \times$ hinge rotations):

$$W_{int} = M_p\,\theta\ (A) + M_p\,(2\theta)\ (C) + M_p\,\theta\ (B) = 4 M_p \theta$$

External work. UDL through a triangular deflection profile, average deflection $\delta/2$, total load $wL$:

$$W_{ext} = wL \times \frac{\delta}{2} = \frac{wL\delta}{2}$$

With $\delta = \theta L/2$: $W_{ext} = \dfrac{wL}{2}\cdot\dfrac{\theta L}{2} = \dfrac{wL^2\theta}{4}$.

Step 3 — Equate $W_{int} = W_{ext}$:

$$4 M_p \theta = \frac{wL^2\theta}{4} \Rightarrow w_c = \frac{16 M_p}{L^2}$$

(Equivalent to $M_p = \dfrac{w_c L^2}{16}$ for a fixed-ended beam under UDL.)

Step 4 — Numerical values.

$$w_c = \frac{16 \times 150}{8^2} = \frac{2400}{64} = 37.5\text{ kN/m}$$ $$W_c = w_c L = 37.5 \times 8 = 300\text{ kN}$$

Step 5 — Load factor at working UDL $= 20$ kN/m.

$$\lambda = \frac{w_c}{w_{working}} = \frac{37.5}{20} = 1.875$$

Results (bold):

• $w_c = 37.5\text{ kN/m}$
• Total collapse load $W_c = 300\text{ kN}$
• Load factor $\lambda = 1.875$

Three-moment equation (no settlement, constant $EI$):

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

Simple ends → $M_A = M_C = 0$.

For a span with UDL $w$, the free BMD is parabolic. Area $A = \tfrac{2}{3}\cdot\tfrac{wL^2}{8}\cdot L = \tfrac{wL^3}{12}$; centroid at mid-span $\bar x = L/2$. Hence

$$\frac{6 A \bar x}{L} = \frac{6}{L}\cdot\frac{wL^3}{12}\cdot\frac{L}{2} = \frac{wL^3}{4}$$

With $L_1 = L_2 = L = 5$ m, $w = 12$ kN/m:

$$2 M_B (5 + 5) = -\left(\frac{wL^3}{4} + \frac{wL^3}{4}\right) = -\frac{wL^3}{2}$$ $$20 M_B = -\frac{12 \times 125}{2} = -\frac{1500}{2} = -750$$ $$M_B = -37.5\text{ kN·m}$$

Result: $M_B = 37.5\text{ kN·m}$ (hogging over support $B$).

Cross-check (two equal spans, equal UDL): $M_B = \dfrac{wL^2}{8} = \dfrac{12 \times 25}{8} = 37.5\text{ kN·m}$ ✓.

Portal method assumptions:

1. A point of contraflexure forms at the mid-height of every column.
2. A point of contraflexure forms at the mid-span of the beam.
3. Interior columns carry twice the shear of exterior columns. For a single bay there are two exterior columns sharing the shear equally.

Step 1 — Column shears. Storey shear $= 40\text{ kN}$ shared equally by 2 columns:

$$V_{col} = \frac{40}{2} = 20\text{ kN each}$$

Step 2 — Column end moments. Hinge at mid-height ($h/2 = 2$ m):

$$M_{col} = V_{col} \times \frac{h}{2} = 20 \times 2 = 40\text{ kN·m}$$

Each column has $40\text{ kN·m}$ at top and $40\text{ kN·m}$ at base.

Step 3 — Axial forces in columns (overturning). Overturning moment of the lateral load about the base $= 40 \times 4 = 160\text{ kN·m}$, resisted by an axial couple over the bay width $= 6$ m:

$$N = \frac{40 \times 4}{6} = \frac{160}{6} = 26.67\text{ kN}$$

• Leeward column: $+26.67\text{ kN}$ (compression)
• Windward column: $-26.67\text{ kN}$ (tension)

Results (bold):

• Column shear $= 20\text{ kN}$ each
• Column end moments $= 40\text{ kN·m}$ (top and bottom)
• Column axial force $= 26.67\text{ kN}$ (one tension, one compression)

Beam end moment (joint equilibrium) $= 40\text{ kN·m}$ at each end, balancing the column top moment.

Definitions.

• Static indeterminacy ($D_s$): the number of unknown forces/moments (reactions + internal forces) in excess of the available independent equilibrium equations. If $D_s > 0$ the structure is statically indeterminate (redundant).
• Kinematic indeterminacy ($D_k$): the number of independent unknown joint displacement components (degrees of freedom) of the structure.

(a) Rigid plane frame. $D_s = 3m + r - 3j$. Reaction count: fixed support $r = 3$, each hinged support $r = 2$ → $r = 3 + 2 + 2 = 7$. With $m = 3$, $j = 4$:

$$D_s = 3(3) + 7 - 3(4) = 9 + 7 - 12 = 4$$

$D_s = 4$ (indeterminate to the 4th degree).

(b) Plane truss. $D_s = m + r - 2j$.

$$D_s = 11 + 3 - 2(7) = 14 - 14 = 0$$

$D_s = 0$ → statically determinate (and stable if properly arranged).

Summary (bold): frame is 4° indeterminate; truss is just-rigid / determinate.

Definition. The shape factor $S$ is the ratio of the plastic moment $M_p$ to the yield moment $M_y$, equivalently the ratio of the plastic section modulus $Z_p$ to the elastic section modulus $Z_e$:

$$S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e}$$

It measures the reserve strength of a section beyond first yield.

(a) Rectangle $b\times d$. Elastic modulus: $Z_e = \dfrac{b d^2}{6}$. Plastic modulus: each half-area $bd/2$ acts at lever arm $d/4$ →

$$Z_p = 2 \times \left(\frac{bd}{2}\right)\times\frac{d}{4} = \frac{b d^2}{4}$$ $$S = \frac{Z_p}{Z_e} = \frac{bd^2/4}{bd^2/6} = \frac{6}{4} = 1.5$$

$S_{rect} = 1.5$.

(b) Solid circle diameter $D$. Elastic modulus: $Z_e = \dfrac{\pi D^3}{32}$. Plastic modulus: $Z_p = 2\cdot\dfrac{\pi D^2}{8}\cdot\dfrac{2D}{3\pi} = \dfrac{D^3}{6}$.

$$S = \frac{Z_p}{Z_e} = \frac{D^3/6}{\pi D^3/32} = \frac{32}{6\pi} = \frac{16}{3\pi} = 1.698 \approx 1.70$$

$S_{circle} = 1.70$.

Comment. The circular section has a larger shape factor (1.70 vs 1.50): more of its material lies near the neutral axis and is mobilised only after the extreme fibre yields, giving more reserve between first yield and full plasticity. For strength-per-weight efficiency, however, the I-section (shape factor ≈ 1.12–1.15) is best because material is concentrated in the flanges far from the neutral axis.

Step 1 — Slope-deflection equations with settlement. With no external load, both ends fixed ($\theta_A = \theta_B = 0$), and chord rotation $\psi = \Delta/L$:

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

Both ends carry equal magnitude for pure settlement of a fixed beam.

Step 2 — Substitute values. $\Delta = 8\text{ mm} = 0.008\text{ m}$, $L = 5\text{ m}$, $EI = 30000\text{ kN·m}^2$.

$$M = \frac{6 \times 30000 \times 0.008}{5^2} = \frac{1440}{25} = 57.6\text{ kN·m}$$

Step 3 — Result.

• $M_{AB} = 57.6\text{ kN·m}$
• $M_{BA} = 57.6\text{ kN·m}$ (Both of the same sense as the chord rotation; magnitude $= 6EI\Delta/L^2$.)

Result (bold): $|M_{AB}| = |M_{BA}| = 57.6\text{ kN·m}$.

Comparison table.

Rotational element stiffness matrix. For a prismatic beam element with end rotations $\theta_i, \theta_j$ and end moments $M_i, M_j$ (translations restrained):

$$\begin{Bmatrix} M_i \\ M_j \end{Bmatrix} = \frac{EI}{L}\begin{bmatrix} 4 & 2 \\ 2 & 4 \end{bmatrix}\begin{Bmatrix} \theta_i \\ \theta_j \end{Bmatrix}$$

Application. End $i$ held against rotation ($\theta_i = 0$); moment $M_0$ applied at end $j$, so $M_j = M_0$. From the second row:

$$M_j = \frac{EI}{L}(2\theta_i + 4\theta_j) = \frac{EI}{L}(0 + 4\theta_j) = M_0$$ $$\boxed{\theta_j = \frac{M_0 L}{4EI}}$$

Bold key answer: stiffness method primary unknowns are displacements with $[K]\{d\}=\{F\}$; element matrix $\dfrac{EI}{L}\begin{bmatrix}4&2\\2&4\end{bmatrix}$; the guided-end rotation is $\theta_j = \dfrac{M_0 L}{4EI}$.