BE Civil Engineering (IOE, TU) Theory of Structures I (IOE, CE 554) Question Paper 2078 Nepal

Support reactions

By symmetry, $R_A = R_B = \dfrac{wL}{2} = \dfrac{8 \times 6}{2} = 24\text{ kN}$.

Bending moment at a section distance $x$ from $A$

$$M(x) = R_A x - \frac{w x^2}{2} = 24x - 4x^2 \quad(\text{kN}\cdot\text{m})$$

Differential equation of the elastic curve

$$EI\frac{d^2 y}{dx^2} = M(x) = 24x - 4x^2$$

Integrating once (slope):

$$EI\frac{dy}{dx} = 12x^2 - \frac{4x^3}{3} + C_1$$

Integrating again (deflection):

$$EI\,y = 4x^3 - \frac{x^4}{3} + C_1 x + C_2$$

Boundary conditions

At $x = 0$, $y = 0 \Rightarrow C_2 = 0$.

By symmetry the slope is zero at mid-span $x = L/2 = 3\text{ m}$:

$$0 = 12(3)^2 - \frac{4(3)^3}{3} + C_1 = 108 - 36 + C_1 \Rightarrow C_1 = -72$$

(Check with $y=0$ at $x=6$: $4(216) - \frac{1296}{3} - 72(6) = 864 - 432 - 432 = 0$ \u2713)

(a) Equation of the elastic curve

$$\boxed{EI\,y = 4x^3 - \frac{x^4}{3} - 72x}$$

(b) Maximum deflection

Occurs at mid-span $x = 3\text{ m}$ (zero slope, by symmetry):

$$EI\,y_{max} = 4(3)^3 - \frac{(3)^4}{3} - 72(3) = 108 - 27 - 216 = -135\text{ kN}\cdot\text{m}^3$$ $$y_{max} = \frac{-135}{24000} = -5.625\times10^{-3}\text{ m} = -5.625\text{ mm}$$

The negative sign indicates a downward deflection. This matches the standard result $y_{max} = \dfrac{5wL^4}{384EI} = \dfrac{5(8)(6^4)}{384(24000)} = \dfrac{51840}{9216000} = 5.625\times10^{-3}\text{ m}$. ✓

Maximum deflection $= 5.625\text{ mm}$ (downward) at mid-span.

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

$$EI\,\theta_A = C_1 = -72 \Rightarrow \theta_A = \frac{-72}{24000} = -3.0\times10^{-3}\text{ rad}$$

Slope at $A = 3.0\times10^{-3}\text{ rad} = 0.172^{\circ}$ (clockwise). By symmetry $\theta_B = +3.0\times10^{-3}\text{ rad}$.

Geometry / coordinates

$A(0,0),\ B(4,0),\ C(8,0),\ D(12,0),\ E(4,3),\ F(8,3)$.

Member $AE$: $\Delta x = 4,\ \Delta y = 3$, length $=\sqrt{4^2+3^2}=5\text{ m}$. Direction cosines: $\cos\alpha = 4/5 = 0.8$, $\sin\alpha = 3/5 = 0.6$.

Support reactions

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

Taking moments about $A$ (roller reaction $R_D$ vertical):

$$\sum M_A = 0:\ R_D(12) - 20(4) - 20(8) = 0 \Rightarrow R_D = \frac{80 + 160}{12} = 20\text{ kN}\ (\uparrow)$$

Vertical equilibrium: $R_{Ay} + R_D = 40 \Rightarrow R_{Ay} = 20\text{ kN}\ (\uparrow)$.

No horizontal loads $\Rightarrow R_{Ax} = 0$.

Joint A (members $AB$ horizontal, $AE$ inclined at $\alpha$):

Vertical: $R_{Ay} + F_{AE}\sin\alpha = 0$ (assuming $AE$ tension pulling away from joint, its vertical component is $+F_{AE}\sin\alpha$ upward only if it points up-right; here $E$ is up-right of $A$).

$$20 + F_{AE}(0.6) = 0 \Rightarrow F_{AE} = -33.33\text{ kN}$$

$F_{AE} = 33.33\text{ kN (compression)}$

Horizontal: $F_{AB} + F_{AE}\cos\alpha = 0 \Rightarrow F_{AB} = -(-33.33)(0.8) = 26.67\text{ kN}$

$F_{AB} = 26.67\text{ kN (tension)}$

Joint B (members $AB$, $BC$ horizontal; $BE$ vertical):

No external load at $B$. Vertical: $F_{BE} = 0$.

$F_{BE} = 0$ (zero-force member)

Horizontal: $F_{BC} = F_{AB} = 26.67\text{ kN (tension)}$ (no other horizontal member except chord).

Joint E (members $AE$, $EF$ horizontal-ish, $BE$ vertical, $EC$ diagonal; external load $20\text{ kN}\downarrow$):

$AE$ direction from $E$ toward $A$ is down-left: components $(-0.8,-0.6)$ scaled by force. With $F_{AE} = -33.33$ (compression, pushing the joint away from $A$, i.e. up-right $(+0.8,+0.6)$ on joint E): contribution $(+26.67, +20.0)$.

Vertical at $E$: contribution of $AE$ $= +20.0$; load $= -20$; $BE = 0$; $EC$ provides the balance. $EC$: from $E(4,3)$ to $C(8,0)$, $\Delta x = 4,\ \Delta y = -3$, length $5$, cosines $(0.8,-0.6)$.

$$\sum F_y:\ 20.0 - 20 + F_{EC}(-0.6) = 0 \Rightarrow F_{EC} = 0$$

Horizontal at $E$: $AE$ contributes $+26.67$ (rightward), $EF$ horizontal, $EC$ contributes $F_{EC}(0.8)=0$.

$$\sum F_x:\ 26.67 + F_{EF} + 0 = 0 \Rightarrow F_{EF} = -26.67\text{ kN}$$

$F_{EF} = 26.67\text{ kN (compression)}$

Summary

The results are symmetric, so $F_{FD}=33.33$ (C), $F_{CD}=26.67$ (T), $F_{CF}=0$.

Geometry

Parabolic axis: $y = \dfrac{4h}{L^2}x(L - x) = \dfrac{4(6)}{24^2}x(24 - x) = \dfrac{24}{576}x(24-x) = \dfrac{x(24-x)}{24}$.

Vertical reactions

$$\sum M_A = 0:\ V_B(24) - 60(6) = 0 \Rightarrow V_B = \frac{360}{24} = 15\text{ kN}\ (\uparrow)$$ $$\sum F_y = 0:\ V_A = 60 - 15 = 45\text{ kN}\ (\uparrow)$$

Horizontal thrust (use the crown hinge $C$ at $x = 12\text{ m}$, where the bending moment is zero). Consider the right segment $C$–$B$ (no load on it):

$$M_C = 0:\ V_B(12) - H(h) = 0 \Rightarrow 15(12) - H(6) = 0 \Rightarrow H = \frac{180}{6} = 30\text{ kN}$$

By horizontal equilibrium $H_A = H_B = H = 30\text{ kN}$ (thrust).

(a) Reactions: $V_A = 45\text{ kN}\uparrow,\ V_B = 15\text{ kN}\uparrow,\ H_A = H_B = 30\text{ kN}$ (inward thrust).

(b) Bending moment under the load at $x = 6\text{ m}$.

Rise of arch axis at $x = 6$:

$$y = \frac{6(24-6)}{24} = \frac{6\times18}{24} = \frac{108}{24} = 4.5\text{ m}$$

The load is AT the section, so the beam moment (from left, taking the section) uses only $V_A$:

$$M_{beam} = V_A(6) = 45 \times 6 = 270\text{ kN}\cdot\text{m}$$

Arch bending moment $= M_{beam} - H\,y$:

$$M_{x=6} = 270 - 30(4.5) = 270 - 135 = 135\text{ kN}\cdot\text{m}$$

Bending moment under the load $= 135\text{ kN}\cdot\text{m}$ (sagging, tension on the inner/intrados side).

Verification at crown $x=12$: $M = V_A(12) - W(12-6) - H\,y_C = 45(12) - 60(6) - 30(6) = 540 - 360 - 180 = 0$ ✓ (crown hinge).

Let $a = 4\text{ m}$ (distance $AC$), $b = 6\text{ m}$ (distance $CB$), span $L = 10\text{ m}$.

(a) Influence lines (ILs)

IL for $R_A$: ordinate $= \dfrac{L - x}{L}$ for a unit load at distance $x$ from $A$. It is a straight line from $1.0$ at $A$ ($x=0$) to $0$ at $B$ ($x=10$). At $C$: $R_A = (10-4)/10 = 0.6$.

IL for shear at $C$ ($V_C$): Two straight segments.

• For unit load in $AC$ ($0\le x<4$): $V_C = -\dfrac{x}{L}$, rising from $0$ at $A$ to $-a/L = -0.4$ just left of $C$.
• For unit load in $CB$ ($4< x\le10$): $V_C = +\dfrac{L-x}{L}$, from $+b/L = +0.6$ just right of $C$ down to $0$ at $B$. Discontinuity (jump) of $1.0$ at $C$. Peak positive ordinate $= +0.6$; peak negative ordinate $= -0.4$.

IL for bending moment at $C$ ($M_C$): Triangle with apex under $C$. Peak ordinate $= \dfrac{ab}{L} = \dfrac{4\times6}{10} = 2.4\text{ m}$. Zero at both supports, linear in between.

(b) Single moving load $W = 50\text{ kN}$

Maximum bending moment at $C$ occurs when the load is exactly at $C$ (peak IL ordinate):

$$M_{C,max} = W \times \frac{ab}{L} = 50 \times 2.4 = 120\text{ kN}\cdot\text{m}$$

Maximum shear at $C$: place the load at the larger IL ordinate. Positive: $50 \times 0.6 = +30\text{ kN}$ (load just right of $C$). Negative: $50 \times (-0.4) = -20\text{ kN}$ (load just left of $C$).

Maximum bending moment at $C = 120\text{ kN}\cdot\text{m}$; maximum absolute shear at $C = 30\text{ kN}$ (positive).

(c) UDL $w = 12\text{ kN/m}$ for maximum positive shear at $C$

For maximum positive shear, load only the portion of the beam where the IL is positive, i.e. segment $CB$ (length $b = 6\text{ m}$). The IL is a triangle over $CB$ with peak $+0.6$ at $C$ and $0$ at $B$.

Area of positive IL region $= \dfrac{1}{2} \times b \times (+0.6) = \dfrac{1}{2}\times 6 \times 0.6 = 1.8\text{ m}$.

$$V_{C,max}^{+} = w \times (\text{positive IL area}) = 12 \times 1.8 = 21.6\text{ kN}$$

Maximum positive shear at $C$ (UDL) $= 21.6\text{ kN}$.

(a) Horizontal tension

For a parabolic cable carrying a horizontal UDL, taking moments about the support for one half, or using the standard relation:

$$H = \frac{wL^2}{8f} = \frac{5 \times 40^2}{8 \times 4} = \frac{5 \times 1600}{32} = \frac{8000}{32} = 250\text{ kN}$$

Horizontal tension $H = 250\text{ kN}$.

(b) Maximum tension

The vertical reaction at each support (supports at same level) $= \dfrac{wL}{2} = \dfrac{5\times40}{2} = 100\text{ kN}$.

Maximum tension occurs at the supports (steepest slope):

$$T_{max} = \sqrt{H^2 + V^2} = \sqrt{250^2 + 100^2} = \sqrt{62500 + 10000} = \sqrt{72500} = 269.26\text{ kN}$$

Maximum tension $T_{max} = 269.26\text{ kN}$, occurring at the supports.

The minimum tension equals $H = 250\text{ kN}$ at the lowest point (mid-span).

(c) Length of cable (parabolic approximation, supports at same level):

$$S = L\left(1 + \frac{8}{3}\frac{f^2}{L^2} - \frac{32}{5}\frac{f^4}{L^4} + \cdots\right)$$

Using the leading two terms:

$$\frac{f}{L} = \frac{4}{40} = 0.1,\quad \frac{f^2}{L^2} = 0.01,\quad \frac{f^4}{L^4} = 0.0001$$ $$S = 40\left(1 + \frac{8}{3}(0.01) - \frac{32}{5}(0.0001)\right) = 40\left(1 + 0.026667 - 0.00064\right)$$ $$S = 40 \times 1.026027 = 41.041\text{ m}$$

Length of cable $\approx 41.04\text{ m}$.

(Using only the first correction term gives $S = 40(1.026667) = 41.067\text{ m}$, about $41.07\text{ m}$ — either is acceptable.)

Definitions

• Statically determinate: A structure whose support reactions and internal member forces can be found using only the equations of static equilibrium ($\sum F_x = 0,\ \sum F_y = 0,\ \sum M = 0$). The number of unknowns equals the number of independent equilibrium equations.
• Statically indeterminate: The unknowns exceed the available equilibrium equations; extra (compatibility) equations involving deformations are required. The surplus is the degree of static indeterminacy (DSI).
• Stability: A structure is stable if it can maintain its geometry (resist a general system of loads) without rigid-body motion or collapse, i.e. it has an adequate number and arrangement of members/supports. Improper arrangement (e.g. concurrent or parallel reactions) causes geometric instability even if the count is satisfied.

Classification formulae

Plane truss: compare $m + r$ with $2j$.

• $m + r < 2j$ → unstable
• $m + r = 2j$ → statically determinate (if stable arrangement)
• $m + r > 2j$ → indeterminate, with $\text{DSI} = (m + r) - 2j$

Plane frame (rigid joints): $\text{DSI} = (3m + r) - 3j$ (and add internal condition releases as needed).

Classification of the given truss ($m = 11,\ r = 3,\ j = 7$):

$$m + r = 11 + 3 = 14,\qquad 2j = 2 \times 7 = 14$$

Since $m + r = 2j$, the truss is statically determinate (and stable, assuming a proper member/support arrangement).

Bending moment diagram (BMD)

Measuring $x$ from the free end $B$, $M = -P x$; at the fixed end $A$, $M_A = -P L = -15 \times 3 = -45\text{ kN}\cdot\text{m}$. The BMD is a triangle with maximum magnitude $45\text{ kN}\cdot\text{m}$ at $A$ and zero at $B$.

Moment-area theorems (reference tangent at the fixed support $A$, where slope and deflection are zero).

Area of the $M/EI$ diagram (triangle):

$$A = \frac{1}{2}\times L \times \frac{PL}{EI} = \frac{1}{2}\times 3 \times \frac{45}{EI} = \frac{67.5}{EI}$$

Theorem 1 — slope at $B$ equals the area of the $M/EI$ diagram between $A$ and $B$:

$$\theta_B = \frac{A}{1} = \frac{67.5}{EI} = \frac{67.5}{9000} = 7.5\times10^{-3}\text{ rad}$$

This equals the standard result $\dfrac{PL^2}{2EI} = \dfrac{15(3^2)}{2(9000)} = \dfrac{135}{18000} = 7.5\times10^{-3}\text{ rad}$. ✓

Theorem 2 — deflection at $B$ equals the first moment of the $M/EI$ area about $B$. The centroid of the triangle (apex at $A$, zero at $B$) is at distance $\bar{x} = \frac{2}{3}L = 2\text{ m}$ from $B$:

$$\delta_B = A \times \bar{x} = \frac{67.5}{EI}\times 2 = \frac{135}{EI} = \frac{135}{9000} = 15\times10^{-3}\text{ m} = 15\text{ mm}$$

This equals the standard result $\dfrac{PL^3}{3EI} = \dfrac{15(3^3)}{3(9000)} = \dfrac{405}{27000} = 15\times10^{-3}\text{ m}$. ✓

Slope at free end $\theta_B = 7.5\times10^{-3}\text{ rad}$; deflection $\delta_B = 15\text{ mm}$ (downward).

Unit load method (virtual work) for trusses

To find the deflection $\delta$ at a joint in a given direction:

1. Compute the real member forces $P$ due to the actual applied loads.
2. Remove the real loads and apply a single unit virtual load ($1$) at the joint in the direction of the required deflection; compute the virtual member forces $u$.
3. The deflection is obtained from the virtual work equation:

$$1\cdot\delta = \sum \frac{P\,u\,L}{A E}$$

The sign of $\delta$ follows the assumed direction of the unit load (positive = same direction).

Computation

$E = 200\text{ GPa} = 200\times10^{6}\text{ kN/m}^2$. Convert areas to m$^2$.

Compute $\dfrac{P u L}{A E}$ for each member (forces in kN, $L$ in m, $A$ in m$^2$):

Sum:

$$\delta_C = (4.80 + 2.34 + 2.50)\times10^{-4} = 9.64\times10^{-4}\text{ m}$$

Vertical deflection at $C$, $\delta_C = 9.64\times10^{-4}\text{ m} = 0.964\text{ mm}$ (downward, same sense as the unit load).

Conjugate beam theorems

1. The slope at any section of the real beam equals the shear force at the corresponding section of the conjugate beam.
2. The deflection at any section of the real beam equals the bending moment at the corresponding section of the conjugate beam.

The conjugate beam is loaded with the $M/EI$ diagram of the real beam.

Support conversion rules

(A simply supported beam remains simply supported in its conjugate.)

Application

Real beam: central load $P = 40\text{ kN}$, span $L = 8\text{ m}$. Maximum bending moment at mid-span:

$$M_{max} = \frac{PL}{4} = \frac{40\times8}{4} = 80\text{ kN}\cdot\text{m}$$

The BMD is a triangle, peak $80\text{ kN}\cdot\text{m}$ at centre, zero at supports. The conjugate beam carries the $M/EI$ diagram (a triangle of peak $80/EI$).

Total load on conjugate beam (area of $M/EI$ triangle):

$$W^* = \frac{1}{2}\times L \times \frac{M_{max}}{EI} = \frac{1}{2}\times 8 \times \frac{80}{EI} = \frac{320}{EI}$$

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

Deflection at mid-span $=$ bending moment of conjugate beam at the centre. Take the left half: reaction $\dfrac{160}{EI}$ at distance $4\text{ m}$, minus the moment of the left-half triangular load (area $=\frac{1}{2}\times4\times\frac{80}{EI}=\frac{160}{EI}$, centroid at $\frac{2}{3}\times4 = 2.667\text{ m}$ from the support, i.e. $1.333\text{ m}$ from mid-span):

$$\delta_{mid} = \frac{160}{EI}(4) - \frac{160}{EI}(1.333) = \frac{160}{EI}(4 - 1.333) = \frac{160\times2.667}{EI} = \frac{426.7}{EI}$$ $$\delta_{mid} = \frac{426.7}{20000} = 0.02133\text{ m} = 21.33\text{ mm}$$

This matches the standard result $\dfrac{PL^3}{48EI} = \dfrac{40(8^3)}{48(20000)} = \dfrac{20480}{960000} = 0.02133\text{ m}$. ✓

Mid-span deflection $= 21.33\text{ mm}$ (downward).

Loads

• UDL on $AB$: $10\text{ kN/m}\times 6\text{ m} = 60\text{ kN}$, acting at $3\text{ m}$ from $A$ (mid-span of $AB$).
• Point load at $C$: $15\text{ kN}$, at $8\text{ m}$ from $A$ ($6 + 2$).

Reactions ($R_A$ at $A$, $R_B$ at $B$, both upward):

$$\sum M_A = 0:\ R_B(6) - 60(3) - 15(8) = 0 \Rightarrow R_B = \frac{180 + 120}{6} = \frac{300}{6} = 50\text{ kN}$$ $$\sum F_y = 0:\ R_A + R_B = 60 + 15 = 75 \Rightarrow R_A = 75 - 50 = 25\text{ kN}$$

Reactions: $R_A = 25\text{ kN}\uparrow,\ R_B = 50\text{ kN}\uparrow$.

Bending moment over $B$ (take the cantilever overhang to the right of $B$, only the $15\text{ kN}$ load at $C$, $2\text{ m}$ away):

$$M_B = -15\times 2 = -30\text{ kN}\cdot\text{m}$$

Bending moment over $B = -30\text{ kN}\cdot\text{m}$ (hogging).

Location of maximum sagging moment in $AB$

Shear in $AB$ at distance $x$ from $A$: $V(x) = R_A - 10x = 25 - 10x$. Maximum sagging moment where $V = 0$:

$$25 - 10x = 0 \Rightarrow x = 2.5\text{ m from }A$$

Maximum sagging moment:

$$M_{max} = R_A x - \frac{10 x^2}{2} = 25(2.5) - 5(2.5)^2 = 62.5 - 31.25 = 31.25\text{ kN}\cdot\text{m}$$

Maximum sagging moment $= 31.25\text{ kN}\cdot\text{m}$ at $2.5\text{ m}$ from $A$.

Derivation of horizontal thrust for a unit load at distance $a$ from $A$

Let the unit load (value $1$) act at distance $a$ from the left support $A$ ($b = L - a$ from $B$).

Vertical reactions:

$$V_A = \frac{L - a}{L} = \frac{b}{L},\qquad V_B = \frac{a}{L}$$

The horizontal thrust is found from the condition that the bending moment at the crown $C$ (mid-span, $x = L/2$, rise $h$) is zero.

Case 1 — load on the left half ($a \le L/2$): Consider the right segment $C$–$B$ (unloaded):

$$M_C = V_B\cdot\frac{L}{2} - H\cdot h = 0 \Rightarrow H = \frac{V_B L}{2h} = \frac{(a/L)L}{2h} = \frac{a}{2h}$$

Case 2 — load on the right half ($a \ge L/2$): by symmetry $H = \dfrac{b}{2h} = \dfrac{L-a}{2h}$.

So the influence line for $H$ consists of two straight lines, rising from $0$ at each support to a peak under the crown.

Maximum (peak) IL ordinate — occurs when the load is at the crown, $a = L/2$:

$$H_{max,IL} = \frac{a}{2h}\Big|_{a=L/2} = \frac{L/2}{2h} = \frac{L}{4h}$$

Numerical evaluation ($L = 30\text{ m},\ h = 5\text{ m}$):

$$\text{Peak IL ordinate} = \frac{L}{4h} = \frac{30}{4\times5} = \frac{30}{20} = 1.5$$

Horizontal thrust for a $100\text{ kN}$ load at mid-span:

$$H = 100 \times 1.5 = 150\text{ kN}$$

Peak IL ordinate for horizontal thrust $= 1.5$; horizontal thrust for $100\text{ kN}$ at crown $= 150\text{ kN}$.

Check directly: load at crown gives $V_A = V_B = 50\text{ kN}$; using right segment $M_C = 50(15) - H(5) = 0 \Rightarrow H = 750/5 = 150\text{ kN}$. ✓