BE Civil Engineering (IOE, TU) Theory of Structures I (IOE, CE 554) Question Paper 2077 Nepal

Reactions. Taking moments about $A$:

$$R_B = \frac{Pa}{L} = \frac{40\times 4}{6} = 26.667\ \text{kN}, \qquad R_A = P - R_B = 40 - 26.667 = 13.333\ \text{kN}.$$

Let $x$ be measured from $A$.

Segment 1 ($0 \le x \le a = 4$ m): Bending moment $M_1 = R_A x = 13.333x$.

$$EI\,y'' = 13.333x$$ $$EI\,y' = 6.667x^2 + C_1$$ $$EI\,y = 2.222x^3 + C_1 x + C_2$$

Segment 2 ($4 \le x \le 6$ m): $M_2 = R_A x - P(x-a) = 13.333x - 40(x-4)$.

$$EI\,y'' = 13.333x - 40(x-4)$$ $$EI\,y' = 6.667x^2 - 20(x-4)^2 + C_3$$ $$EI\,y = 2.222x^3 - 6.667(x-4)^3 + C_3 x + C_4$$

Boundary / continuity conditions.

• At $x=0$, $y=0 \Rightarrow C_2 = 0$.
• Slope continuity at $x=4$ (the $(x-4)$ terms vanish): $C_1 = C_3$.
• Deflection continuity at $x=4$ (the $(x-4)^3$ term vanishes): $C_2 = C_4 = 0$.
• At $x=6$, $y=0$ (segment 2):

$$2.222(6)^3 - 6.667(2)^3 + C_3(6) = 0$$ $$2.222(216) - 6.667(8) + 6C_3 = 0$$ $$480.0 - 53.33 + 6C_3 = 0 \Rightarrow C_3 = -71.11$$

Hence $C_1 = C_3 = -71.11\ \text{kN·m}^2$.

(a) Elastic-curve equations:

$$EI\,y_1 = 2.222x^3 - 71.11x \qquad (0\le x\le 4)$$ $$EI\,y_2 = 2.222x^3 - 6.667(x-4)^3 - 71.11x \qquad (4\le x\le 6)$$

(b) Deflection under the load ($x = a = 4$ m):

$$EI\,y = 2.222(64) - 71.11(4) = 142.2 - 284.4 = -142.2\ \text{kN·m}^3$$ $$y = \frac{-142.2}{25000} = -5.69\times 10^{-3}\ \text{m} = -5.69\ \text{mm}$$

Check against $y_C = \dfrac{Pa^2b^2}{3EIL} = \dfrac{40(16)(4)}{3(25000)(6)} = \dfrac{2560}{450000} = 5.69\times10^{-3}\ \text{m}$. ✓

Deflection under the load $= \mathbf{5.69\ mm}$ (downward).

(c) Slope at $A$ ($x=0$):

$$EI\,y' = 6.667x^2 - 71.11 \Rightarrow EI\,\theta_A = -71.11$$ $$\theta_A = \frac{-71.11}{25000} = -2.844\times 10^{-3}\ \text{rad}$$

Slope at $A = \mathbf{2.84\times 10^{-3}\ rad}$ (clockwise), about $0.163^\circ$.

For a cantilever fixed at $A$, the tangent at $A$ is horizontal, so the slope and deflection at $B$ come directly from the area of the $M/EI$ diagram between $A$ and $B$ and its moment about $B$ (Mohr's theorems).

• $\theta_B = \dfrac{1}{EI}\,(\text{Area of } M\text{ diagram between }A\text{ and }B)$
• $\delta_B = \dfrac{1}{EI}\,(\text{Moment of that area about }B)$

We superpose the two loadings (magnitudes used; both produce hogging moment).

(i) Point load $W = 30$ kN at $B$. $M$ varies linearly from $0$ at $B$ to $WL = 30(4) = 120\ \text{kN·m}$ at $A$ (triangle).

• Area $A_1 = \tfrac12 (L)(WL) = \tfrac12 (4)(120) = 240\ \text{kN·m}^2$.
• Centroid from $B$: $\bar{x}_1 = \tfrac{2}{3}L = 2.667\ \text{m}$.
• Moment about $B$: $A_1\bar{x}_1 = 240 \times 2.667 = 640\ \text{kN·m}^3$.

(Standard checks: $WL^2/2 = 30(16)/2 = 240$ ✓; $WL^3/3 = 30(64)/3 = 640$ ✓.)

(ii) UDL $w = 15$ kN/m. $M$ varies parabolically from $0$ at $B$ to $wL^2/2 = 15(16)/2 = 120\ \text{kN·m}$ at $A$ (parabola, vertex at $B$).

• Area $A_2 = \tfrac13 (L)(wL^2/2) = \tfrac13 (4)(120) = 160\ \text{kN·m}^2$.
• Centroid from $B$: $\bar{x}_2 = \tfrac34 L = 3.0\ \text{m}$.
• Moment about $B$: $A_2\bar{x}_2 = 160 \times 3.0 = 480\ \text{kN·m}^3$.

(Standard checks: $wL^3/6 = 15(64)/6 = 160$ ✓; $wL^4/8 = 15(256)/8 = 480$ ✓.)

Total slope at $B$:

$$\theta_B = \frac{A_1 + A_2}{EI} = \frac{240 + 160}{40000} = \frac{400}{40000} = 1.00\times 10^{-2}\ \text{rad}$$

Total deflection at $B$:

$$\delta_B = \frac{640 + 480}{EI} = \frac{1120}{40000} = 2.80\times 10^{-2}\ \text{m} = 28.0\ \text{mm}$$

Slope at $B = \mathbf{0.0100\ rad}$; deflection at $B = \mathbf{28.0\ mm}$ (downward).

(a) Determinacy. Members $m = 7$ ($AB,BC,AC,BD,CD,CE,DE$); reactions $r = 3$ (pin $=2$, roller $=1$); joints $j = 5$ ($A,B,C,D,E$).

$$m + r = 7 + 3 = 10, \qquad 2j = 2(5) = 10.$$

Since $m + r = 2j$, the truss is statically determinate (and stable). ✓

(b) Reactions. Total downward load $= 20 + 10 = 30\ \text{kN}$. Moments about $A$ (CCW +), loads at $C(x=3)$ and $D(x=4.5)$:

$$\sum M_A = 0:\quad R_E(6) - 20(3) - 10(4.5) = 0$$ $$6R_E = 60 + 45 = 105 \Rightarrow R_E = 17.5\ \text{kN (up)}$$ $$R_A = 30 - 17.5 = 12.5\ \text{kN (up)}, \qquad H_A = 0.$$

(c) Method of joints. Member $AB$: from $A(0,0)$ to $B(1.5,2)$, length $=\sqrt{1.5^2+2^2}=\sqrt{6.25}=2.5\ \text{m}$, so $\cos\theta = 0.6$, $\sin\theta = 0.8$.

Joint $A$ (members $AB$ inclined up-right, $AC$ horizontal right; $R_A=12.5$ up). Take tension positive (member force pulling away from joint).

$$\sum F_y = 0:\quad 12.5 + F_{AB}(0.8) = 0 \Rightarrow F_{AB} = -15.625\ \text{kN}$$

Negative ⇒ compression. $F_{AB} = 15.63\ \text{kN (Compression)}$.

$$\sum F_x = 0:\quad F_{AC} + F_{AB}(0.6) = 0 \Rightarrow F_{AC} = -(-15.625)(0.6) = 9.375\ \text{kN}$$

Positive ⇒ tension. $F_{AC} = 9.38\ \text{kN (Tension)}$.

Joint $C$ (members $AC$ left-horizontal, $CE$ right-horizontal, $BC$ inclined up-left, $CD$ inclined up-right; external $20$ kN down). $BC$ goes from $C(3,0)$ to $B(1.5,2)$: same direction cosines $0.6, 0.8$. By the symmetry of the geometry about $C$, resolve vertically at $C$. The vertical components come only from $BC$ and $CD$. Using $\sum F_y = 0$ at $C$ with the known left-side result and joint-A equilibrium feeding $F_{BC}$:

$$\sum F_y = 0\ \text{at }C:\quad F_{BC}(0.8) + F_{CD}(0.8) - 20 = 0.$$

To isolate $F_{BC}$, resolve vertical equilibrium of joint $B$ (members $AB$, $BC$, $BD$; $BD$ horizontal, no external load):

$$\sum F_y = 0\ \text{at }B:\quad -F_{AB}(0.8) - F_{BC}(0.8) = 0 \Rightarrow F_{BC} = -F_{AB} = -15.625\ \text{kN}$$

Negative ⇒ compression. $F_{BC} = 15.63\ \text{kN (Compression)}$.

Summary: $F_{AB} = 15.63$ kN (C), $F_{AC} = 9.38$ kN (T), $F_{BC} = 15.63$ kN (C).

The parabolic axis is $y = \dfrac{4h}{L^2}x(L-x) = \dfrac{4(6)}{24^2}x(24-x) = \dfrac{x(24-x)}{24}$.

(a) Vertical reactions. Moments about $A$ (load at $x=6$ m):

$$\sum M_A = 0:\quad V_B(24) - 80(6) = 0 \Rightarrow V_B = \frac{480}{24} = 20\ \text{kN}$$ $$V_A = 80 - 20 = 60\ \text{kN}$$

(b) Horizontal thrust. Bending moment at the crown hinge $C$ ($x=12$ m) is zero. Using the right-hand portion (no load between $C$ and $B$):

$$M_C = 0:\quad V_B(12) - H(h) = 0 \Rightarrow 20(12) - H(6) = 0 \Rightarrow H = \frac{240}{6} = 40\ \text{kN}$$

(c) Bending moment under the load ($x = 6$ m). Rise of the axis there:

$$y = \frac{6(24-6)}{24} = \frac{108}{24} = 4.5\ \text{m}$$

Arch BM $=$ (equivalent simple-beam moment) $-\,H\,y$. Beam moment at the load using the left part:

$$M_{\text{beam}} = V_A(6) = 60(6) = 360\ \text{kN·m}$$ $$M = M_{\text{beam}} - H\,y = 360 - 40(4.5) = 360 - 180 = 180\ \text{kN·m}$$

Results: $V_A = \mathbf{60\ kN}$, $V_B = \mathbf{20\ kN}$, $H = \mathbf{40\ kN}$, BM under the load $= \mathbf{180\ kN·m}$ (sagging).

(a) Influence lines (ILs).

IL for $R_A$: straight line, ordinate $1$ at $A$ and $0$ at $B$; ordinate at distance $x$ from $A$ is $(L-x)/L$.

IL for shear at $C$ ($a = 4$ m, $b = 6$ m):

• Unit load between $A$ and $C$: ordinate $= -x/L$, reaching $-a/L = -0.4$ just left of $C$.
• Unit load between $C$ and $B$: ordinate $= +(L-x)/L$, equal to $+b/L = +0.6$ just right of $C$, falling to $0$ at $B$.
• A jump of $1$ at $C$; peak positive $=+0.6$, peak negative $=-0.4$.

IL for bending moment at $C$: triangle, apex under $C$ with peak $= \dfrac{ab}{L} = \dfrac{4\times 6}{10} = 2.4\ \text{m}$; zero at both supports.

(b) UDL $w = 12$ kN/m.

Maximum positive bending moment at $C$: BM influence line is wholly positive, so load the entire span. $M_C = w \times (\text{area of IL})$. Area $= \tfrac12 \times L \times \dfrac{ab}{L} = \tfrac12 \times 10 \times 2.4 = 12.0\ \text{m}^2$.

$$M_{C,\max} = 12 \times 12.0 = 144\ \text{kN·m}$$

Maximum positive bending moment at $C = \mathbf{144\ kN·m}$.

Maximum positive shear at $C$: load only the positive part, segment $C$–$B$ (length $b = 6$ m). Positive area $= \tfrac12 \times b \times \dfrac{b}{L} = \tfrac12 \times 6 \times 0.6 = 1.8\ \text{m}$.

$$V_{C,\max}^{+} = 12 \times 1.8 = 21.6\ \text{kN}$$

Maximum positive shear at $C = \mathbf{21.6\ kN}$.

(For reference, max negative shear loads $A$–$C$: area $=\tfrac12(4)(0.4)=0.8\,\text{m}$ giving $-9.6$ kN.)

Real beam. Reactions $= P/2 = 25$ kN each; maximum BM at mid-span $= PL/4 = 50(8)/4 = 100\ \text{kN·m}$. The $M$ diagram is a triangle (0 at supports, $100$ at centre).

Conjugate beam. Same span, simply supported, loaded with $M/EI$ (triangular, peak $=100/EI$ at centre). Total elastic load $= \dfrac{1}{EI}\left(\tfrac12 \times 8 \times 100\right) = \dfrac{400}{EI}$. By symmetry each conjugate reaction $= \dfrac{200}{EI}$.

Maximum deflection $=$ bending moment of the conjugate beam at mid-span. Using the left half: the conjugate reaction $200/EI$ acts at the left support; the elastic load on the left half is a triangle of area $\tfrac12 \times 4 \times (100/EI) = 200/EI$ with centroid $\tfrac23(4)=2.667$ m from the support, i.e. $4-2.667 = 1.333$ m from mid-span.

$$M'_{mid} = \frac{200}{EI}(4) - \frac{200}{EI}(1.333) = \frac{800 - 266.7}{EI} = \frac{533.3}{EI}$$ $$\delta_{\max} = \frac{533.3}{30000} = 0.01778\ \text{m} = 17.78\ \text{mm}$$

Check: $\delta = \dfrac{PL^3}{48EI} = \dfrac{50(512)}{48(30000)} = \dfrac{25600}{1.44\times10^6} = 0.01778\ \text{m}$ ✓.

Maximum deflection $= \mathbf{17.78\ mm}$ (downward).

General expression. Apply a virtual unit load at the joint, in the required direction. By virtual work the deflection is

$$\Delta = \sum_{i=1}^{n} \frac{P_i\,u_i\,L_i}{A_i E_i}$$

for each member $i$: $P_i$ = axial force due to the real loading; $u_i$ = axial force due to the virtual unit load; $L_i$ = length; $A_i$ = cross-sectional area; $E_i$ = Young's modulus. The term $P_iL_i/(A_iE_i)$ is the real member elongation; multiplying by $u_i$ and summing gives the internal virtual work, equal to external virtual work $1\cdot\Delta$.

Numerical contribution. $A = 1500\ \text{mm}^2 = 1.5\times10^{-3}\ \text{m}^2$; $E = 200\ \text{GPa} = 200\times10^{6}\ \text{kN/m}^2$.

$$AE = (1.5\times10^{-3})(200\times10^{6}) = 3.0\times10^{5}\ \text{kN}$$ $$\frac{PL}{AE} = \frac{120\times 4}{3.0\times10^{5}} = \frac{480}{300000} = 1.6\times10^{-3}\ \text{m} = 1.6\ \text{mm}$$ $$\frac{PuL}{AE} = (1.6\times10^{-3})(0.75) = 1.2\times10^{-3}\ \text{m}$$

Member contribution $= \mathbf{1.2\ mm}$ (in the direction of the unit load).

For a horizontally distributed udl the cable profile is parabolic and $H$ is constant.

(a) Horizontal tension (supports at the same level):

$$H = \frac{wL^2}{8d} = \frac{10\times 40^2}{8\times 4} = \frac{16000}{32} = 500\ \text{kN}$$

$H = \mathbf{500\ kN}$.

(b) Maximum tension at a support, where the vertical component equals the reaction $V = wL/2$:

$$V = \frac{wL}{2} = \frac{10\times 40}{2} = 200\ \text{kN}$$ $$T_{\max} = \sqrt{H^2 + V^2} = \sqrt{500^2 + 200^2} = \sqrt{290000} = 538.5\ \text{kN}$$

$T_{\max} = \mathbf{538.5\ kN}$.

(c) Inclination at support:

$$\tan\theta = \frac{V}{H} = \frac{200}{500} = 0.4 \Rightarrow \theta = \tan^{-1}(0.4) = 21.8^\circ$$

(Also $\tan\theta = 4d/L = 16/40 = 0.4$ ✓.) $\theta = \mathbf{21.8^\circ}$ from the horizontal.

Loads. UDL total $= 8\times 8 = 64$ kN at centroid $x = 4$ m from $A$; point load $15$ kN at $C$ ($x = 8$ m). $B$ is at $x = 6$ m.

(a) Reactions. Moments about $A$ (CCW +):

$$\sum M_A = 0:\quad R_B(6) - 64(4) - 15(8) = 0$$ $$6R_B = 256 + 120 = 376 \Rightarrow R_B = 62.667\ \text{kN}$$ $$R_A = (64 + 15) - 62.667 = 79 - 62.667 = 16.333\ \text{kN}$$

$R_A = 16.33\ \text{kN}$, $R_B = 62.67\ \text{kN}$ (both up).

(b) Shear at $B$. Just left of $B$ ($x = 6^-$): $V = R_A - w(6) = 16.333 - 48 = -31.667\ \text{kN}$. Just right of $B$: $V = -31.667 + 62.667 = +31.0\ \text{kN}$ (balances overhang load $w(2)+15 = 16+15 = 31$ kN ✓).

Bending moment at $B$ (from the overhang side):

$$M_B = -\big[w(2)(1) + 15(2)\big] = -\big[8(2)(1) + 30\big] = -46\ \text{kN·m}$$

$M_B = -46\ \text{kN·m}$ (hogging).

Maximum sagging moment in $AB$. Shear zero at $x = R_A/w = 16.333/8 = 2.042\ \text{m}$ from $A$.

$$M_{\max} = R_A x - w\frac{x^2}{2} = 16.333(2.042) - 4(2.042)^2 = 33.35 - 16.68 = 16.68\ \text{kN·m}$$

Maximum sagging moment $= \mathbf{16.68\ kN·m}$ at $x = 2.04\ m$ from $A$.

Reactions. Supports at $A(x=0)$ and $B(x=12)$.

$$\sum M_A = 0:\quad R_B(12) - 20(3) - 40(6) = 0 \Rightarrow 12R_B = 60 + 240 = 300 \Rightarrow R_B = 25\ \text{kN}$$ $$R_A = (20+40) - 25 = 35\ \text{kN}$$

Section through the second panel (cut top chord, diagonal, bottom chord between $x=3$ and $x=6$). Take the left portion; external vertical forces: $R_A = 35$ up and $20$ down at $x=3$. Net vertical (panel shear) on the left part:

$$V = R_A - 20 = 35 - 20 = 15\ \text{kN (upward)}$$

The two chord forces are horizontal, so only the diagonal balances this shear. At $45^\circ$ its vertical component is $F_d\sin45^\circ = F_d/\sqrt2$:

$$\frac{F_d}{\sqrt2} = 15 \Rightarrow F_d = 15\sqrt2 = 21.21\ \text{kN}$$

Nature. In a Pratt truss the diagonals are arranged to carry panel shear in tension under gravity loading; the diagonal must pull the cut faces together to resist the net upward shear of the left part, so it is in tension.

Force in the second-panel diagonal $= \mathbf{21.21\ kN\ (Tension)}$.

(a) Definitions.

• Static indeterminacy (SI): number of unknown forces (reactions + internal member forces) in excess of the independent static equilibrium equations. SI $>0$ ⇒ indeterminate; $=0$ ⇒ determinate; $<0$ ⇒ unstable.
• Kinematic indeterminacy (KI): number of independent unknown joint displacement components (degrees of freedom) not restrained by supports.

Standard expressions:

• Plane truss: $D_s = (m + r) - 2j$.
• Plane rigid frame: $D_s = (3m + r) - 3j - c$, where $c$ = number of internal release (condition) equations.
• Plane beam: $D_s = r - 3 - c$.

Here $m$ = members, $r$ = reaction components, $j$ = joints, $c$ = internal conditions.

(b)(i) Portal frame, two fixed supports. Fixed supports: $r = 2\times3 = 6$; $m = 3$, $j = 4$, $c = 0$.

$$D_s = 3m + r - 3j - c = 3(3) + 6 - 3(4) - 0 = 9 + 6 - 12 = 3.$$

Degree of static indeterminacy $= \mathbf{3}$.

(b)(ii) Continuous beam, 4 supports (1 pin + 3 rollers). $r = 2 + 3(1) = 5$, $c = 0$.

$$D_s = r - 3 - c = 5 - 3 - 0 = 2.$$

Degree of static indeterminacy $= \mathbf{2}$.