BE Civil Engineering (IOE, TU) Theory of Structures II (IOE, CE 602) Question Paper 2076 Nepal

Step 1 — Choose the redundant

The structure is statically indeterminate to the first degree. Remove the prop at $B$ to obtain the primary (released) cantilever fixed at $A$, free at $B$. The redundant is the vertical prop reaction $R_B$ (upward positive).

Compatibility: net vertical deflection at $B$ = 0.

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

where $\delta_{B0}$ = downward deflection at $B$ on the primary structure due to applied loads, and $\delta_{BB}$ = deflection at $B$ due to unit upward load at $B$.

Step 2 — Deflection at free end of cantilever due to loads ($\delta_{B0}$)

Standard cantilever (free end deflection, downward positive):

• UDL over whole span: $\dfrac{wL^4}{8EI}$
• Point load $P$ at mid-span ($a=L/2$): tip deflection $= \dfrac{Pa^2}{6EI}(3L-a)$

UDL term:

$$\frac{wL^4}{8EI}=\frac{20\times 6^4}{8EI}=\frac{20\times1296}{8EI}=\frac{25920}{8EI}=\frac{3240}{EI}$$

Point-load term with $P=30$, $a=3$, $L=6$:

$$\frac{Pa^2(3L-a)}{6EI}=\frac{30\times 3^2\times(18-3)}{6EI}=\frac{30\times 9\times 15}{6EI}=\frac{4050}{6EI}=\frac{675}{EI}$$ $$\delta_{B0}=\frac{3240+675}{EI}=\frac{3915}{EI}\ (\text{downward})$$

Step 3 — Flexibility $\delta_{BB}$

Unit upward load at tip of cantilever gives upward tip deflection:

$$\delta_{BB}=\frac{L^3}{3EI}=\frac{6^3}{3EI}=\frac{216}{3EI}=\frac{72}{EI}$$

Step 4 — Solve compatibility

Taking downward as positive for $\delta_{B0}$ and the redundant lifting $B$ up:

$$\frac{3915}{EI}-R_B\frac{72}{EI}=0\;\Rightarrow\; R_B=\frac{3915}{72}=54.375\,\text{kN}$$ $$\boxed{R_B = 54.38\,\text{kN (upward)}}$$

Step 5 — Remaining reactions and fixing moment at A

Total load $= wL + P = 20\times6 + 30 = 120+30 = 150\,\text{kN}$.

$$R_A = 150 - 54.375 = 95.625\,\text{kN}$$

Fixing moment at $A$ (take moments of all forces about $A$, sagging positive on the beam; $M_A$ is the support moment):

$$M_A = R_B\,L - wL\cdot\frac{L}{2} - P\cdot\frac{L}{2}$$ $$M_A = 54.375\times 6 - 120\times 3 - 30\times 3 = 326.25 - 360 - 90 = -123.75\,\text{kN·m}$$

The negative sign indicates a hogging moment at the fixed end.

$$\boxed{M_A = 123.75\,\text{kN·m (hogging)}}$$

Step 6 — Bending-moment diagram (salient values)

Measuring $x$ from $B$ (prop), with $R_B=54.375$ kN upward:

At mid-span ($x=3$ m, only UDL acting between B and C):

$$M_C = R_B\times 3 - w\times3\times\frac{3}{2} = 54.375\times3 - 20\times3\times1.5 = 163.125 - 90 = +73.125\,\text{kN·m (sagging)}$$

Key ordinates:

• $M_B = 0$ (prop)
• $M_C$ (mid-span) $= +73.13\,\text{kN·m}$ (sagging)
• $M_A = -123.75\,\text{kN·m}$ (hogging)

        +73.13 (sag, mid)
          ___
         /   \
  B_____/     \________
  0              \     |
                  \____|  A
                  -123.75 (hog)

A point of contraflexure lies between $C$ and $A$ where the BMD crosses zero.

Step 1 — Fixed-end moments (FEM)

Span $AB$ (UDL $w=15$, $L=6$):

$$M^F_{AB}=-\frac{wL^2}{12}=-\frac{15\times36}{12}=-45\,\text{kN·m},\quad M^F_{BA}=+45\,\text{kN·m}$$

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

$$M^F_{BC}=-\frac{PL}{8}=-\frac{40\times4}{8}=-20\,\text{kN·m},\quad M^F_{CB}=+20\,\text{kN·m}$$

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

Step 2 — Slope-deflection equations (no settlement, chord rotation = 0)

$\theta_A=0$ (fixed). Unknowns: $\theta_B,\ \theta_C$.

$$M_{AB}=-45+\frac{2EI}{6}(2\theta_A+\theta_B)=-45+\frac{EI}{3}\theta_B$$ $$M_{BA}=+45+\frac{2EI}{6}(2\theta_B+\theta_A)=45+\frac{2EI}{3}\theta_B$$ $$M_{BC}=-20+\frac{2EI}{4}(2\theta_B+\theta_C)=-20+EI\theta_B+\tfrac{EI}{2}\theta_C$$ $$M_{CB}=+20+\frac{2EI}{4}(2\theta_C+\theta_B)=20+EI\theta_C+\tfrac{EI}{2}\theta_B$$

Step 3 — Equilibrium / boundary conditions

At C (free end, no applied moment): $M_{CB}=0$

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

At B (joint equilibrium): $M_{BA}+M_{BC}=0$

$$\big(45+\tfrac{2EI}{3}\theta_B\big)+\big(-20+EI\theta_B+\tfrac{EI}{2}\theta_C\big)=0$$ $$25+\tfrac{5EI}{3}\theta_B+\tfrac{EI}{2}\theta_C=0 \quad (ii)$$

Substitute $(i)$ into $(ii)$. Note $\tfrac{EI}{2}\theta_C=\tfrac12\big(-20-\tfrac{EI}{2}\theta_B\big)=-10-\tfrac{EI}{4}\theta_B$:

$$25+\tfrac{5EI}{3}\theta_B-10-\tfrac{EI}{4}\theta_B=0$$ $$15+EI\theta_B\Big(\tfrac53-\tfrac14\Big)=0,\quad \tfrac53-\tfrac14=\tfrac{20-3}{12}=\tfrac{17}{12}$$ $$EI\theta_B=-\frac{15\times12}{17}=-\frac{180}{17}=-10.588$$

From $(i)$: $EI\theta_C=-20-\tfrac12(-10.588)=-20+5.294=-14.706$

Step 4 — Substitute back for moments

$$M_{AB}=-45+\tfrac13(-10.588)=-45-3.529=-48.53\,\text{kN·m}$$ $$M_{BA}=45+\tfrac23(-10.588)=45-7.059=+37.94\,\text{kN·m}$$ $$M_{BC}=-20+(-10.588)+\tfrac12(-14.706)=-20-10.588-7.353=-37.94\,\text{kN·m}$$ $$M_{CB}=20+(-14.706)+\tfrac12(-10.588)=20-14.706-5.294=0\,\text{kN·m}\;\checkmark$$

Check at B: $M_{BA}+M_{BC}=37.94-37.94=0\;\checkmark$

Step 5 — Results (support moments)

$$\boxed{M_A=48.53\,\text{kN·m (hogging)},\quad M_B=37.94\,\text{kN·m (hogging)},\quad M_C=0}$$

Step 6 — Bending-moment diagram

Free (simply supported) span moments:

• Mid-span $AB$: $\dfrac{wL^2}{8}=\dfrac{15\times36}{8}=67.5\,\text{kN·m}$ (sag)
• Mid-span $BC$: $\dfrac{PL}{4}=\dfrac{40\times4}{4}=40\,\text{kN·m}$ (sag)

Net mid-span (free moment minus average end moment):

• $AB$ mid: $67.5-\tfrac{48.53+37.94}{2}=67.5-43.24=+24.26\,\text{kN·m (sag)}$
• $BC$ mid: $40-\tfrac{37.94+0}{2}=40-18.97=+21.03\,\text{kN·m (sag)}$

  -48.53        -37.94        0
   A______________B___________C
      \  +24.26 /    \ +21.03/
       \______/       \_____/

Step 1 — Fixed-end moments

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

$$M^F_{AB}=-\frac{wL^2}{12}=-\frac{24\times25}{12}=-50\,\text{kN·m},\quad M^F_{BA}=+50\,\text{kN·m}$$

Span BC (point load $P=60$ at $a=1$, $b=3$, $L=4$):

$$M^F_{BC}=-\frac{Pab^2}{L^2}=-\frac{60\times1\times9}{16}=-\frac{540}{16}=-33.75\,\text{kN·m}$$ $$M^F_{CB}=+\frac{Pa^2b}{L^2}=+\frac{60\times1\times3}{16}=+\frac{180}{16}=+11.25\,\text{kN·m}$$

Step 2 — Distribution factors at B

Both far ends fixed, so use stiffness $k=4EI/L$ (relative).

$$k_{BA}=\frac{4EI}{5}=0.8EI,\qquad k_{BC}=\frac{4EI}{4}=1.0EI,\qquad \sum k=1.8EI$$ $$DF_{BA}=\frac{0.8}{1.8}=0.4444,\qquad DF_{BC}=\frac{1.0}{1.8}=0.5556$$

A and C are fixed ⇒ no distribution there (DF = 0); they only receive carry-over.

Step 3 — Moment distribution table

Unbalanced moment at B $=M^F_{BA}+M^F_{BC}=50+(-33.75)=+16.25$. Balancing moment $=-16.25$.

Balance computation: $-16.25\times0.4444=-7.222$ (to BA); $-16.25\times0.5556=-9.028$ (to BC). Carry-over to A: $\tfrac12(-7.222)=-3.611$; to C: $\tfrac12(-9.028)=-4.514$.

Since A and C are fixed, no further balancing is required (single iteration converges exactly).

Check at B: $42.78+(-42.78)=0\;\checkmark$

Step 4 — Results

$$\boxed{M_A=53.61\,\text{kN·m (hog)},\ M_B=42.78\,\text{kN·m (hog)},\ M_C=6.74\,\text{kN·m}}$$

Final end moments (clockwise-positive convention):

• $M_{AB}=-53.61$, $M_{BA}=+42.78$, $M_{BC}=-42.78$, $M_{CB}=+6.74$ kN·m.

Step 5 — Note on BMD

Free-span sagging maxima: $AB$ mid $=\tfrac{wL^2}{8}=75$ kN·m; $BC$ under load $=\tfrac{Pab}{L}=\tfrac{60\times1\times3}{4}=45$ kN·m. Superpose the negative support-moment line $(M_A,M_B,M_C)$ on these to obtain the net diagram, hogging over $A$ and $B$.

Step 1 — Model and active DOF

With $A$ and $C$ fixed, the only free joint displacement is the rotation $\theta_B$. All translations are restrained (no chord rotation). So this reduces to a $1\times1$ stiffness problem.

Step 2 — Fixed-end moments (equivalent joint load)

Span AB (central $P=48$, $L=4$):

$$M^F_{AB}=-\frac{PL}{8}=-\frac{48\times4}{8}=-24\,\text{kN·m},\quad M^F_{BA}=+24\,\text{kN·m}$$

Span BC (UDL $w=30$, $L=4$):

$$M^F_{BC}=-\frac{wL^2}{12}=-\frac{30\times16}{12}=-40\,\text{kN·m},\quad M^F_{CB}=+40\,\text{kN·m}$$

Net fixed-end moment at joint B (sum of member ends meeting at B):

$$M^F_B=M^F_{BA}+M^F_{BC}=24+(-40)=-16\,\text{kN·m}$$

Equivalent joint load (restraining moment reversed): $P_B=-M^F_B=+16\,\text{kN·m}$.

Step 3 — Structure stiffness for $\theta_B$

For a prismatic beam element, the rotational stiffness at the near joint (far end fixed) is $4EI/L$.

$$K_{BB}=\frac{4EI}{L_{AB}}+\frac{4EI}{L_{BC}}=\frac{4EI}{4}+\frac{4EI}{4}=EI+EI=2EI$$

Step 4 — Solve $K\,\theta_B = P_B$

$$2EI\,\theta_B = 16\;\Rightarrow\;\theta_B = \frac{16}{2EI}=\frac{8}{EI}\,\text{rad}$$ $$\boxed{\theta_B = \frac{8}{EI}\ (\text{positive} = \text{clockwise})}$$

Step 5 — Member-end moments

Member-end moment $=$ FEM $+$ stiffness terms $\times$ displacements. With $\theta_A=\theta_C=0$, near-end coefficient $4EI/L$ and carry-over $2EI/L$:

$$M_{AB}=M^F_{AB}+\frac{2EI}{4}\theta_B=-24+\frac{EI}{2}\cdot\frac{8}{EI}=-24+4=-20\,\text{kN·m}$$ $$M_{BA}=M^F_{BA}+\frac{4EI}{4}\theta_B=+24+EI\cdot\frac{8}{EI}=24+8=+32\,\text{kN·m}$$ $$M_{BC}=M^F_{BC}+\frac{4EI}{4}\theta_B=-40+8=-32\,\text{kN·m}$$ $$M_{CB}=M^F_{CB}+\frac{2EI}{4}\theta_B=+40+\frac{EI}{2}\cdot\frac{8}{EI}=40+4=+44\,\text{kN·m}$$

Check joint B equilibrium: $M_{BA}+M_{BC}=32-32=0\;\checkmark$

Step 6 — Results

$$\boxed{M_{AB}=-20,\ M_{BA}=+32,\ M_{BC}=-32,\ M_{CB}=+44\ \text{(all kN·m)};\quad \theta_B=\tfrac{8}{EI}}$$

Support moments (magnitudes): $M_A=20$, $M_B=32$, $M_C=44$ kN·m.

(a) Assumptions of plastic analysis

1. Material is ideally elastic–perfectly plastic (no strain hardening).
2. The plastic moment $M_p$ is reached and maintained at a plastic hinge while rotation continues.
3. Sections remain plane; plane bending only (axial/shear interaction neglected in simple theory).
4. Connections develop full $M_p$.
5. Deformations up to collapse are small; collapse occurs when enough hinges form a mechanism.
6. Loads increase proportionally (proportional loading).

(b) Collapse mechanism

A propped cantilever under UDL is indeterminate to degree 1; it needs $1+1 = 2$ plastic hinges to form a mechanism:

• one hinge at the fixed end $A$ (max negative moment),
• one hinge in the span at the point of maximum sagging moment, distance $x$ from the propped end $B$.

Virtual-work (kinematic) approach

Let the span hinge be at distance $x$ from $B$. Give it a virtual deflection $\delta$.

Rotation of segment $B$–hinge: $\theta_1=\dfrac{\delta}{x}$. Rotation of segment hinge–$A$: $\theta_2=\dfrac{\delta}{L-x}$. Rotation at the span hinge: $\theta_1+\theta_2$. Rotation at hinge $A$: $\theta_2$.

Internal work (hinges at $A$ and at span point):

$$W_{int}=M_p\,\theta_2 + M_p(\theta_1+\theta_2)=M_p\Big(\theta_1+2\theta_2\Big)=M_p\left(\frac{\delta}{x}+\frac{2\delta}{L-x}\right)$$

External work (UDL over the whole span moving through the triangular deflection profile; triangle area $=\tfrac12 L\,\delta$):

$$W_{ext}=w\cdot\frac{1}{2}L\,\delta$$

Equate:

$$w\cdot\tfrac12 L\,\delta = M_p\,\delta\left(\frac{1}{x}+\frac{2}{L-x}\right)\;\Rightarrow\; w=\frac{2M_p}{L}\left(\frac{1}{x}+\frac{2}{L-x}\right)$$

(c) Locate hinge — minimise $w$ (the true collapse load is the minimum)

Let $f(x)=\dfrac{1}{x}+\dfrac{2}{L-x}$. Set $f'(x)=0$:

$$-\frac{1}{x^2}+\frac{2}{(L-x)^2}=0\;\Rightarrow\;2x^2=(L-x)^2\;\Rightarrow\;\sqrt2\,x=L-x$$ $$x(\sqrt2+1)=L\;\Rightarrow\;x=\frac{L}{1+\sqrt2}=L(\sqrt2-1)=8\times0.4142=3.314\,\text{m from }B$$

Collapse load value

With $x=3.314$ m, $L-x=4.686$ m:

$$\frac1x+\frac{2}{L-x}=\frac{1}{3.314}+\frac{2}{4.686}=0.3018+0.4268=0.7286\,\text{m}^{-1}$$ $$w_u=\frac{2M_p}{L}\times0.7286=\frac{2\times180}{8}\times0.7286=45\times0.7286=32.79\,\text{kN/m}$$

Closed-form check: the standard result is $w_u=\dfrac{M_p(6+4\sqrt2)}{L^2}=\dfrac{180\times11.657}{64}=\dfrac{2098.2}{64}=32.79\,\text{kN/m}$. ✓

$$\boxed{w_u = 32.79\,\text{kN/m},\quad \text{span hinge at } 3.31\,\text{m from the propped end }B}$$

Definitions

• Static indeterminacy (DSI): the number of independent force/reaction components in excess of the equations of static equilibrium; i.e. the number of redundants that must be released to render the structure statically determinate.
• Kinematic indeterminacy (DKI): the number of independent, unknown joint displacement components (the degrees of freedom) — translations and rotations not prevented by supports.

Degree of static indeterminacy (plane rigid frame)

$$DSI = (3m + r) - 3j$$ $$DSI = (3\times8 + 4) - 3\times6 = (24+4) - 18 = 28 - 18 = 6$$ $$\boxed{DSI = 6}$$

Degree of kinematic indeterminacy (plane frame)

Each free joint of a plane rigid frame has 3 possible displacements ($u, v, \theta$).

$$DKI = 3j - (\text{support restraints})$$

The 4 reaction components correspond to 4 restrained displacement components.

$$DKI = 3\times6 - 4 = 18 - 4 = 14$$ $$\boxed{DKI = 14 \ (\text{with axial deformation considered})}$$

Note: If members were taken as axially rigid (inextensible), the number of independent joint translations would be reduced by the number of independent axial-length constraints, lowering the DKI.

Theorem of three moments

For three consecutive supports $1$–$2$–$3$ over two spans of lengths $L_1$ and $L_2$ (constant $EI$, no settlement):

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

where the $A\bar x/L$ terms are the moment-area contributions of the free-bending-moment diagrams. For a span under UDL the standard right-hand term per span is $\dfrac{wL^3}{4}$, so:

$$M_A L_1 + 2M_B(L_1+L_2) + M_C L_2 = -\frac{w_1 L_1^3}{4} - \frac{w_2 L_2^3}{4}$$

Application

Ends simply supported: $M_A = 0$, $M_C = 0$. Equal spans $L_1=L_2=L=5$, $w=16$.

$$2M_B(L+L) = -\frac{wL^3}{4} - \frac{wL^3}{4}$$ $$4M_B L = -\frac{2wL^3}{4} = -\frac{wL^3}{2}$$ $$M_B = -\frac{wL^3}{2\times 4L} = -\frac{wL^2}{8}$$ $$M_B = -\frac{16\times 5^2}{8} = -\frac{16\times25}{8} = -\frac{400}{8} = -50\,\text{kN·m}$$ $$\boxed{M_B = 50\,\text{kN·m (hogging)}}$$

Check: matches the standard result $M_B = wL^2/8$ for two equal UDL spans.

Assumptions of the portal method

1. A point of contraflexure (inflection) occurs at the mid-height of each column.
2. A point of contraflexure occurs at the mid-span of each beam (girder).
3. The horizontal shear carried by each interior column is twice that of each exterior column (each interior column is shared between two adjacent portal bays).

The method suits frames that are relatively low and wide, where lateral response is dominated by shear (portal) action.

Column shear distribution

Let the exterior-column shear $=V$. Then interior-column shear $=2V$. For a two-bay frame: exterior + interior + exterior columns.

$$V + 2V + V = H \;\Rightarrow\; 4V = 60 \;\Rightarrow\; V = 15\,\text{kN}$$

• Exterior columns: $V = 15\,\text{kN}$ each
• Interior column: $2V = 30\,\text{kN}$

$$\boxed{V_{ext}=15\,\text{kN (each)},\quad V_{int}=30\,\text{kN}}$$

Check: $15 + 30 + 15 = 60\,\text{kN} = H\;\checkmark$

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 between first yield and full plastification.

(a) Rectangular section $b\times d$

Elastic modulus:

$$Z_e=\frac{I}{y}=\frac{bd^3/12}{d/2}=\frac{bd^2}{6}$$

Plastic modulus (two halves, each area $bd/2$, centroid lever arm $d/4$):

$$Z_p=2\left(\frac{bd}{2}\cdot\frac{d}{4}\right)=\frac{bd^2}{4}$$ $$S=\frac{Z_p}{Z_e}=\frac{bd^2/4}{bd^2/6}=\frac{6}{4}=1.5$$ $$\boxed{S_{rect}=1.5}$$

(b) Solid circular section, diameter $D$

Elastic modulus:

$$Z_e=\frac{\pi D^3}{32}$$

Plastic modulus of a solid circle:

$$Z_p=\frac{D^3}{6}$$ $$S=\frac{Z_p}{Z_e}=\frac{D^3/6}{\pi D^3/32}=\frac{32}{6\pi}=\frac{32}{18.850}=1.698\approx1.70$$ $$\boxed{S_{circle}=1.70}$$

Comment

A high shape factor (circle 1.70 > rectangle 1.50) means a larger fraction of the area lies near the neutral axis and is mobilised only after first yield, giving greater post-yield reserve. By contrast an I-section (shape factor ≈ 1.12–1.15) places material at the extreme fibres and is structurally more economical even though its shape factor is lower. A high shape factor signals large reserve, not necessarily efficient material use.

Concept

For a fixed–fixed beam, a relative support settlement $\Delta$ (chord rotation $\psi=\Delta/L$) with both end rotations held at zero induces equal end moments of magnitude:

$$M_A = M_B = \frac{6EI\,\Delta}{L^2}$$

This follows from the slope-deflection equations with $\theta_A=\theta_B=0$:

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

Computation

Given $EI=24\,000\,\text{kN·m}^2$, $\Delta=12\,\text{mm}=0.012\,\text{m}$, $L=6\,\text{m}$.

$$M=\frac{6\times24\,000\times0.012}{6^2}=\frac{6\times24\,000\times0.012}{36}$$

Numerator: $6\times24\,000=144\,000$; $144\,000\times0.012=1\,728$.

$$M=\frac{1\,728}{36}=48\,\text{kN·m}$$

Result

$$\boxed{M_A = M_B = 48\,\text{kN·m}}$$

Sense: With $B$ settling below $A$, both end moments have magnitude $48\,\text{kN·m}$. The deflected shape is antisymmetric with a single point of contraflexure at mid-span, so the mid-span moment is zero (no transverse load).

Beam element stiffness matrix

Degrees of freedom ordered as $\{v_1,\ \theta_1,\ v_2,\ \theta_2\}$ (transverse displacement and rotation at node 1, then node 2):

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

Physical meaning of a column

Each column gives the nodal forces/moments required to produce a unit value of the corresponding DOF while all other DOFs are held at zero.

Example — column 2 (corresponding to $\theta_1 = 1$, all other DOFs zero) gives the end actions:

• Force at node 1: $\dfrac{6EI}{L^2}$
• Moment at node 1: $\dfrac{4EI}{L}$
• Force at node 2: $-\dfrac{6EI}{L^2}$
• Moment at node 2: $\dfrac{2EI}{L}$

These match the slope-deflection actions for a unit near-end rotation: near-end stiffness $4EI/L$ and carry-over moment $2EI/L$ at the far end.

Two key properties

1. Symmetry: $[k]=[k]^T$ (a consequence of the Maxwell–Betti reciprocal theorem; $k_{ij}=k_{ji}$).
2. Singularity / positive semi-definite: the unconstrained element matrix is singular ($\det[k]=0$) because it admits rigid-body modes (rigid translation and rigid rotation produce no internal forces); it becomes positive-definite only after sufficient supports are applied.