BE Civil Engineering (IOE, TU) Theory of Structures I (IOE, CE 554) Question Paper 2080 Nepal

Step 1 — Reactions

Loads: UDL of $10\text{ kN/m}$ over $0$ to $3\text{ m}$ (resultant $= 30\text{ kN}$ at $x = 1.5\text{ m}$), and a point load $30\text{ kN}$ at $x = 4\text{ m}$.

Taking moments about $A$ ($\circlearrowleft +$):

$$R_B(6) = 30(1.5) + 30(4) = 45 + 120 = 165$$ $$R_B = 27.5\text{ kN}, \qquad R_A = 60 - 27.5 = 32.5\text{ kN}$$

Step 2 — Bending-moment expression (Macaulay form)

Measure $x$ from $A$. The UDL ends at $x=3$; to use Macaulay brackets we extend it to the cut and subtract an equal upward UDL from $x=3$ onward.

$$M(x) = 32.5x - \frac{10}{2}x^2 + \frac{10}{2}\langle x-3\rangle^2 - 30\langle x-4\rangle$$

Step 3 — Integrate ($EI\,y'' = M$)

$$EI\,\frac{dy}{dx} = \frac{32.5}{2}x^2 - \frac{10}{6}x^3 + \frac{10}{6}\langle x-3\rangle^3 - \frac{30}{2}\langle x-4\rangle^2 + C_1$$ $$EI\,y = \frac{32.5}{6}x^3 - \frac{10}{24}x^4 + \frac{10}{24}\langle x-3\rangle^4 - \frac{30}{6}\langle x-4\rangle^3 + C_1 x + C_2$$

Step 4 — Boundary conditions

At $x=0$, $y=0$ (drop bracket terms): $C_2 = 0$.

At $x=6$, $y=0$:

$$\frac{32.5}{6}(216) - \frac{10}{24}(1296) + \frac{10}{24}(3)^4 - \frac{30}{6}(2)^3 + 6C_1 = 0$$ $$1170 - 540 + 33.75 - 40 + 6C_1 = 0$$ $$623.75 + 6C_1 = 0 \Rightarrow C_1 = -103.958\text{ kN}\cdot\text{m}^2$$

Step 5 — Slope at A

$$EI\,\theta_A = C_1 = -103.958$$ $$\theta_A = \frac{-103.958}{20000} = -5.198\times10^{-3}\text{ rad}$$

$\boxed{\theta_A = 5.20\times10^{-3}\text{ rad (clockwise)}}$

Step 6 — Deflection at mid-span ($x = 3$)

At $x=3$, only the first two terms and $C_1x$ survive (brackets $\langle x-3\rangle$ and $\langle x-4\rangle$ are zero):

$$EI\,y_C = \frac{32.5}{6}(27) - \frac{10}{24}(81) + (-103.958)(3)$$ $$= 146.25 - 33.75 - 311.875 = -199.375\text{ kN}\cdot\text{m}^3$$ $$y_C = \frac{-199.375}{20000} = -9.969\times10^{-3}\text{ m}$$

$\boxed{y_C = 9.97\text{ mm downward}}$

Step 1 — Reactions

Total vertical load $= 40 + 20 = 60\text{ kN}$.

Moments about $A$ ($\circlearrowleft +$): load at $C$ acts at $x=4$, load at $D$ at $x=6$.

$$R_E(8) = 40(4) + 20(6) = 160 + 120 = 280 \Rightarrow R_E = 35\text{ kN }(\uparrow)$$ $$R_{Ay} = 60 - 35 = 25\text{ kN }(\uparrow), \qquad R_{Ax} = 0$$

Step 2 — Joint A (method of joints)

Members at $A$: $AB$ (to $(2,3)$) and $AC$ (horizontal).

Unit vector along $AB$: length $=\sqrt{2^2+3^2}=\sqrt{13}=3.6056$; direction $(\tfrac{2}{\sqrt{13}}, \tfrac{3}{\sqrt{13}}) = (0.5547, 0.8321)$.

Vertical equilibrium ($\Sigma F_y=0$), taking tension positive:

$$25 + F_{AB}(0.8321) = 0 \Rightarrow F_{AB} = -30.05\text{ kN}$$

$F_{AB} = 30.05\text{ kN (Compression)}$

Horizontal equilibrium ($\Sigma F_x=0$):

$$F_{AC} + F_{AB}(0.5547) = 0 \Rightarrow F_{AC} = -(-30.05)(0.5547) = 16.67\text{ kN}$$

$F_{AC} = 16.67\text{ kN (Tension)}$

Step 3 — Joint C

Members at $C$: $AC$ (to left, horizontal), $CE$ (to right, horizontal), $BC$ (to $B=(2,3)$, i.e. up-left), $CD$ (to $D=(6,3)$, up-right). External load $40\text{ kN}$ down.

$BC$: from $C(4,0)$ to $B(2,3)$: components $(-2,3)$, length $\sqrt{13}$, unit $(-0.5547, 0.8321)$. $CD$: from $C(4,0)$ to $D(6,3)$: components $(2,3)$, unit $(0.5547, 0.8321)$.

$\Sigma F_y=0$ at $C$:

$$F_{BC}(0.8321) + F_{CD}(0.8321) - 40 = 0$$ $$F_{BC} + F_{CD} = 48.07 \quad (i)$$

$\Sigma F_x=0$ at $C$:

$$-F_{AC} + F_{CE} - F_{BC}(0.5547) + F_{CD}(0.5547) = 0 \quad (ii)$$

Two equations, three unknowns ($F_{CE}$ also unknown), so resolve $BC$ via the section method below; here record relation $(i)$.

Step 4 — Method of sections for CD

Cut through $BD$, $CD$, $CE$ between the two halves and consider the right portion (joints $D$, $E$). Forces: $R_E=35\text{ kN}\uparrow$, load $20\text{ kN}$ down at $D$, and cut members $BD$ (horizontal at $y=3$), $CD$, $CE$ (horizontal at $y=0$).

Take moments about joint $E=(8,0)$ to eliminate $CE$ and $R_E$ — but $BD$ and $CD$ remain. Instead take moments about the intersection of $BD$ and $CE$ extended; simpler: resolve vertically on the right portion. Only $CD$ has a vertical component (BD and CE are horizontal):

$$\Sigma F_y = 0:\quad 35 - 20 + F_{CD}(0.8321)\cdot(\text{sign}) = 0$$

Member $CD$ pulls the right portion toward $C$ (down-left) if in tension, giving vertical component $-0.8321 F_{CD}$:

$$35 - 20 - 0.8321\,F_{CD} = 0 \Rightarrow F_{CD} = \frac{15}{0.8321} = 18.03\text{ kN}$$

$F_{CD} = 18.03\text{ kN (Tension)}$

Step 5 — Back-substitute for BC

From $(i)$: $F_{BC} = 48.07 - 18.03 = 30.04\text{ kN}$.

$F_{BC} = 30.04\text{ kN (Tension)}$

Summary

Step 1 — Vertical reactions

Span $L = 40\text{ m}$, load $120\text{ kN}$ at $x = 10\text{ m}$ from $A$.

Moments about $A$:

$$V_B(40) = 120(10) = 1200 \Rightarrow V_B = 30\text{ kN}$$ $$V_A = 120 - 30 = 90\text{ kN}$$

Step 2 — Horizontal thrust (crown hinge condition)

Bending moment at crown hinge $C$ (mid-span, $x=20$) $=0$. Using the left part:

$$M_C = V_A(20) - 120(20-10) - H(8) = 0$$ $$90(20) - 120(10) - 8H = 0$$ $$1800 - 1200 - 8H = 0 \Rightarrow 8H = 600 \Rightarrow H = 75\text{ kN}$$

$\boxed{H = 75\text{ kN}}$, $V_A = 90\text{ kN}$, $V_B = 30\text{ kN}$.

Step 3 — Rise of arch axis at load point

Parabolic axis: $y = \dfrac{4h}{L^2}x(L-x)$ with $h=8$, $L=40$:

$$y = \frac{4(8)}{40^2}x(40-x) = \frac{32}{1600}x(40-x) = 0.02\,x(40-x)$$

At $x=10$: $y = 0.02(10)(30) = 6\text{ m}$.

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

The load is at the section, so consider the moment just to the left (the load contributes no moment about its own point):

$$M = V_A(10) - H\,y = 90(10) - 75(6) = 900 - 450 = 450\text{ kN}\cdot\text{m}$$

$\boxed{M_{x=10} = 450\text{ kN}\cdot\text{m (sagging)}}$

(Check: just to the right, $M = V_A(10) - 120(0) - H y =$ same, since the load arm is zero at its own location.)

Step 1 — Member lengths

• $AB$: from $(0,0)$ to $(3,0)$ → $L_{AB}=3\text{ m}$
• $BC$: from $(3,0)$ to $(3,4)$ → $L_{BC}=4\text{ m}$
• $CA$: from $(3,4)$ to $(0,0)$ → $L_{CA}=\sqrt{3^2+4^2}=5\text{ m}$

Step 2 — Real member forces ($P$) under the $50\text{ kN}$ load at $C$

Reactions: Moments about $A$: $R_{By}(3) = 50(3) \Rightarrow R_{By}=50\text{ kN}\uparrow$. Then $R_{Ay} = 50-50 = 0$. Horizontal: $R_{Ax}$ balances any horizontal; here $\Sigma F_x=0$ gives $R_{Ax} = 0$ after solving joints.

Joint C ($50\text{ kN}$ down). Members $CB$ (vertical, down to $B$) and $CA$ (to $A$, direction $(-3,-4)/5=(-0.6,-0.8)$).

$$\Sigma F_y: -50 - F_{CB} - 0.8F_{CA}=0$$ $$\Sigma F_x: -0.6F_{CA}=0 \Rightarrow F_{CA}=0$$

Then $F_{CB} = -50\text{ kN}$ (compression).

Joint A: members $AB$ (horizontal) and $AC$ ($=0$). $\Sigma F_x: F_{AB} + 0 = 0 \Rightarrow F_{AB}=0$.

So: $P_{AB}=0$, $P_{BC}=-50\text{ kN}$, $P_{CA}=0\text{ kN}$.

Step 3 — Virtual forces ($k$) under unit vertical load at $C$

A unit downward load at $C$ produces forces proportional to the real case (same load direction and point):

$$k_{AB}=0,\quad k_{BC}=-1,\quad k_{CA}=0$$

Step 4 — Apply $\delta_C = \sum \dfrac{P\,k\,L}{AE}$

$$\sum PkL = 200\text{ kN}^2\cdot\text{m}$$ $$\delta_C = \frac{200}{1.5\times10^5} = 1.333\times10^{-3}\text{ m}$$

$\boxed{\delta_C = 1.33\text{ mm downward}}$

Step 1 — Horizontal tension

For a parabolic cable with UDL $w$ over span $L$ and central dip $d$:

$$H = \frac{wL^2}{8d} = \frac{15 \times 60^2}{8 \times 6} = \frac{15 \times 3600}{48} = \frac{54000}{48} = 1125\text{ kN}$$

$\boxed{H = 1125\text{ kN}}$

Step 2 — Maximum tension

The vertical reaction at each support (supports at same level) $= \dfrac{wL}{2} = \dfrac{15\times60}{2} = 450\text{ kN}$.

Maximum tension occurs at the supports:

$$T_{max} = \sqrt{H^2 + V^2} = \sqrt{1125^2 + 450^2} = \sqrt{1265625 + 202500} = \sqrt{1468125} = 1211.7\text{ kN}$$

$\boxed{T_{max} = 1211.7\text{ kN at the supports}}$

(Minimum tension $= H = 1125\text{ kN}$ at the lowest point / mid-span.)

Step 3 — Cable length (approximate parabola)

For a parabolic cable, level supports:

$$L_{cable} \approx L\left(1 + \frac{8}{3}\left(\frac{d}{L}\right)^2\right)$$ $$= 60\left(1 + \frac{8}{3}\left(\frac{6}{60}\right)^2\right) = 60\left(1 + \frac{8}{3}(0.01)\right)$$ $$= 60\left(1 + 0.026667\right) = 60(1.026667) = 61.60\text{ m}$$

$\boxed{L_{cable} \approx 61.60\text{ m}}$

(a) Definitions

Static indeterminacy (degree of statical indeterminacy, DSI): the number of independent force/reaction unknowns in excess of the available independent equations of static equilibrium. If $DSI>0$ the structure is statically indeterminate and equilibrium alone cannot determine all forces.

Kinematic indeterminacy (degree of freedom, DKI): the number of independent joint displacement components (translations and rotations) that are unknown, i.e. not restrained by supports. It governs the displacement (stiffness) method.

(b) (i) Plane truss

For a plane truss: $DSI = (m + r) - 2j$.

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

Determinate (and stable, assuming proper arrangement).

(b) (ii) Plane rigid frame (portal, fixed–fixed, one closed loop, no hinge)

For a rigid plane frame, $DSI = 3C - r_r$, where $C$ = number of closed loops and $r_r$ = released conditions (internal hinges etc.). With one closed loop and no releases:

$$DSI = 3(1) - 0 = 3$$

Equivalently by $DSI = (3m + r) - 3j - r_c$: a single-bay portal has $m=3$, $j=4$, $r=6$ (two fixed supports $\times 3$):

$$DSI = (3\times3 + 6) - 3\times4 = 15 - 12 = 3$$

Statically indeterminate to the 3rd degree (internally + externally), and stable.

(b) (iii) Classification summary

(a) Influence lines

ILD for reaction $R_A$: For a unit load at distance $x$ from $A$, $R_A = \dfrac{L-x}{L} = \dfrac{10-x}{10}$.

• Ordinate $= 1$ at $A$ ($x=0$)
• Ordinate $= 0$ at $B$ ($x=10$) A straight line falling from $1$ to $0$.

ILD for bending moment at $D$ ($a=4\text{ m}$, $b=6\text{ m}$): triangular shape, zero at both supports, peak under $D$.

• Peak ordinate $= \dfrac{ab}{L} = \dfrac{4\times6}{10} = 2.4\text{ m}$ at $x=4$.
• Left limb rises $0 \to 2.4$ over $x=0\to4$; right limb falls $2.4\to0$ over $x=4\to10$.

ILD R_A:        1.0 ____
                     \____ 
                          \____ 0
                A             B

ILD M_D:            /\  (peak 2.4 at D)
                   /  \___
                  /       \___
                A    D        B

(b) Maximum effects from $80\text{ kN}$ moving load

Max reaction at $A$: maximum ILD ordinate is $1.0$ (load at $A$).

$$R_{A,max} = 80 \times 1.0 = 80\text{ kN}$$

Max bending moment at $D$: maximum ordinate is $2.4\text{ m}$ (load at $D$).

$$M_{D,max} = 80 \times 2.4 = 192\text{ kN}\cdot\text{m}$$

$\boxed{R_{A,max} = 80\text{ kN}, \quad M_{D,max} = 192\text{ kN}\cdot\text{m}}$

Step 1 — Bending-moment diagram

For a cantilever with end load $W=25\text{ kN}$, $L=4\text{ m}$, the BM at the fixed end is $-WL = -100\text{ kN}\cdot\text{m}$, varying linearly to $0$ at the free end. The $M/EI$ diagram is a triangle with peak $\dfrac{WL}{EI}$ at $A$ and zero at $B$.

Step 2 — Mohr's first theorem (slope at B)

Slope at $B$ relative to the tangent at fixed end $A$ (which is horizontal) $=$ area of $M/EI$ diagram between $A$ and $B$:

$$\theta_B = \frac{1}{EI}\times\left(\tfrac{1}{2}\times L \times WL\right) = \frac{WL^2}{2EI}$$ $$\theta_B = \frac{25 \times 4^2}{2 \times 12000} = \frac{400}{24000} = 0.016667\text{ rad}$$

$\boxed{\theta_B = 0.01667\text{ rad} = 16.67\times10^{-3}\text{ rad}}$

Step 3 — Mohr's second theorem (deflection at B)

Deflection at $B$ $=$ first moment of the $M/EI$ area about $B$. The triangle's centroid is at $\dfrac{2}{3}L$ from the apex ($A$), i.e. at distance $\dfrac{2L}{3}$ from $B$ measured from $A$... taking moment of the area about $B$, the centroid lies at $\dfrac{2}{3}L$ from $B$:

$$\delta_B = \frac{1}{EI}\times\text{Area}\times\bar{x}_{from B} = \frac{1}{EI}\left(\tfrac{1}{2}WL\cdot L\right)\left(\tfrac{2}{3}L\right) = \frac{WL^3}{3EI}$$ $$\delta_B = \frac{25 \times 4^3}{3 \times 12000} = \frac{25 \times 64}{36000} = \frac{1600}{36000} = 0.04444\text{ m}$$

$\boxed{\delta_B = 44.4\text{ mm downward}}$

Step 1 — Real beam BMD

Reactions $= 20\text{ kN}$ each. Maximum BM at centre $= \dfrac{WL}{4} = \dfrac{40\times8}{4} = 80\text{ kN}\cdot\text{m}$. The BMD is a triangle, peak $80$ at mid-span, zero at supports.

Step 2 — Conjugate beam loading

The conjugate beam (same span, simply supported) is loaded with the $M/EI$ diagram: a triangle of peak $\dfrac{80}{EI}$ at mid-span.

Total elastic load $W^* =$ area of $M/EI$ triangle $= \dfrac{1}{2}\times L \times \dfrac{M_{max}}{EI} = \dfrac{1}{2}\times 8 \times \dfrac{80}{EI} = \dfrac{320}{EI}$.

By symmetry, each conjugate reaction $= \dfrac{1}{2}W^* = \dfrac{160}{EI}$.

Step 3 — Slope at left support

Slope in real beam $=$ shear in conjugate beam at that point $=$ conjugate reaction:

$$\theta_A = \frac{160}{EI} = \frac{160}{16000} = 0.01\text{ rad}$$

$\boxed{\theta_A = 0.0100\text{ rad} = 10\times10^{-3}\text{ rad}}$

(Check: standard result $\theta_A = \dfrac{WL^2}{16EI} = \dfrac{40\times64}{16\times16000}=0.01$ ✓)

Step 4 — Deflection at mid-span

Deflection in real beam $=$ bending moment in conjugate beam at mid-span. Take the left half: conjugate reaction $\dfrac{160}{EI}$ at support, and the half-triangle elastic load.

Half-triangle area (left half) $= \dfrac{1}{2}\times 4 \times \dfrac{80}{EI} = \dfrac{160}{EI}$, with centroid at $\dfrac{4}{3}\text{ m}$ from mid-span (i.e. $\dfrac{2}{3}\times4$ from the support, so $4 - \dfrac{8}{3} = \dfrac{4}{3}$ from centre).

$$M^*_{mid} = \underbrace{\frac{160}{EI}\times 4}_{\text{reaction moment}} - \underbrace{\frac{160}{EI}\times \frac{4}{3}}_{\text{load moment}} = \frac{640}{EI} - \frac{213.33}{EI} = \frac{426.67}{EI}$$ $$\delta_{mid} = \frac{426.67}{16000} = 0.026667\text{ m}$$

$\boxed{\delta_{mid} = 26.7\text{ mm downward}}$

(Check: standard $\delta = \dfrac{WL^3}{48EI} = \dfrac{40\times512}{48\times16000} = 0.02667$ ✓)

Step 1 — Reactions

Total length $AC = 8\text{ m}$. UDL total $= 12\times8 = 96\text{ kN}$ acting at $4\text{ m}$ from $A$. Point load $20\text{ kN}$ at $C$ ($x=8$).

Moments about $A$:

$$R_B(6) = 96(4) + 20(8) = 384 + 160 = 544 \Rightarrow R_B = 90.667\text{ kN}$$ $$R_A = (96 + 20) - 90.667 = 116 - 90.667 = 25.333\text{ kN}$$

$R_A = 25.33\text{ kN}, \quad R_B = 90.67\text{ kN}$

Step 2 — Shear and moment at B

Just left of $B$ ($x=6^-$):

$$V = R_A - 12(6) = 25.333 - 72 = -46.667\text{ kN}$$

Just right of $B$ ($x=6^+$): the overhang carries UDL over $2\text{ m}$ plus point load:

$$V_{6^+} = 12(2) + 20 = 44\text{ kN (to the right of B, supporting the overhang)}$$

Bending moment at $B$ (compute from the overhang, right side — simpler):

$$M_B = -\left[12(2)(1) + 20(2)\right] = -(24 + 40) = -64\text{ kN}\cdot\text{m}$$

(hogging).

Check from left: $M_B = R_A(6) - 12(6)(3) = 25.333(6) - 216 = 152 - 216 = -64\text{ kN}\cdot\text{m}$ ✓

Step 3 — Maximum sagging moment in span AB

Shear in $AB$: $V(x) = R_A - 12x = 25.333 - 12x$. Set $V=0$:

$$x = \frac{25.333}{12} = 2.111\text{ m from } A$$

Maximum sagging moment there:

$$M_{max} = R_A x - 12\frac{x^2}{2} = 25.333(2.111) - 6(2.111)^2$$ $$= 53.48 - 6(4.456) = 53.48 - 26.74 = 26.74\text{ kN}\cdot\text{m}$$

$\boxed{M_{max} = 26.74\text{ kN}\cdot\text{m (sagging) at } x = 2.11\text{ m from } A}$

(a) Muller-Breslau principle

Statement: The influence line for any force quantity (reaction, shear, or bending moment) in a structure is, to a scale, the deflected shape of the structure obtained by removing the restraint corresponding to that quantity and giving a unit displacement (or rotation) in the direction of the released action.

Usefulness: It lets one sketch the shape of an influence line quickly without point-by-point computation, which is especially valuable for indeterminate structures and for locating critical load positions for moving loads.

(b) Influence line for moment at $x=9\text{ m}$

For a simply supported beam of span $L=12\text{ m}$, the ILD for bending moment at a section distance $a$ from the left ($a=9$, $b=L-a=3$) is triangular with peak ordinate:

$$\eta_{peak} = \frac{a\,b}{L} = \frac{9 \times 3}{12} = \frac{27}{12} = 2.25\text{ m}$$

The peak occurs under the section ($x=9$), zero at both supports.

Bending moment from the $60\text{ kN}$ load at $x=9$

Load placed at the panel point (peak of ILD):

$$M_{x=9} = 60 \times 2.25 = 135\text{ kN}\cdot\text{m}$$

$\boxed{\eta_{peak} = 2.25\text{ m}, \quad M = 135\text{ kN}\cdot\text{m}}$