BE Civil Engineering (IOE, TU) Theory of Structures II (IOE, CE 602) Question Paper 2079 Nepal

Step 1 — Choose redundant and primary structure. Remove the prop at $B$, taking $R_B$ (vertical, upward) as the redundant $X_1$. The primary structure is a cantilever fixed at $A$, free at $B$, carrying the UDL.

Step 2 — Compatibility condition. Vertical deflection at $B$ must be zero:

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

where $\delta_{B0}$ is the downward deflection at $B$ of the cantilever under the UDL, and $\delta_{BB}$ is the upward deflection at $B$ per unit upward load.

Step 3 — Deflections (standard cantilever formulae).

$$\delta_{B0} = \frac{wL^4}{8EI} \;(\downarrow), \qquad \delta_{BB} = \frac{L^3}{3EI}\;(\uparrow \text{ per unit load})$$

Step 4 — Solve compatibility. Taking upward as positive and recognising $\delta_{B0}$ acts downward:

$$R_B\cdot\frac{L^3}{3EI} = \frac{wL^4}{8EI}\;\Rightarrow\; R_B = \frac{3wL}{8}$$ $$R_B = \frac{3(20)(6)}{8} = \frac{360}{8} = \mathbf{45\ kN\ (\uparrow)}$$

Step 5 — Reaction at A. Total load $= wL = 20\times6 = 120\ \text{kN}$.

$$R_A = 120 - 45 = \mathbf{75\ kN\ (\uparrow)}$$

Step 6 — Fixed-end moment at A. Taking moments about $A$ (sagging positive):

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

This is a hogging moment at the fixed end: $\mathbf{M_A = 90\ kN\cdot m\ (hogging)}$. (Equivalent standard result $M_A = wL^2/8 = 720/8 = 90$ kN·m.)

Step 7 — Maximum sagging moment. Measuring $x$ from $B$ (prop end), shear: $V(x) = R_B - wx = 45 - 20x$. Zero shear at $x = 45/20 = 2.25\ \text{m}$ from $B$.

$$M_{max} = R_B x - \frac{wx^2}{2} = 45(2.25) - \frac{20(2.25)^2}{2} = 101.25 - 50.625 = \mathbf{50.625\ kN\cdot m\ (sagging)}$$

Step 8 — Bending moment diagram (described).

  M_A = 90 kN·m (hogging) at A
        |\
        | \___ curve crosses zero, then sags
        |     \___/  +50.625 kN·m at 2.25 m from B
  A=====================B  (M_B = 0 at prop)

The BMD is hogging near $A$ (peak $90$ kN·m), crosses zero at $x = 3.75\ \text{m}$ from $B$, and is sagging with maximum $50.625$ kN·m at $2.25$ m from $B$; zero at the prop $B$.

Step 1 — Fixed-end moments (FEM). Span $AB$ (UDL $w=12$, $L=6$):

$$M^F_{AB} = -\frac{wL^2}{12} = -\frac{12(6)^2}{12} = -36\ \text{kN·m}, \quad M^F_{BA} = +36\ \text{kN·m}$$

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

$$M^F_{BC} = -\frac{PL}{8} = -\frac{40(4)}{8} = -20\ \text{kN·m}, \quad M^F_{CB} = +20\ \text{kN·m}$$

(Sign convention: clockwise moment on member end positive.)

Step 2 — Slope-deflection equations. No settlement, $\theta_A = 0$ (fixed). Unknowns: $\theta_B,\ \theta_C$.

$$M_{AB} = -36 + \frac{2EI}{6}(2\theta_A+\theta_B) = -36 + \frac{EI}{3}\theta_B$$ $$M_{BA} = +36 + \frac{2EI}{6}(2\theta_B+\theta_A) = 36 + \frac{2EI}{3}\theta_B$$ $$M_{BC} = -20 + \frac{2EI}{4}(2\theta_B+\theta_C) = -20 + EI\theta_B + \tfrac{EI}{2}\theta_C$$ $$M_{CB} = +20 + \frac{2EI}{4}(2\theta_C+\theta_B) = 20 + EI\theta_C + \tfrac{EI}{2}\theta_B$$

Step 3 — Equilibrium (joint) equations. At $B$: $M_{BA}+M_{BC}=0$. At $C$ (simple end): $M_{CB}=0$.

Joint $C$: $20 + EI\theta_C + \tfrac12 EI\theta_B = 0 \Rightarrow EI\theta_C = -20 - \tfrac12 EI\theta_B$.

Joint $B$: $36 + \tfrac23 EI\theta_B - 20 + EI\theta_B + \tfrac12 EI\theta_C = 0$

$$16 + \tfrac{5}{3}EI\theta_B + \tfrac12 EI\theta_C = 0$$

Substitute $EI\theta_C$:

$$16 + \tfrac{5}{3}EI\theta_B + \tfrac12(-20 - \tfrac12 EI\theta_B) = 0$$ $$16 + \tfrac{5}{3}EI\theta_B - 10 - \tfrac14 EI\theta_B = 0$$ $$6 + \left(\tfrac{5}{3}-\tfrac14\right)EI\theta_B = 0,\qquad \tfrac{5}{3}-\tfrac14 = \tfrac{17}{12}$$ $$EI\theta_B = -\frac{6\times12}{17} = -\frac{72}{17} = -4.235\ \text{kN·m}^2$$ $$EI\theta_C = -20 - \tfrac12(-4.235) = -20 + 2.118 = -17.882\ \text{kN·m}^2$$

Step 4 — Back-substitute for moments.

$$M_{AB} = -36 + \tfrac13(-4.235) = -36 - 1.412 = \mathbf{-37.41\ kN\cdot m}$$ $$M_{BA} = 36 + \tfrac23(-4.235) = 36 - 2.824 = \mathbf{+33.18\ kN\cdot m}$$ $$M_{BC} = -20 + (-4.235) + \tfrac12(-17.882) = -20 - 4.235 - 8.941 = \mathbf{-33.18\ kN\cdot m}$$ $$M_{CB} = 20 + (-17.882) + \tfrac12(-4.235) = 20 - 17.882 - 2.118 = \mathbf{0\ kN\cdot m}\ \checkmark$$

Check $B$: $M_{BA}+M_{BC} = 33.18 - 33.18 = 0$ ✓.

Step 5 — BMD. Support moments (hogging): $M_A = 37.41$, $M_B = 33.18$, $M_C = 0$ kN·m. Span $AB$ free moment $wL^2/8 = 12(36)/8 = 54$; net mid-span sag $\approx 54 - (37.41+33.18)/2 = 18.70$ kN·m. Span $BC$ free moment $PL/4 = 40(4)/4 = 40$; net mid sag $\approx 40 - 33.18/2 = 23.41$ kN·m.

  hog 37.41   hog 33.18      0
     A------------B-----------C
        \__sag__/    \_sag__/
        +18.70        +23.41

Step 1 — Fixed-end moments. Span $AB$: $M^F_{AB} = -\dfrac{wL^2}{12} = -\dfrac{24(5)^2}{12} = -50$ kN·m; $M^F_{BA}=+50$ kN·m. Span $BC$: $M^F_{BC} = -\dfrac{PL}{8} = -\dfrac{60(5)}{8} = -37.5$ kN·m; $M^F_{CB}=+37.5$ kN·m.

Step 2 — Distribution factors. Ends $A$ and $C$ are simple supports → use modified stiffness $\tfrac34\cdot\tfrac{4EI}{L}=\tfrac{3EI}{L}$ for the spans framing into $B$. With equal spans, $K_{BA}=K_{BC}=\dfrac{3EI}{5}$.

$$DF_{BA}=DF_{BC}=0.5$$

Step 3 — Release the simple ends first. Release $A$: balance $M^F_{AB} = -50$ → add $+50$ at $A$, carry over $+25$ to $M_{BA}$. Release $C$: balance $M^F_{CB} = +37.5$ → add $-37.5$ at $C$, carry over $-18.75$ to $M_{BC}$. Updated end moments at $B$:

$$M_{BA}=50+25 = 75,\qquad M_{BC} = -37.5 - 18.75 = -56.25$$

Step 4 — Balance joint B (reduced stiffness; A, C pinned so no further carry-over). Unbalanced moment at $B = 75 + (-56.25) = +18.75$ kN·m. Distribute $-18.75$: $0.5\times(-18.75) = -9.375$ to each side.

$$M_{BA}=75 - 9.375 = \mathbf{+65.625\ kN\cdot m}$$ $$M_{BC}=-56.25 - 9.375 = \mathbf{-65.625\ kN\cdot m}$$

Check: $M_{BA}+M_{BC}=0$ ✓. Final $A$ and $C$ moments are zero.

$$\boxed{M_B = 65.625\ \text{kN·m (hogging)}}$$

Step 5 — Reactions. Span $AB$ (simple beam + end moment $M_B$ at B): UDL reaction $24\times5/2 = 60$ kN each; moment correction $= M_B/L = 65.625/5 = 13.125$ kN.

$$R_A = 60 - 13.125 = 46.875\ \text{kN},\quad R_{B,AB} = 60 + 13.125 = 73.125\ \text{kN}$$

Span $BC$ (point load + end moment $M_B$ at B): reaction $60/2 = 30$ kN each; correction $= 65.625/5 = 13.125$ kN.

$$R_C = 30 - 13.125 = 16.875\ \text{kN},\quad R_{B,BC} = 30 + 13.125 = 43.125\ \text{kN}$$ $$R_B = 73.125 + 43.125 = 116.25\ \text{kN}$$

Check total: $46.875+116.25+16.875 = 180$ kN $= 24(5)+60 = 180$ kN ✓.

Final: $\mathbf{R_A=46.875\ kN,\ R_B=116.25\ kN,\ R_C=16.875\ kN,\ M_B=65.625\ kN\cdot m}$.

Step 1 — Degrees of freedom. Only joint $C$ is free → DOFs: $u$ ($x$-displacement), $v$ ($y$-displacement).

Step 2 — Member orientations and axial stiffness $k=EA/L$.

• Member 1 ($AC$): along $x$, $\theta_1 = 0^\circ$, $k_1 = EA/L_1 = 2000/4 = 500$ kN/m. $c=\cos0=1,\ s=\sin0=0$.
• Member 2 ($BC$, vertical): along $y$, $\theta_2 = 90^\circ$, $k_2 = EA/L_2 = 2000/3 = 666.67$ kN/m. $c=0,\ s=1$.

Step 3 — Member contribution to joint $C$ stiffness. For a bar, the $2\times2$ stiffness at the free joint is $k\begin{bmatrix} c^2 & cs \\ cs & s^2 \end{bmatrix}$. Member 1: $500\begin{bmatrix}1&0\\0&0\end{bmatrix} = \begin{bmatrix}500&0\\0&0\end{bmatrix}$. Member 2: $666.67\begin{bmatrix}0&0\\0&1\end{bmatrix} = \begin{bmatrix}0&0\\0&666.67\end{bmatrix}$.

Step 4 — Assembled structure stiffness matrix at C.

$$[K] = \begin{bmatrix} 500 & 0 \\ 0 & 666.67 \end{bmatrix}\ \text{kN/m}$$

Step 5 — Load vector and solve. $\{F\} = \begin{Bmatrix} 50 \\ 0 \end{Bmatrix}$ kN.

$$u = \frac{50}{500} = 0.10\ \text{m},\quad v = \frac{0}{666.67} = 0\ \text{m}$$ $$\boxed{u = 0.10\ \text{m (}\to\text{)},\quad v = 0}$$

Step 6 — Member forces $N = k\,(c\,u + s\,v)$ (support ends fixed). Member 1: $N_1 = 500(1\times0.10 + 0) = 50$ kN → tension.

$$\mathbf{N_1 = +50\ kN\ (tension)}$$

Member 2: $N_2 = 666.67(0\times0.10 + 1\times0) = 0$ kN.

$$\mathbf{N_2 = 0\ kN}$$

Step 7 — Equilibrium check at C. $\sum F_x$: member 1 force $50$ kN balances applied $50$ kN ✓. $\sum F_y = 0$ ✓. The vertical member is a zero-force member under this horizontal loading.

Step 1 — Indeterminacy and hinges. A fixed-fixed beam has degree of static indeterminacy $= 2$. Number of plastic hinges for a mechanism $= D_s + 1 = 2 + 1 = \mathbf{3}$, forming at both ends ($A$, $B$) and at mid-span.

Step 2 — Beam mechanism (virtual work). Give the mid-span a virtual deflection $\Delta$. Each half rotates by $\theta = \Delta/(L/2) = 2\Delta/L$.

• End hinge $A$: rotation $\theta$; end hinge $B$: rotation $\theta$; central hinge: total relative rotation $2\theta$.

Internal work:

$$W_{int} = M_p\theta + M_p(2\theta) + M_p\theta = 4M_p\theta$$

External work by UDL (load moves through average deflection $\Delta/2$ over length $L$):

$$W_{ext} = w_c L \frac{\Delta}{2}$$

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

Step 3 — Equate and solve.

$$\frac{w_c L^2 \theta}{4} = 4 M_p \theta \Rightarrow w_c = \frac{16 M_p}{L^2}$$ $$w_c = \frac{16(150)}{(8)^2} = \frac{2400}{64} = \mathbf{37.5\ kN/m}$$

Step 4 — Load factor.

$$\lambda = \frac{w_c}{w_{working}} = \frac{37.5}{18} = \mathbf{2.083}$$

Step 5 — Summary.

• Degree of static indeterminacy $= 2$.
• Plastic hinges at collapse $= 3$ (at $A$, mid-span, $B$).
• Collapse load $w_c = 37.5$ kN/m; load factor $\lambda = 2.083$.

Verification: $w_cL^2/8 = 37.5(64)/8 = 300 = 2M_p = 2(150) = 300$ ✓.

Three-moment theorem (no settlement, constant $EI$): For three consecutive supports $1,2,3$ on spans $L_1$ (1–2) and $L_2$ (2–3):

$$M_1 L_1 + 2M_2(L_1+L_2) + M_3 L_2 = -6\left(\frac{A_1\bar{x}_1}{L_1} + \frac{A_2\bar{x}_2}{L_2}\right)$$

For a span carrying UDL the right-side term per span reduces to $\dfrac{wL^3}{4}$.

Application. Supports $A,B,C$. $M_A = M_C = 0$. $L_1=L_2=6$ m, $w=15$ kN/m.

$$2M_B(6+6) = -\left(\frac{wL_1^3}{4} + \frac{wL_2^3}{4}\right)$$ $$24 M_B = -2\times\frac{15(6)^3}{4} = -2\times\frac{15\times216}{4} = -2\times810 = -1620$$ $$M_B = \frac{-1620}{24} = \mathbf{-67.5\ kN\cdot m}$$

The moment over central support $B$ is $\mathbf{67.5\ kN\cdot m\ (hogging)}$.

Check: symmetric two-span result $M_B = -wL^2/8 = -15(36)/8 = -67.5$ kN·m ✓.

Portal method — assumptions:

1. A point of contraflexure (zero moment) at the mid-height of every column.
2. A point of contraflexure at the mid-span of every beam.
3. Interior column shear is twice that of an exterior column. For a single-bay frame both columns are exterior and share the storey shear equally.

Step 1 — Column shears. Storey shear $= H = 40$ kN over two equal columns:

$$V_{col} = \frac{40}{2} = \mathbf{20\ kN\ per\ column}$$

Step 2 — Column-base moments. Hinge at mid-height: $M = V_{col}\times(h/2)$:

$$M_{base} = 20 \times \frac{4}{2} = \mathbf{40\ kN\cdot m\ per\ column}$$

Step 3 — Column axial forces (overturning). Overturning moment $= H\times h = 40\times4 = 160$ kN·m, resisted by the axial couple over bay width $b$:

$$N = \frac{H h}{b} = \frac{160}{6} = \mathbf{26.67\ kN}$$

Leeward column $+26.67$ kN (compression); windward column $26.67$ kN (tension).

Summary: each column shear $=20$ kN, base moment $=40$ kN·m, axial $=\pm26.67$ kN.

Definitions.

• Static indeterminacy ($D_s$): number of force/reaction components in excess of the available equilibrium equations (number of redundants).
• Kinematic indeterminacy ($D_k$): number of independent unknown joint displacement components (degrees of freedom).

(a) Continuous beam, 4 supports (1 fixed + 3 rollers), no internal hinge. Reactions: fixed $=2$, three rollers $=3$ → $r = 5$. For a beam (vertical loads), $D_s = r - 2 = 5 - 2 = \mathbf{3}$? Using the general plane count $D_s = r - 3 = 5 - 3 = \mathbf{2}$ (treating it as a planar structure with 3 equations). $D_s = 2$. Kinematic (neglecting axial): unknown rotations occur at the three roller joints (fixed end rotation $=0$) → $D_k = \mathbf{3}$.

(b) Single-bay two-storey rigid plane frame, fixed bases. Members $m = 6$ (4 columns + 2 beams). Rigid joints $j = 6$ (4 free + 2 fixed bases). Reactions $r_e = 2\times3 = 6$.

$$D_s = 3m + r_e - 3j = 3(6) + 6 - 3(6) = 18 + 6 - 18 = \mathbf{6}$$

Kinematic (general, with axial): 4 free rigid joints × 3 DOF $= \mathbf{12}$. Neglecting axial deformation: rotations (4) + storey sways (2) $= \mathbf{6}$.

Summary table:

General slope-deflection equation:

$$M_{AB} = \frac{2EI}{L}\left(2\theta_A + \theta_B - 3\frac{\Delta}{L}\right) + M^F_{AB}$$

where $\frac{2EI}{L}$ is the stiffness term; $\theta_A,\theta_B$ are the end rotations (clockwise +); $\frac{\Delta}{L}=\psi$ is the chord rotation from relative settlement; $M^F_{AB}$ is the fixed-end moment from transverse loads.

Given: no load → $M^F=0$; $\theta_A=\theta_B=0$; $\Delta = 0.012$ m; $L=5$ m; $EI = 1.2\times10^4$ kN·m².

Chord rotation: $\psi = \dfrac{0.012}{5} = 0.0024$ rad.

End moments (equal for pure settlement, no rotation):

$$M_{AB} = M_{BA} = \frac{2EI}{L}\left(-3\frac{\Delta}{L}\right) = -\frac{6EI\Delta}{L^2}$$ $$= -\frac{6(1.2\times10^4)(0.012)}{(5)^2} = -\frac{864}{25} = \mathbf{-34.56\ kN\cdot m}$$

Both end moments $= -34.56$ kN·m (magnitude $\mathbf{34.56\ kN\cdot m}$ at each end), confirming the standard fixed-fixed settlement result $M = 6EI\Delta/L^2$.

Definitions.

• Plastic moment of resistance ($M_p$): the moment at which the entire cross-section has yielded (fully plastic stress block), $M_p = f_y Z_p$.
• Shape factor ($S$): ratio of plastic to yield moment, $S = M_p/M_y = Z_p/Z$; reserve strength between first yield and full plastification, depending only on shape.

Given: $b=200$ mm, $d=400$ mm, $f_y = 250$ N/mm².

Step 1 — Elastic section modulus.

$$Z = \frac{bd^2}{6} = \frac{200(400)^2}{6} = \frac{32{,}000{,}000}{6} = 5.333\times10^6\ \text{mm}^3$$

Step 2 — Plastic section modulus.

$$Z_p = \frac{bd^2}{4} = \frac{200(400)^2}{4} = \frac{32{,}000{,}000}{4} = 8.0\times10^6\ \text{mm}^3$$

Step 3 — Shape factor.

$$S = \frac{Z_p}{Z} = \frac{8.0\times10^6}{5.333\times10^6} = \mathbf{1.5}$$

(For any rectangle $S=1.5$.)

Step 4 — Yield moment.

$$M_y = f_y Z = 250\times5.333\times10^6 = 1.333\times10^9\ \text{N·mm} = \mathbf{1333.3\ kN\cdot m}$$

($1\ \text{kN·m} = 10^6\ \text{N·mm}$).

Step 5 — Plastic moment.

$$M_p = f_y Z_p = 250\times8.0\times10^6 = 2.0\times10^9\ \text{N·mm} = \mathbf{2000\ kN\cdot m}$$

Check: $M_p/M_y = 2000/1333.3 = 1.5 = S$ ✓.

Basic matrix relations.

• Flexibility (force) method: $[F]\{Q\} = -\{\delta_L\}$ — solve for redundant forces $\{Q\}$ enforcing compatibility ($\{\delta_L\}$ = displacements at redundant locations on the released structure under load).
• Stiffness (displacement) method: $\{P\} = [K]\{d\}$ — solve $\{d\} = [K]^{-1}\{P\}$ for joint displacements, then recover member forces.

Comparison table.

Key remark. Modern structural-analysis software and the finite element method use the stiffness method because $[K]$ assembles systematically from element stiffness matrices and the kinematically determinate base structure is unique — ideal for automation.