BE Civil Engineering (IOE, TU) Theory of Structures II (IOE, CE 602) Question Paper 2077 Nepal

Method of consistent deformation

The propped cantilever is statically indeterminate to the first degree ($D_s = 1$). Remove the vertical prop at $B$ and take the redundant as $R_B$ (vertical reaction at $B$).

Primary (released) structure: cantilever fixed at A, free at B

Step 1 — Deflection at B due to UDL only ($\delta_{B0}$), downward positive:

$$\delta_{B0} = \frac{wL^4}{8EI} = \frac{20 \times 6^4}{8EI} = \frac{20 \times 1296}{8EI} = \frac{25920}{8EI} = \frac{3240}{EI}\ (\downarrow)$$

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

$$\delta_{BB} = \frac{L^3}{3EI} = \frac{6^3}{3EI} = \frac{216}{3EI} = \frac{72}{EI}\ (\uparrow\text{ per unit load})$$

Compatibility

Net vertical deflection at the prop $B$ is zero:

$$\delta_{B0} - R_B\,\delta_{BB} = 0 \;\Rightarrow\; R_B = \frac{\delta_{B0}}{\delta_{BB}} = \frac{3240/EI}{72/EI} = 45\text{ kN}$$ $$\boxed{R_B = 45\text{ kN }(\uparrow)}$$

This matches the standard result $R_B = \tfrac{3}{8}wL = \tfrac{3}{8}(20)(6) = 45\text{ kN}$.

Reaction and moment at A

Total load $= wL = 20\times6 = 120\text{ kN}$.

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

Fixed-end moment at $A$ (taking moments about $A$, sagging +):

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

Equivalently $M_A = \tfrac{wL^2}{8} = \tfrac{20\times36}{8}=90$ kN·m, hogging.

Maximum sagging moment

Shear is zero at distance $x$ from $B$: $R_B - wx = 0 \Rightarrow x = 45/20 = 2.25\text{ m}$ from $B$.

$$M_{max} = R_B x - \frac{w x^2}{2} = 45(2.25) - \frac{20(2.25)^2}{2} = 101.25 - 50.625 = 50.625\text{ kN·m}$$ $$\boxed{M_{max,sag} = 50.625\text{ kN·m at }2.25\text{ m from }B}$$

Bending moment diagram (described)

  A(-90)                                   B(0)
   |\                                       /
   | \___                              ____/
   |     \____                    ____/
   |          \___           ___/  +50.625 (at 2.25 m from B)
   |              \___ 0 ___/
   hogging near A      sagging near B

Point of contraflexure where $M=0$: from $B$, $R_B x - \tfrac{w x^2}{2}=0 \Rightarrow x=\tfrac{2R_B}{w}=4.5$ m, i.e. 1.5 m from A.

Slope-deflection analysis

Unknown rotations: $\theta_B$ and $\theta_C$ ($\theta_A=0$ fixed, no settlement so $\psi=0$). End $C$ is a simple support so $M_{CB}=0$.

Fixed-end moments (sign: clockwise on member end positive)

Span AB (central point load $P=40$ kN, $L=5$ m):

$$\text{FEM}_{AB} = -\frac{PL}{8} = -\frac{40\times5}{8} = -25\text{ kN·m},\quad \text{FEM}_{BA} = +25\text{ kN·m}$$

Span BC (UDL $w=15$ kN/m, $L=4$ m):

$$\text{FEM}_{BC} = -\frac{wL^2}{12} = -\frac{15\times16}{12} = -20\text{ kN·m},\quad \text{FEM}_{CB} = +20\text{ kN·m}$$

Slope-deflection equations

Using $M_{nf} = \tfrac{2EI}{L}(2\theta_n+\theta_f) + \text{FEM}_{nf}$. Let $a = EI\theta_B,\; b=EI\theta_C$.

$$M_{AB} = \tfrac{2EI}{5}(\theta_B) - 25 = 0.4a - 25$$ $$M_{BA} = \tfrac{2EI}{5}(2\theta_B) + 25 = 0.8a + 25$$ $$M_{BC} = \tfrac{2EI}{4}(2\theta_B+\theta_C) - 20 = a + 0.5b - 20$$ $$M_{CB} = \tfrac{2EI}{4}(2\theta_C+\theta_B) + 20 = b + 0.5a + 20$$

Equilibrium / boundary conditions

(i) Joint B: $M_{BA}+M_{BC}=0$:

$$0.8a + 25 + a + 0.5b - 20 = 0 \;\Rightarrow\; 1.8a + 0.5b + 5 = 0 \quad(1)$$

(ii) End C (simple support): $M_{CB}=0$:

$$b + 0.5a + 20 = 0 \;\Rightarrow\; b = -0.5a - 20 \quad(2)$$

Substitute (2) into (1):

$$1.8a + 0.5(-0.5a-20) + 5 = 0 \Rightarrow 1.8a - 0.25a - 10 + 5 = 0 \Rightarrow 1.55a = 5$$ $$a = EI\theta_B = 3.2258\;(\text{kN·m}^2,\ \text{per }EI)$$ $$b = -0.5(3.2258) - 20 = -21.613$$

Final moments

$$M_{AB} = 0.4(3.2258) - 25 = 1.290 - 25 = -23.71\text{ kN·m}$$ $$M_{BA} = 0.8(3.2258) + 25 = 2.581 + 25 = +27.58\text{ kN·m}$$ $$M_{BC} = 3.226 + 0.5(-21.613) - 20 = 3.226 - 10.806 - 20 = -27.58\text{ kN·m}$$ $$M_{CB} = -21.613 + 0.5(3.2258) + 20 = -21.613 + 1.613 + 20 = 0\ \checkmark$$ $$\boxed{M_{AB} = -23.71\text{ kN·m},\quad M_{BA}=+27.58\text{ kN·m},\quad M_{BC}=-27.58\text{ kN·m},\quad M_{CB}=0}$$

(Joint B balances: $27.58 - 27.58 = 0$ ✓)

Reaction at B

Span AB (P=40 at mid-span, $M_{AB}=-23.71$, $M_{BA}=+27.58$). Reaction at B from this span (simple-beam part + end-moment correction):

$$R_B^{AB} = \frac{40}{2} + \frac{M_{BA}+M_{AB}}{L} = 20 + \frac{27.58 + (-23.71)}{5} = 20 + \frac{3.87}{5} = 20 + 0.774 = 20.77\text{ kN}$$

Span BC (UDL 15 kN/m over 4 m, $M_{BC}=-27.58$, $M_{CB}=0$):

$$R_B^{BC} = \frac{15\times4}{2} + \frac{(-M_{BC})+(-M_{CB})}{L} = 30 + \frac{27.58 + 0}{4} = 30 + 6.90 = 36.90\text{ kN}$$ $$\boxed{R_B = R_B^{AB} + R_B^{BC} = 20.77 + 36.90 = 57.67\text{ kN}}$$

Moment distribution method

Fixed-end moments (clockwise positive)

Span AB ($w=12$ kN/m, $L=6$ m):

$$\text{FEM}_{AB} = -\frac{wL^2}{12} = -\frac{12\times36}{12} = -36\text{ kN·m},\quad \text{FEM}_{BA} = +36\text{ kN·m}$$

Span BC ($P=60$ kN, $L=4$ m):

$$\text{FEM}_{BC} = -\frac{PL}{8} = -\frac{60\times4}{8} = -30\text{ kN·m},\quad \text{FEM}_{CB} = +30\text{ kN·m}$$

Stiffness and distribution factors

Both far ends ($A$ and $C$) are fixed, so use $k = \tfrac{4EI}{L}$ for the members meeting at $B$ (only joint $B$ rotates).

$$k_{BA} = \frac{4EI}{6} = 0.6667EI,\qquad k_{BC} = \frac{4EI}{4} = 1.0EI$$ $$\sum k = 1.6667EI$$ $$DF_{BA} = \frac{0.6667}{1.6667} = 0.40,\qquad DF_{BC} = \frac{1.0}{1.6667} = 0.60$$

Joints $A$ and $C$ are fixed: $DF = 0$ (they only receive carry-over). Carry-over factor $= \tfrac12$.

Unbalanced moment at B

$$M_B^{unbal} = \text{FEM}_{BA} + \text{FEM}_{BC} = 36 + (-30) = +6\text{ kN·m}$$

Balancing moment $= -6$ kN·m distributed: $BA: -6(0.40)=-2.4$; $BC: -6(0.60)=-3.6$.

Distribution table (single cycle suffices — only B is free)

Results

$$\boxed{M_{AB} = -37.2\text{ kN·m},\; M_{BA} = +33.6\text{ kN·m},\; M_{BC} = -33.6\text{ kN·m},\; M_{CB} = +28.2\text{ kN·m}}$$

Check at B: $M_{BA}+M_{BC} = 33.6 - 33.6 = 0$ ✓ (joint balanced).

Since only one joint is free, a single balance plus carry-over gives the exact answer.

Direct stiffness method for an axial bar element

Geometry / direction cosines

Node 1 $(0,0)$, node 2 $(3,4)$. Length:

$$L = \sqrt{3^2+4^2} = 5\text{ m}$$ $$c = \cos\theta = \frac{3}{5} = 0.6,\qquad s = \sin\theta = \frac{4}{5} = 0.8$$

Axial stiffness

$$A = 1200\text{ mm}^2 = 1200\times10^{-6}\text{ m}^2,\quad E = 200\times10^9\text{ Pa}$$ $$\frac{AE}{L} = \frac{(1200\times10^{-6})(200\times10^9)}{5} = \frac{2.4\times10^8}{5} = 4.8\times10^7\text{ N/m} = 48000\text{ kN/m}$$

Global element stiffness matrix

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

With $c^2=0.36,\ s^2=0.64,\ cs=0.48$ and $\tfrac{AE}{L}=48000$ kN/m:

$$[k] = 48000\begin{bmatrix} 0.36 & 0.48 & -0.36 & -0.48 \\ 0.48 & 0.64 & -0.48 & -0.64 \\ -0.36 & -0.48 & 0.36 & 0.48 \\ -0.48 & -0.64 & 0.48 & 0.64 \end{bmatrix}\text{ kN/m}$$

Apply boundary conditions

Node 1 pinned: $u_1=v_1=0$. Free DOFs: $u_2, v_2$.

Applied force at node 2 is 50 kN along the bar axis (direction $(c,s)=(0.6,0.8)$):

$$F_{x2} = 50(0.6) = 30\text{ kN},\quad F_{y2} = 50(0.8) = 40\text{ kN}$$

A single bar resists only motion along its axis, so the axial elongation is:

$$\delta = \frac{N\,L}{AE} = \frac{50\text{ kN}}{48000\text{ kN/m}} = 1.0417\times10^{-3}\text{ m} = 1.042\text{ mm}$$

Project onto global axes:

$$u_2 = \delta\,c = 1.0417\times10^{-3}(0.6) = 6.25\times10^{-4}\text{ m} = 0.625\text{ mm}$$ $$v_2 = \delta\,s = 1.0417\times10^{-3}(0.8) = 8.333\times10^{-4}\text{ m} = 0.833\text{ mm}$$ $$\boxed{u_2 = 0.625\text{ mm},\quad v_2 = 0.833\text{ mm}}$$

Axial force

Member axial force $= \tfrac{AE}{L}\big[(u_2-u_1)c + (v_2-v_1)s\big]$:

$$N = 48000\big[(6.25\times10^{-4})(0.6) + (8.333\times10^{-4})(0.8)\big]$$ $$= 48000\big[3.75\times10^{-4} + 6.667\times10^{-4}\big] = 48000(1.0417\times10^{-3}) = 50\text{ kN (tension)}$$ $$\boxed{N = +50\text{ kN (tension)}}$$

The result is self-consistent: the recovered member force equals the applied axial force, confirming the formulation.

Plastic collapse of a propped cantilever under UDL

A propped cantilever under UDL collapses by a combined mechanism: a plastic hinge at the fixed end plus a sagging plastic hinge at the point of maximum sagging moment. Two hinges convert this (one-degree indeterminate) beam into a mechanism.

Collapse load by the mechanism (kinematic) method

The sagging hinge forms at $x = (\sqrt{2}-1)L = 0.4142L$ from the propped end. Equating external work of the UDL to internal work at the two hinges gives the classical result:

$$w_c = \frac{11.657\,M_p}{L^2}$$

Using $M_p = 150$ kN·m, $L = 8$ m:

$$w_c = \frac{11.657 \times 150}{8^2} = \frac{1748.55}{64} = 27.32\text{ kN/m}$$ $$\boxed{w_c \approx 27.3\text{ kN/m}}$$

Location of the sagging hinge: $x = 0.4142\times8 = 3.31\text{ m}$ from the prop.

Shape factor of the rectangle

For a rectangular section $b\times d$:

• Elastic section modulus: $Z = \dfrac{bd^2}{6}$
• Plastic section modulus: $Z_p = \dfrac{bd^2}{4}$

$$\text{Shape factor } v = \frac{Z_p}{Z} = \frac{bd^2/4}{bd^2/6} = \frac{6}{4} = 1.5$$ $$\boxed{v = 1.5}$$

Check consistency of M_p

$$Z_p = \frac{bd^2}{4} = \frac{200\times400^2}{4}\text{ mm}^3 = \frac{200\times160000}{4} = 8.0\times10^{6}\text{ mm}^3$$ $$M_p = f_y Z_p = 250\text{ MPa}\times8.0\times10^6\text{ mm}^3 = 2.0\times10^9\text{ N·mm} = 2000\text{ kN·m}$$

The section's actual $M_p = 2000$ kN·m, which is far larger than the assumed 150 kN·m. Therefore the assumed $M_p = 150$ kN·m is not consistent with a solid $200\times400$ mm steel section at $f_y=250$ MPa; that value would correspond to a much smaller or weaker section. With the true $M_p=2000$ kN·m, $w_c = \tfrac{11.657\times2000}{64} = 364.3$ kN/m.

Three-moment theorem (Clapeyron)

For three consecutive supports $A$, $B$, $C$ of a continuous beam (spans $L_1=AB$, $L_2=BC$), constant $EI$, no settlement:

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

where $A_1\bar{x}_1$, $A_2\bar{x}_2$ are the moments of the free (simply-supported) bending-moment-diagram areas about the respective outer supports. For a span under UDL $w$, the corresponding term is $\dfrac{6A\bar{x}}{L} = \dfrac{wL^3}{4}$.

Application

End supports are simple ⇒ $M_A = 0$, $M_C = 0$. Symmetric loading, $L_1=L_2=L=6$ m, $w=10$ kN/m.

$$0 + 2M_B(6+6) + 0 = -\left(\frac{wL_1^3}{4} + \frac{wL_2^3}{4}\right)$$ $$24 M_B = -\left(\frac{10\times6^3}{4} + \frac{10\times6^3}{4}\right) = -\left(\frac{10\times216}{4}\times2\right) = -(540\times2) = -1080$$ $$M_B = \frac{-1080}{24} = -45\text{ kN·m}$$ $$\boxed{M_B = -45\text{ kN·m (hogging)}}$$

Check: For two equal UDL spans, the standard result is $M_B = -\tfrac{wL^2}{8} = -\tfrac{10\times36}{8} = -45$ kN·m ✓.

Portal method (approximate lateral analysis)

Assumptions

1. A point of contraflexure (hinge) exists at the mid-height of each column.
2. A point of contraflexure exists at the mid-span of each beam.
3. Interior columns carry twice the shear of exterior columns. For a single bay there are two exterior columns of equal shear.

Column shears

Total storey shear $= 40$ kN, shared by 2 equal exterior columns:

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

Column base moments

Hinge at mid-height ⇒ moment arm $= h/2 = 1.5$ m:

$$M_{base} = V_{col}\times\frac{h}{2} = 20\times1.5 = 30\text{ kN·m}$$ $$\boxed{M_{base} = 30\text{ kN·m at each column base}}$$

(The moment at the top of each column is equal in magnitude.)

Axial forces in columns (overturning)

The lateral load creates an overturning moment about the base, resisted by an axial couple across the bay width:

$$\text{Overturning moment} = 40\times3 = 120\text{ kN·m}$$ $$P_{axial} = \frac{120}{6} = 20\text{ kN}$$

Windward (left) column: tension $20$ kN; leeward (right) column: compression $20$ kN.

$$\boxed{\text{Axial force} = 20\text{ kN (one column tension, the other compression)}}$$

Definitions

Static indeterminacy ($D_s$): the number of unknown forces/reactions in excess of the number of independent equilibrium equations; the structure cannot be analysed by statics alone when $D_s>0$.

Kinematic indeterminacy ($D_k$): the number of independent joint displacement (degree-of-freedom) components needed to fully describe the deformed shape — i.e. the number of unknown nodal displacements/rotations.

(a) Continuous beam, 5 supports, one end fixed, others simple, no internal hinge

Reactions: fixed end gives 2 ($V, M$); each of the other 4 simple supports gives 1 ⇒ $r = 2 + 4 = 6$. Equilibrium equations for a beam = 3.

$$D_s = r - 3 = 6 - 3 = 3$$ $$\boxed{D_s = 3}$$

(b) Rigid plane frame: m = 3, j = 4, r = 3

For a plane rigid frame:

$$D_s = (3m + r) - 3j = (3\times3 + 3) - 3\times4 = (9+3) - 12 = 0$$ $$\boxed{D_s = 0\ \text{(statically determinate)}}$$

(c) Plane truss: m = 11, j = 7, r = 3

For a plane (pin-jointed) truss:

$$D_s = (m + r) - 2j = (11 + 3) - 2\times7 = 14 - 14 = 0$$ $$\boxed{D_s = 0\ \text{(statically determinate, just-rigid)}}$$

Stiffness factors at joint O

For a prismatic member with the far end fixed, rotational stiffness $= \dfrac{4EI}{L}$. For a member with the far end pinned (simple support), use the modified stiffness $= \dfrac{3EI}{L}$, because a pinned far end develops no carry-over moment; the reduced stiffness accounts for this directly.

Member stiffnesses (take $EI$ common)

$$k_{OA} = \frac{4EI}{4} = 1.000\,EI$$ $$k_{OB} = \frac{4EI}{6} = 0.6667\,EI$$ $$k_{OC} = \frac{3EI}{5} = 0.600\,EI\quad(\text{modified, far end pinned})$$ $$\sum k = 1.000 + 0.6667 + 0.600 = 2.2667\,EI$$

Distribution factors

$$DF_{OA} = \frac{1.000}{2.2667} = 0.441$$ $$DF_{OB} = \frac{0.6667}{2.2667} = 0.294$$ $$DF_{OC} = \frac{0.600}{2.2667} = 0.265$$

Check: $0.441 + 0.294 + 0.265 = 1.000$ ✓

$$\boxed{DF_{OA}=0.441,\quad DF_{OB}=0.294,\quad DF_{OC}=0.265}$$

Why modified stiffness for OC

Using $3EI/L$ for the pin-ended member means no carry-over is applied to the pinned end $C$ during distribution. This avoids repeatedly re-balancing the (always-zero) moment at the simple support, converges in fewer cycles, and gives the exact result for that member directly.

Shape factor of a symmetric I-section

Depth $D=300$, flange $b=150$, $t_f=20$, web $t_w=10$ (mm). Web clear depth $d_w = D - 2t_f = 300 - 40 = 260$ mm.

Moment of inertia about major axis ($I_x$)

Use $I = \dfrac{b D^3}{12} - \dfrac{(b - t_w)\,d_w^3}{12}$ (full bounding rectangle minus the two side voids):

$$\frac{b D^3}{12} = \frac{150\times300^3}{12} = \frac{150\times27{,}000{,}000}{12} = \frac{4.05\times10^9}{12} = 3.375\times10^8\text{ mm}^4$$ $$260^3 = 17{,}576{,}000;\quad \frac{(150-10)\times260^3}{12} = \frac{140\times17{,}576{,}000}{12} = \frac{2.46064\times10^9}{12} = 2.0505\times10^8\text{ mm}^4$$ $$I_x = 3.375\times10^8 - 2.0505\times10^8 = 1.3245\times10^8\text{ mm}^4$$

Elastic section modulus

$$Z = \frac{I_x}{D/2} = \frac{1.3245\times10^8}{150} = 8.830\times10^5\text{ mm}^3$$

Plastic section modulus

About the equal-area axis (centroid, at mid-depth by symmetry), $Z_p = \sum (\text{area}\times\text{distance of its centroid from the axis})$ for the upper half, doubled:

• Flange area $= b\,t_f = 150\times20 = 3000\text{ mm}^2$; centroid distance $= \dfrac{D}{2}-\dfrac{t_f}{2} = 150-10 = 140$ mm.
• Half web area $= t_w\times\dfrac{d_w}{2} = 10\times130 = 1300\text{ mm}^2$; centroid distance $= \dfrac{d_w}{4} = 65$ mm.

$$Z_p = 2\big[A_f y_f + A_{hw} y_{hw}\big] = 2\big[3000\times140 + 1300\times65\big]$$ $$= 2\big[420{,}000 + 84{,}500\big] = 2\times504{,}500 = 1{,}009{,}000\text{ mm}^3 = 1.009\times10^6\text{ mm}^3$$

Shape factor

$$v = \frac{Z_p}{Z} = \frac{1.009\times10^6}{8.830\times10^5} = 1.143$$ $$\boxed{Z = 8.83\times10^5\text{ mm}^3,\quad Z_p = 1.009\times10^6\text{ mm}^3,\quad v \approx 1.14}$$

This is typical for rolled I-sections (shape factor ≈ 1.10–1.18).

Flexibility method vs. Stiffness method

Four key points of difference

1. Unknowns: flexibility solves for redundant forces; stiffness solves for nodal displacements.
2. Indeterminacy used: flexibility size $= D_s$; stiffness size $= D_k$.
3. Equation type: flexibility uses compatibility equations; stiffness uses equilibrium equations.
4. Selection of unknowns: flexibility needs a non-unique choice of redundants; stiffness is fully systematic, requiring no such choice.

Why stiffness is preferred for computers

• The global stiffness matrix is assembled directly and systematically from element stiffness matrices via a simple connectivity (address) mapping — ideal for programming.
• No analyst judgement is needed to select redundants, so the procedure is uniform for beams, frames and trusses.
• The resulting $[K]$ is symmetric, banded and sparse, allowing efficient storage and solution.
• It handles highly indeterminate structures without difficulty, since only $D_k$ governs and assembly stays routine.

For these reasons virtually all general-purpose structural analysis / FEM software is built on the direct stiffness method.