BE Civil Engineering (IOE, TU) Theory of Structures I (IOE, CE 554) Question Paper 2079 Nepal

Step 1 — Support reactions

Span $L = 8\,\text{m}$. Point load $P = 40\,\text{kN}$ at $x = 3\,\text{m}$. UDL $w = 10\,\text{kN/m}$ over $x = 4\,\text{m}$ to $x = 8\,\text{m}$ (length $4\,\text{m}$, resultant $40\,\text{kN}$ at $x = 6\,\text{m}$).

Taking moments about $A$:

$$R_B \times 8 = 40\times 3 + 40\times 6 = 120 + 240 = 360 \Rightarrow R_B = 45\,\text{kN}$$ $$R_A = (40+40) - 45 = 35\,\text{kN}$$

Step 2 — Macaulay bending moment expression

Measuring $x$ from $A$ (extend the UDL to the end so the bracket acts only beyond $x=4$):

$$EI\frac{d^2y}{dx^2} = M = 35x - 40\langle x-3\rangle - \tfrac{10}{2}\langle x-4\rangle^2$$

Step 3 — Integrate twice

$$EI\frac{dy}{dx} = \tfrac{35x^2}{2} - \tfrac{40}{2}\langle x-3\rangle^2 - \tfrac{10}{6}\langle x-4\rangle^3 + C_1$$ $$EIy = \tfrac{35x^3}{6} - \tfrac{40}{6}\langle x-3\rangle^3 - \tfrac{10}{24}\langle x-4\rangle^4 + C_1 x + C_2$$

Step 4 — Boundary conditions

At $x=0$, $y=0$: all brackets negative (dropped) $\Rightarrow C_2 = 0$.

At $x=8$, $y=0$:

$$0 = \tfrac{35(8)^3}{6} - \tfrac{40}{6}(5)^3 - \tfrac{10}{24}(4)^4 + 8C_1$$

• $\dfrac{35\times 512}{6} = 2986.667$
• $\dfrac{40}{6}\times 125 = 833.333$
• $\dfrac{10}{24}\times 256 = 106.667$

$$0 = 2986.667 - 833.333 - 106.667 + 8C_1 = 2046.667 + 8C_1 \Rightarrow C_1 = -255.833\,\text{kN·m}^2$$

Step 5 — (a) Slope at A ($x=0$)

$$EI\,\theta_A = C_1 = -255.833 \Rightarrow \theta_A = \frac{-255.833}{30000} = -0.008528\,\text{rad}$$

$\theta_A = -8.53\times 10^{-3}\,\text{rad}$ (rotation directed downward into the span).

Step 6 — (b) Deflection under the 40 kN load ($x=3$)

Brackets $\langle x-3\rangle$, $\langle x-4\rangle$ are zero/negative at $x=3$:

$$EIy_{x=3} = \tfrac{35(3)^3}{6} + C_1(3) = 157.5 - 767.5 = -610.0$$ $$y_{x=3} = \frac{-610.0}{30000} = -0.02033\,\text{m}$$

$y_{x=3} = -20.3\,\text{mm}$ (downward).

Geometry

Span $L=30\,\text{m}$, rise $h=6\,\text{m}$, parabola from $A$:

$$y = \frac{4h}{L^2}x(L-x) = \frac{24}{900}x(30-x) = 0.02667\,x(30-x)$$

Step 1 — Vertical reactions

Load $120\,\text{kN}$ at $x=10\,\text{m}$. Moments about $A$:

$$V_B\times 30 = 120\times 10 = 1200 \Rightarrow V_B = 40\,\text{kN},\quad V_A = 120 - 40 = 80\,\text{kN}$$

Step 2 — Horizontal thrust (crown hinge, right segment unloaded)

$M_C = 0$ using right part $C\!-\!B$ ($x=15$, $y=6$):

$$V_B\times 15 - H\times 6 = 0 \Rightarrow H = \frac{40\times 15}{6} = 100\,\text{kN}$$

Step 3 — (a) Reactions

• $V_A = 80\,\text{kN}\uparrow$, $V_B = 40\,\text{kN}\uparrow$, $H_A = H_B = 100\,\text{kN}$ (inward).
• $R_A=\sqrt{80^2+100^2}=\sqrt{16400}=128.1\,\text{kN}$.

Step 4 — (b) Bending moment under the load ($x=10$)

$y_{10} = 0.02667\times 10\times 20 = 5.333\,\text{m}$. Using the left segment:

$$M = V_A\times 10 - H\times y_{10} = 80\times 10 - 100\times 5.333 = 800 - 533.3 = 266.7\,\text{kN·m}$$

$M = +266.7\,\text{kN·m}$ (sagging).

Step 5 — Slope of axis at $x=10$

$$\frac{dy}{dx} = 0.02667(30-2x) = 0.02667(10)=0.2667$$ $$\theta = \tan^{-1}(0.2667) = 14.93^\circ,\quad \cos\theta = 0.9662,\ \sin\theta = 0.2577$$

Step 6 — (c) Normal thrust and radial shear (just left of load)

Left-side forces: $H=100\,\text{kN}$ horizontal, net vertical shear $V = V_A = 80\,\text{kN}$.

$$N = H\cos\theta + V\sin\theta = 100(0.9662) + 80(0.2577) = 96.62 + 20.62 = 117.2\,\text{kN}$$ $$Q = V\cos\theta - H\sin\theta = 80(0.9662) - 100(0.2577) = 77.30 - 25.77 = 51.5\,\text{kN}$$

$N = 117.2\,\text{kN}$ (compression), $Q = 51.5\,\text{kN}$.

Geometry

$A=(0,0)$, $C=(4,0)$, $E=(8,0)$, $D=(2,3)$, $F=(6,3)$. Diagonal $CF$: from $C(4,0)$ to $F(6,3)$, length $\sqrt{2^2+3^2}=\sqrt{13}=3.606\,\text{m}$; direction cosines $\cos = 2/3.606 = 0.5547$ (horiz), $\sin = 3/3.606 = 0.8321$ (vert). Top chord $DF$ horizontal, length $4\,\text{m}$.

Step 1 — (a) Reactions

By symmetry with central load $60\,\text{kN}$ at $C$:

$$V_A = V_E = 30\,\text{kN}\ \uparrow,\quad H_A = 0$$

Step 2 — (b) Section through $DF$, $CF$, $CE$; take the RIGHT portion

Right free body carries only $V_E = 30\,\text{kN}\uparrow$ at $E$ (the $60\,\text{kN}$ at $C$ is excluded from the right side).

Force in $CF$ (vertical equilibrium)

$CF$ is the only cut member with a vertical component. A tensile $CF$ pulls the right body toward $C$ (down-left), vertical component downward:

$$\sum F_y = 0:\ 30 - F_{CF}(0.8321) = 0 \Rightarrow F_{CF} = \frac{30}{0.8321} = 36.05\,\text{kN}$$

$F_{CF} = 36.1\,\text{kN (Tension)}$

Force in $DF$ (moments about $C$)

About $C=(4,0)$: $CF$ and $CE$ pass through $C$ (no moment). $V_E=30$ up at $E(8,0)$, arm $4\,\text{m}$. $DF$ horizontal at height $3\,\text{m}$.

$$\sum M_C = 0:\ 30\times 4 = F_{DF}\times 3 \Rightarrow F_{DF} = \frac{120}{3} = 40\,\text{kN}$$

The top chord is in compression (it must push back against the sagging tendency). $F_{DF} = 40\,\text{kN (Compression)}$

Force in $CE$ (moments about $F$)

About $F=(6,3)$: $V_E=30$ up at $E(8,0)$, horizontal arm $2\,\text{m}$. $CE$ horizontal at $y=0$, perpendicular distance $3\,\text{m}$.

$$\sum M_F = 0:\ 30\times 2 = F_{CE}\times 3 \Rightarrow F_{CE} = \frac{60}{3} = 20\,\text{kN}$$

Bottom chord is in tension. $F_{CE} = 20\,\text{kN (Tension)}$

Summary

Step 1 — Real moment $M$

Measure $x$ from the free end $C$ ($0\le x\le 6$), real load $w=12\,\text{kN/m}$:

$$M = -\frac{w x^2}{2} = -6x^2\ \text{kN·m}$$

Step 2 — Virtual system

Apply a unit downward load at $C$. Virtual moment at distance $x$ from $C$:

$$m = -1\cdot x = -x\ \text{kN·m}$$

Step 3 — Integral

$$\delta_C = \frac{1}{EI}\int_0^6 (-6x^2)(-x)\,dx = \frac{1}{EI}\int_0^6 6x^3\,dx = \frac{6}{EI}\left[\frac{x^4}{4}\right]_0^6$$ $$= \frac{6}{EI}\cdot\frac{1296}{4} = \frac{6\times 324}{EI} = \frac{1944}{EI}$$

Step 4 — Substitute $EI$

$$\delta_C = \frac{1944}{24000} = 0.081\,\text{m}$$

$\delta_C = 0.081\,\text{m} = 81\,\text{mm}$ (downward).

Check: $\delta = \dfrac{wL^4}{8EI} = \dfrac{12\times 6^4}{8\times 24000} = \dfrac{15552}{192000} = 0.081\,\text{m}\ \checkmark$

Step 1 — ILD for $R_A$

Unit load at $x$ from $A$: $R_A = \dfrac{12-x}{12}$. Ordinate $1$ at $A$, $0$ at $B$; at $C$ ($x=4$): $R_A = 8/12 = 0.667$.

Step 2 — ILD for shear at $C$ ($a=4$, $b=8$)

• Just left of $C$: $V_C = -a/L = -4/12 = -0.333$.
• Just right of $C$: $V_C = +b/L = +8/12 = +0.667$.

Line from $0$ at $A$ to $-0.333$ just left of $C$; jumps to $+0.667$ just right of $C$; down to $0$ at $B$.

Step 3 — ILD for moment at $C$

Peak at $C$: $\dfrac{ab}{L} = \dfrac{4\times 8}{12} = 2.667\,\text{m}$. Triangle: $0$ at $A$, $2.667$ at $C$, $0$ at $B$.

Step 4 — (b) Max positive BM at $C$ (UDL over whole span)

Area of moment ILD $= \tfrac{1}{2}\times 12 \times 2.667 = 16.0\,\text{m}^2$.

$$M_C^{max} = 15\times 16.0 = 240\,\text{kN·m}$$

$M_C^{max} = 240\,\text{kN·m}$

Step 5 — Max positive shear at $C$ (UDL only over positive part $C$–$B$)

Area of positive triangle $= \tfrac{1}{2}\times 8\times 0.667 = 2.667\,\text{m}$.

$$V_C^{+max} = 15\times 2.667 = 40.0\,\text{kN}$$

$V_C^{+max} = 40.0\,\text{kN}$

(For reference, max negative shear: $15\times\tfrac{1}{2}\times 4\times 0.333 = 10.0\,\text{kN}$.)

Conditions

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

• $D_s = 0$: determinate (stable if properly arranged).
• $D_s > 0$: indeterminate to degree $D_s$.
• $D_s < 0$: deficient / unstable (mechanism).

Plane rigid frame: $D_s = (3m + r) - 3j$.

Stability also needs proper member arrangement and non-concurrent, non-parallel reaction components; the count is necessary but not sufficient.

(i) Truss: $m=21,\ r=3,\ j=12$

$$D_s = (21 + 3) - 2(12) = 24 - 24 = 0$$

Statically determinate (stable if properly arranged).

(ii) Frame: $m=8,\ r=4,\ j=6$

$$D_s = (3\times 8 + 4) - 3(6) = 28 - 18 = 10$$

Statically indeterminate to the $10^{\text{th}}$ degree.

Step 1 — $M/EI$ diagram

Fixed end $A$, free end $B$, $L=4\,\text{m}$, $P=20\,\text{kN}$ at $B$. BM at $A = -PL = -80\,\text{kN·m}$, zero at $B$. The $M/EI$ diagram is a triangle, peak $80/EI$ at $A$, zero at $B$.

Step 2 — Slope at $B$ (1st moment-area theorem)

Reference tangent at fixed end $A$ (slope = 0). Slope at $B$ = area of $M/EI$ between $A$ and $B$:

$$\theta_B = \frac{1}{EI}\cdot\tfrac{1}{2}\cdot L\cdot (PL) = \frac{1}{EI}\cdot\tfrac{1}{2}\cdot 4\cdot 80 = \frac{160}{EI} = \frac{160}{16000} = 0.01\,\text{rad}$$

$\theta_B = 0.01\,\text{rad}$

Step 3 — Deflection at $B$ (2nd theorem)

Deflection = first moment of $M/EI$ area about $B$. Centroid of the triangle (peak at $A$) lies at $\tfrac{2L}{3}=2.667\,\text{m}$ from $B$:

$$\delta_B = \frac{1}{EI}\cdot 160 \cdot 2.667 = \frac{426.7}{EI} = \frac{426.7}{16000} = 0.02667\,\text{m}$$

$\delta_B = 26.7\,\text{mm}$ (downward).

Check: $\theta_B = \dfrac{PL^2}{2EI}=0.01$ rad; $\delta_B = \dfrac{PL^3}{3EI}=\dfrac{20\times64}{3\times16000}=0.02667$ m. $\checkmark$

Step 1 — Horizontal tension

Parabolic cable, UDL $w$, span $L$, central dip $d$:

$$H = \frac{wL^2}{8d} = \frac{5\times 40^2}{8\times 4} = \frac{8000}{32} = 250\,\text{kN}$$

(a) $H = 250\,\text{kN}$

Step 2 — Vertical reaction at each support

Total load $W = wL = 200\,\text{kN}$; each support $V = 100\,\text{kN}$.

Step 3 — Maximum tension (at supports)

$$T_{max} = \sqrt{H^2 + V^2} = \sqrt{250^2 + 100^2} = \sqrt{72500} = 269.3\,\text{kN}$$

(b) $T_{max} = 269.3\,\text{kN}$

Step 4 — Maximum slope at supports

$$\tan\theta_{max} = \frac{V}{H} = \frac{100}{250} = 0.4 \Rightarrow \theta_{max} = \tan^{-1}(0.4) = 21.8^\circ$$

(Also $\tan\theta = \dfrac{4d}{L} = \dfrac{16}{40}=0.4$.) $\theta_{max} = 21.8^\circ$

Principle

Replace the real beam with a conjugate beam of equal length loaded by the real $M/EI$ diagram. Then:

• Shear in conjugate beam = slope of real beam.
• Moment in conjugate beam = deflection of real beam.

Support conversion:

Application

$L=6\,\text{m}$, central $P=30\,\text{kN}$; max BM $=\dfrac{PL}{4}=45\,\text{kN·m}$. The $M/EI$ diagram is a triangle, peak $45/EI$. Total conjugate load:

$$A = \tfrac{1}{2}\times 6\times \frac{45}{EI} = \frac{135}{EI}$$

Conjugate end reaction $= A/2 = \dfrac{67.5}{EI}$ = real slope at support:

$$\theta_A = \frac{67.5}{20000} = 0.003375\,\text{rad}$$

$\theta_A = 3.375\times 10^{-3}\,\text{rad}$

Check: $\theta_A=\dfrac{PL^2}{16EI}=\dfrac{30\times36}{16\times20000}=0.003375$ rad. $\checkmark$

Step 1 — Reactions

UDL over $AB$ = $8\times 6 = 48\,\text{kN}$ at $3\,\text{m}$ from $A$; point load $10\,\text{kN}$ at $C$ ($8\,\text{m}$ from $A$).

Moments about $A$:

$$R_B\times 6 = 48\times 3 + 10\times 8 = 144 + 80 = 224 \Rightarrow R_B = 37.33\,\text{kN}$$ $$R_A = (48 + 10) - 37.33 = 20.67\,\text{kN}$$

Step 2 — Location of max sagging moment in $AB$

Shear from $A$: $V(x) = 20.67 - 8x = 0 \Rightarrow x = \dfrac{20.67}{8} = 2.583\,\text{m}$ from $A$.

Step 3 — Maximum sagging moment

$$M_{max} = R_A x - \tfrac{8x^2}{2} = 20.67\times 2.583 - 4\times 2.583^2 = 53.39 - 26.69 = 26.70\,\text{kN·m}$$

$M_{max} = 26.7\,\text{kN·m}$ (sagging) at $x = 2.58\,\text{m}$ from $A$.

(Hogging moment at $B$ from overhang $= -10\times 2 = -20\,\text{kN·m}$.)

Müller-Breslau principle

The influence line for any force response function (reaction, shear, or bending moment) is, to scale, the deflected shape obtained when the restraint corresponding to that force is removed and a unit displacement (or rotation) is applied in its direction.

Basis / validity

It follows from virtual work / Betti's reciprocal theorem. Removing the relevant restraint and imposing a unit virtual displacement, the external virtual work of a unit moving load at any position equals the ordinate of the resulting deflected shape there; equating internal and external virtual work shows the response equals that ordinate, so the deflected shape is the influence line.

For determinate structures, removing one restraint creates a mechanism, so the deflected shape consists of straight rigid-body segments, giving the familiar straight-line influence lines directly without solving compatibility.

Application: reaction $R_A$

Remove the vertical restraint at $A$ and impose a unit upward displacement at $A$ while $B$ stays pinned. The beam rotates rigidly about $B$: ordinate $1$ at $A$ falling linearly to $0$ at $B$. This straight line (ordinate $\dfrac{L-x}{L}$) is the influence line for $R_A$.