BE Civil Engineering (IOE, TU) Theory of Structures I (IOE, CE 554) Question Paper 2076 Nepal

Step 1: Support reactions

By symmetry of the UDL over span $L$:

$$R_A = R_B = \frac{wL}{2} = \frac{12 \times 6}{2} = 36\,\text{kN}$$

Step 2: Bending moment at distance $x$ from $A$

$$M(x) = R_A x - \frac{w x^2}{2} = 36x - 6x^2 \quad (\text{kN·m})$$

Step 3: Governing differential equation

$$EI\frac{d^2 y}{dx^2} = M(x) = 36x - 6x^2$$

Integrate once (slope):

$$EI\frac{dy}{dx} = 18x^2 - 2x^3 + C_1$$

Integrate again (deflection):

$$EI\,y = 6x^3 - \tfrac{1}{2}x^4 + C_1 x + C_2$$

Step 4: Boundary conditions

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

At $x=L=6,\; y=0$:

$$0 = 6(6)^3 - \tfrac12(6)^4 + 6C_1 = 6(216) - \tfrac12(1296) + 6C_1 = 1296 - 648 + 6C_1$$ $$6C_1 = -648 \Rightarrow C_1 = -108$$

Step 5: Elastic curve

$$\boxed{EI\,y = 6x^3 - \tfrac12 x^4 - 108x} \quad (\text{kN·m}^3)$$

Step 6: Maximum deflection

By symmetry the maximum deflection is at midspan $x = 3\,\text{m}$:

$$EI\,y_{max} = 6(27) - \tfrac12(81) - 108(3) = 162 - 40.5 - 324 = -202.5\,\text{kN·m}^3$$

Check with standard formula $y_{max} = \dfrac{5wL^4}{384EI} = \dfrac{5\times12\times6^4}{384\,EI} = \dfrac{77760}{384\,EI} = \dfrac{202.5}{EI}$ — agrees.

Compute $EI$:

$$EI = (200\times10^{9}\,\text{Pa})(80\times10^{-6}\,\text{m}^4) = 16\times10^{6}\,\text{N·m}^2 = 16{,}000\,\text{kN·m}^2$$ $$y_{max} = \frac{202.5\,\text{kN·m}^3}{16000\,\text{kN·m}^2} = 0.012656\,\text{m} = \mathbf{12.66\,mm\ (downward)\ at\ midspan}$$

Step 7: Slope at A

$$EI\,\theta_A = \left.\frac{dy}{dx}\right|_{x=0} = C_1 = -108\,\text{kN·m}^2$$ $$\theta_A = \frac{-108}{16000} = -0.00675\,\text{rad} = \mathbf{6.75\times10^{-3}\,rad}$$

Standard check: $\theta_A = \dfrac{wL^3}{24EI} = \dfrac{12\times216}{24\times16000} = \dfrac{2592}{384000}=0.00675$ rad — agrees. The negative sign denotes a clockwise rotation at $A$.

Geometry

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

End diagonal $AD$: run $4$, rise $3$, length $\sqrt{4^2+3^2}=5\,\text{m}$; direction cosines $\cos\alpha = 4/5 = 0.8$, $\sin\alpha = 3/5 = 0.6$.

(a) Reactions

Loads $40\,\text{kN}$ at $C\,(x=4)$ and $E\,(x=8)$; total $=80\,\text{kN}$.

$\sum M_A = 0:\; R_B(12) = 40(4) + 40(8) = 160 + 320 = 480 \Rightarrow R_B = 40\,\text{kN}$

$\sum F_y = 0:\; R_A = 80 - 40 = 40\,\text{kN}$ (also clear from symmetry).

$$\mathbf{R_A = R_B = 40\,kN\ (upward)}$$

(b) Method of joints

Joint A (members $AC$ horizontal, $AD$ inclined)

$\sum F_y = 0:\; R_A + F_{AD}\sin\alpha = 0$

$$40 + F_{AD}(0.6) = 0 \Rightarrow F_{AD} = -66.67\,\text{kN}$$

Negative ⇒ member pushes into the joint ⇒ $F_{AD} = 66.67\,kN\ (Compression)$.

$\sum F_x = 0:\; F_{AC} + F_{AD}\cos\alpha = 0$

$$F_{AC} = -(-66.67)(0.8) = +53.33\,\text{kN}$$

$F_{AC} = 53.33\,kN\ (Tension)$

Joint C (members $AC$, $CE$ horizontal, $CD$ vertical, plus $40\,\text{kN}$ downward)

$\sum F_y = 0:\; F_{CD} - 40 = 0 \Rightarrow F_{CD} = +40\,\text{kN}$ $F_{CD} = 40\,kN\ (Tension)$ — the vertical carries the panel load up to joint $D$.

(c) Method of sections

Cut a vertical section through the centre panel slicing $DF$ (top), $DE$ (diagonal) and $CE$ (bottom). Consider the left free body (joints $A$, $C$, $D$): external forces $R_A = 40\,\text{kN}\uparrow$ at $A$ and $40\,\text{kN}\downarrow$ at $C$.

Top chord $DF$ — moments about $E\,(8,0)$

$DF$ acts horizontally at height $3\,\text{m}$; its moment arm about $E$ is $3\,\text{m}$. Both $DE$ and $CE$ pass through $E$ and drop out.

$$\sum M_E = 0:\; R_A(8) - 40(8-4) + F_{DF}(3) = 0$$ $$40(8) - 40(4) + 3F_{DF} = 0 \Rightarrow 320 - 160 + 3F_{DF} = 0$$ $$F_{DF} = \frac{-160}{3} = -53.33\,\text{kN}$$

$F_{DF} = 53.33\,kN\ (Compression)$

Bottom chord $CE$ — moments about $D\,(4,3)$

$CE$ acts horizontally at $y=0$; moment arm about $D$ is $3\,\text{m}$. $DF$ and $DE$ pass through $D$ and drop out.

$$\sum M_D = 0:\; R_A(4) - F_{CE}(3) = 0$$ $$40(4) = 3F_{CE} \Rightarrow F_{CE} = \frac{160}{3} = +53.33\,\text{kN}$$

$F_{CE} = 53.33\,kN\ (Tension)$

Summary

Step 1: Vertical reactions

$\sum M_A = 0:\; V_B \times 20 = 60 \times 5 = 300 \Rightarrow V_B = 15\,\text{kN}$

$\sum F_y = 0:\; V_A = 60 - 15 = 45\,\text{kN}$

$$\mathbf{V_A = 45\,kN,\quad V_B = 15\,kN}$$

Step 2: Horizontal thrust (from crown hinge condition)

The bending moment at the crown hinge $C$ is zero. $C$ is at midspan $x=10\,\text{m}$, rise $y_C = h = 5\,\text{m}$.

Take moments of all forces on the right half about $C$ (right of $C$ there is no load; only $V_B$ and $H$):

$$M_C = V_B(10) - H(5) = 0$$ $$15 \times 10 = 5H \Rightarrow H = \frac{150}{5} = 30\,\text{kN}$$ $$\mathbf{H = 30\,kN}$$

Step 3: Bending moment under the load

Load is at $x = 5\,\text{m}$. Rise of arch axis there:

$$y = \frac{4h}{L^2}x(L-x) = \frac{4(5)}{20^2}\,(5)(20-5) = \frac{20}{400}\times 5 \times 15 = 0.05 \times 75 = 3.75\,\text{m}$$

Bending moment at the section under the load (taking the left part; the load acts at the section so it produces no moment about that point):

$$M = V_A \, x - H \, y = 45(5) - 30(3.75) = 225 - 112.5 = 112.5\,\text{kN·m}$$ $$\mathbf{M = 112.5\,kN·m}$$

Step 4: Normal thrust and radial shear under the load

Slope of the arch axis at $x=5\,\text{m}$:

$$\frac{dy}{dx} = \frac{4h}{L^2}(L-2x) = \frac{20}{400}(20-10) = 0.05\times10 = 0.5 \Rightarrow \theta = \tan^{-1}(0.5) = 26.57^{\circ}$$

so $\cos\theta = 0.8944,\ \sin\theta = 0.4472$.

Consider the part of the arch just to the left of the load (load not yet applied). The internal forces transmitted across the section come from $V_A = 45\,\text{kN}\uparrow$ and $H = 30\,\text{kN}$. Net vertical force on the left part = $V_A = 45\,\text{kN}$ (upward); net horizontal force = $H = 30\,\text{kN}$.

Normal thrust $N$ (along the tangent):

$$N = H\cos\theta + V\sin\theta = 30(0.8944) + 45(0.4472) = 26.83 + 20.12 = 46.96\,\text{kN}$$ $$\mathbf{N \approx 46.96\,kN}$$

Radial shear $Q$ (perpendicular to the tangent):

$$Q = V\cos\theta - H\sin\theta = 45(0.8944) - 30(0.4472) = 40.25 - 13.42 = 26.83\,\text{kN}$$ $$\mathbf{Q \approx 26.83\,kN}$$

Note: For a parabolic three-hinged arch under a uniformly distributed load the bending moment everywhere would be zero (the parabola is the funicular shape). Here the single concentrated load produces a non-zero moment of $112.5\,\text{kN·m}$.

Method

By the unit-load (virtual work) theorem, the vertical deflection at $C$ is:

$$\delta_C = \sum_i n_i N_i \frac{L_i}{A_i E_i}$$

where $N_i$ are real member forces, $n_i$ are forces from a unit load at $C$, and $L_i/A_iE_i$ is the member flexibility.

Member-by-member computation

Compute $n_i N_i \,(L_i/A_iE_i)$ for each member (flexibility in $10^{-6}\,\text{m/kN}$):

Sum

$$\delta_C = (1200 + 750 + 480 + 480 + 300 - 450)\times 10^{-6}\,\text{m} = 2760\times10^{-6}\,\text{m}$$ $$\delta_C = 2.760\times10^{-3}\,\text{m} = \mathbf{2.76\,mm\ (downward)}$$

Comment on sign convention

Tensile member forces are taken positive and compressive forces negative; the same sign convention applies to both the real forces $N_i$ and the virtual forces $n_i$. Each term $n_iN_i(L_i/A_iE_i)$ is positive when the real and virtual forces are of the same sense (both tension or both compression) and negative when of opposite sense. Member 6 contributes a negative term ($-450\times10^{-6}$ m) because it carries real compression ($N_6=-45$) while the unit load puts it in tension ($n_6=+0.5$); this member's elongation tendency under the virtual system partly opposes the joint movement, reducing $\delta_C$. The overall positive total confirms the net deflection at $C$ is in the same direction as the applied unit load, i.e. downward.

Step 1: Horizontal tension

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

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

Step 2: Maximum tension and inclination

The maximum tension occurs at the supports. The vertical reaction at each support carries half the total load:

$$V = \frac{wL}{2} = \frac{5 \times 40}{2} = 100\,\text{kN}$$

Maximum (resultant) tension:

$$T_{max} = \sqrt{H^2 + V^2} = \sqrt{250^2 + 100^2} = \sqrt{62500 + 10000} = \sqrt{72500} = 269.26\,\text{kN}$$ $$\mathbf{T_{max} \approx 269.3\,kN}$$

Inclination to horizontal at the support:

$$\tan\theta = \frac{V}{H} = \frac{100}{250} = 0.4 \Rightarrow \theta = \tan^{-1}(0.4) = 21.80^{\circ}$$ $$\mathbf{\theta \approx 21.8^{\circ}}$$

Check via slope: $\tan\theta = 4d/L = (4\times4)/40 = 0.4$ — agrees.

Step 3: Length of cable

For a parabolic cable supported at equal levels, the approximate length is:

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

With $d/L = 4/40 = 0.1$:

$$\frac{8}{3}\frac{d^2}{L^2} = \frac{8}{3}(0.01) = 0.026667$$ $$\frac{32}{5}\frac{d^4}{L^4} = \frac{32}{5}(0.0001) = 0.00064$$ $$S = 40\,(1 + 0.026667 - 0.00064) = 40\,(1.026027) = 41.041\,\text{m}$$ $$\mathbf{S \approx 41.04\,m}$$

Definitions

• Static determinacy: A structure is statically determinate if all support reactions and internal member forces can be found using only the equations of static equilibrium ($\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$). If additional (compatibility) equations are needed it is statically indeterminate; the excess of unknowns over available equilibrium equations is the degree of static indeterminacy (DSI).
• Kinematic indeterminacy: The number of independent joint displacement components (translations and rotations) that are unknown, i.e. the number of degrees of freedom of the structure.
• Stability: A structure is stable if it can maintain equilibrium and resist any system of applied loads without undergoing rigid-body motion or collapse. Both an adequate number and a proper arrangement of members and reactions are required.

Determinacy equations

(a) Plane truss: Compare $(m + r)$ with $2j$.

• $m + r = 2j$ → determinate
• $m + r > 2j$ → indeterminate, $DSI = (m+r) - 2j$
• $m + r < 2j$ → unstable (deficient)

(b) Plane frame (rigid jointed): With no internal releases, $DSI = (3m + r) - 3j$.

Classification

1. Plane truss ($m=13,\ r=3,\ j=8$)

$$m + r = 13 + 3 = 16,\qquad 2j = 16$$

Since $16 = 16$, the truss is statically determinate (and, with proper arrangement, stable).

2. Plane frame ($m=6,\ r=4,\ j=6$)

$$DSI = (3m + r) - 3j = (3\times6 + 4) - (3\times6) = (18+4) - 18 = 4$$

The frame is statically indeterminate to the 4th degree.

Bending moment diagram

Measuring $x$ from the free end $B$, $M(x) = -Px$. At the fixed end $A$ ($x=L$): $M_A = -PL = -15\times3 = -45\,\text{kN·m}$.

The $M/EI$ diagram is a triangle with peak ordinate $\dfrac{PL}{EI}$ at $A$ and zero at $B$.

Moment-area theorems (reference: fixed end $A$, where slope and deflection are zero)

Theorem 1: Change in slope between $A$ and $B$ = area of the $M/EI$ diagram between $A$ and $B$.

$$\theta_B = \text{Area} = \frac{1}{2}\times L \times \frac{PL}{EI} = \frac{PL^2}{2EI}$$ $$\theta_B = \frac{15 \times 3^2}{2 \times 9000} = \frac{135}{18000} = 0.0075\,\text{rad}$$ $$\mathbf{\theta_B = 7.5\times10^{-3}\,rad}$$

Theorem 2: Deflection of $B$ relative to the tangent at $A$ = first moment of the $M/EI$ area about $B$. The centroid of the triangle (peak at $A$) lies at $\tfrac{2}{3}L$ from $B$.

$$\delta_B = \text{Area}\times \bar{x} = \frac{PL^2}{2EI}\times \frac{2L}{3} = \frac{PL^3}{3EI}$$ $$\delta_B = \frac{15 \times 3^3}{3 \times 9000} = \frac{15\times27}{27000} = \frac{405}{27000} = 0.015\,\text{m}$$ $$\mathbf{\delta_B = 15\,mm\ (downward)}$$

Both results match the standard cantilever formulas $\theta_B = PL^2/2EI$ and $\delta_B = PL^3/3EI$.

Influence lines

Let $a = 4\,\text{m}$ (distance of $C$ from $A$) and $b = L - a = 6\,\text{m}$.

ILD for reaction $R_A$

Straight line: ordinate $= 1$ at $A$ ($x=0$), decreasing linearly to $0$ at $B$ ($x=10$). Ordinate at any $x$ from $A$ is $(L-x)/L$.

ILD for shear force at $C$ ($V_C$)

Two parallel sloping segments separated by a unit jump at $C$:

• From $A$ to $C$: ordinate falls from $0$ at $A$ to $-a/L = -0.4$ just left of $C$.
• At $C$ it jumps by $+1$ to $+b/L = +0.6$ just right of $C$.
• From $C$ to $B$: ordinate falls linearly from $+0.6$ to $0$ at $B$. Maximum positive ordinate $= b/L = 0.6$ (just right of $C$); maximum negative ordinate $= -a/L = -0.4$ (just left of $C$).

ILD for bending moment at $C$ ($M_C$)

A triangle with apex under $C$. Peak ordinate at $C$:

$$\text{peak} = \frac{a\,b}{L} = \frac{4 \times 6}{10} = 2.4\,\text{m}$$

The ordinate rises linearly from $0$ at $A$ to $2.4$ at $C$, then falls linearly to $0$ at $B$.

Maximum bending moment at C

For a single concentrated load, the maximum $M_C$ occurs when the load stands directly over $C$ (the peak of the ILD):

$$M_{C,max} = W \times (\text{peak ordinate}) = 80 \times 2.4 = 192\,\text{kN·m}$$ $$\mathbf{M_{C,max} = 192\,kN·m}$$

Equivalently $M_{C,max} = \dfrac{W a b}{L} = \dfrac{80\times4\times6}{10} = 192\,\text{kN·m}$.

Maximum positive shear force at C

The largest positive ordinate of the $V_C$ influence line is $+b/L = +0.6$, just to the right of $C$. The single load placed there gives:

$$V_{C,max(+)} = W \times 0.6 = 80 \times 0.6 = 48\,\text{kN}$$ $$\mathbf{V_{C,max(+)} = 48\,kN}$$

(The maximum negative shear would be $80\times(-0.4)=-32\,\text{kN}$ with the load just left of $C$.)

Geometry and loads

$A$ at $x=0$, $B$ at $x=6$, $C$ at $x=8$. UDL $10\,\text{kN/m}$ over $AB$ (resultant $60\,\text{kN}$ at $x=3$); point load $20\,\text{kN}$ at $C$ ($x=8$).

Reactions

$\sum M_A = 0:\; R_B(6) - 60(3) - 20(8) = 0$

$$6R_B = 180 + 160 = 340 \Rightarrow R_B = 56.67\,\text{kN}$$

$\sum F_y = 0:\; R_A + R_B = 60 + 20 = 80$

$$R_A = 80 - 56.67 = 23.33\,\text{kN}$$ $$\mathbf{R_A = 23.33\,kN,\quad R_B = 56.67\,kN}$$

Bending moment at B

Easiest from the right (overhang only): the $20\,\text{kN}$ load at $C$ is $2\,\text{m}$ from $B$.

$$M_B = -20 \times 2 = -40\,\text{kN·m}$$ $$\mathbf{M_B = -40\,kN·m\ (hogging)}$$

Shear force diagram (left to right)

• Just right of $A$: $V = +23.33\,\text{kN}$.
• Decreases linearly under the UDL across $AB$; just left of $B$: $V = 23.33 - 10\times6 = -36.67\,\text{kN}$.
• At $B$, $R_B = +56.67$ jumps the shear up: $-36.67 + 56.67 = +20\,\text{kN}$ just right of $B$.
• Constant $+20\,\text{kN}$ along the overhang $BC$ until $C$, where the $20\,\text{kN}$ load brings it to $0$.
• Zero shear within $AB$ occurs at $x = 23.33/10 = 2.333\,\text{m}$ from $A$ (location of maximum sagging moment).

Bending moment diagram

• $M_A = 0$.
• Parabolic over $AB$, peaking at $x = 2.333\,\text{m}$: $M_{max} = 23.33(2.333) - 10(2.333)^2/2 = 54.44 - 27.22 = 27.22\,\text{kN·m}$ (sagging).
• $M_B = -40\,\text{kN·m}$ (hogging) — the diagram crosses zero between the peak and $B$.
• Linear over the overhang from $M_B = -40$ at $B$ to $M_C = 0$ at $C$.

The BMD is sagging (positive) in the interior of $AB$ and hogging (negative) near and over the support $B$.

The conjugate beam method

The conjugate beam is an imaginary beam of the same length as the real beam, loaded with the real beam's $M/EI$ diagram as a distributed load.

Two theorems:

1. The slope at a point in the real beam equals the shear force at the corresponding point in the conjugate beam.
2. The deflection at a point in the real beam equals the bending moment at the corresponding point in the conjugate beam.

Support conversion rules (so the conjugate boundary conditions reproduce the real slope/deflection conditions):

• Real simple support → conjugate simple support.
• Real fixed end → conjugate free end.
• Real free end → conjugate fixed end.
• Real interior support → conjugate internal hinge, and vice versa.

For a simply supported real beam the conjugate beam is also simply supported.

Application

Real beam: SS span $L=8\,\text{m}$, central load $P=24\,\text{kN}$. Maximum bending moment at midspan: $M_{mid} = \dfrac{PL}{4} = \dfrac{24\times8}{4} = 48\,\text{kN·m}$.

The $M/EI$ diagram is a triangle, base $L=8$, peak $\dfrac{48}{EI}$ at midspan. Total area (the conjugate load):

$$W = \frac{1}{2}\times 8 \times \frac{48}{EI} = \frac{192}{EI}$$

By symmetry the conjugate end reactions are each $W/2 = \dfrac{96}{EI}$.

Slope at the left support (= conjugate shear at left support = conjugate reaction)

$$\theta_A = \frac{96}{EI} = \frac{96}{12000} = 0.008\,\text{rad}$$ $$\mathbf{\theta_A = 8.0\times10^{-3}\,rad}$$

Standard check: $\theta_A = PL^2/16EI = 24\times64/(16\times12000)=1536/192000=0.008$ rad.

Deflection at midspan (= conjugate bending moment at midspan)

Take moments about midspan of the left half of the conjugate beam. The left-half load (triangle area over $0$ to $4$ m) $= \tfrac12\times4\times\tfrac{48}{EI}=\tfrac{96}{EI}$, acting at $\tfrac{2}{3}\times4 = 2.667\,\text{m}$ from $A$, i.e. $4-2.667=1.333\,\text{m}$ from midspan.

$$\delta_{mid} = \underbrace{\frac{96}{EI}}_{reaction}\times 4 - \underbrace{\frac{96}{EI}}_{left\ load}\times 1.333 = \frac{384 - 128}{EI} = \frac{256}{EI}$$ $$\delta_{mid} = \frac{256}{12000} = 0.021333\,\text{m} = \mathbf{21.33\,mm\ (downward)}$$

Standard check: $\delta_{mid} = PL^3/48EI = 24\times512/(48\times12000)=12288/576000=0.021333$ m.

Muller-Breslau principle

The influence line for any reaction or internal force (force or moment) in a statically determinate structure has the same shape as the deflected form obtained by removing the restraint corresponding to that function and giving a unit displacement (or unit rotation) in the positive direction of the function. For determinate structures the resulting influence line is composed of straight-line segments, so the principle lets one sketch an ILD quickly without recomputing reactions for many load positions.

Influence line for $M_C$

$a = 3\,\text{m}$ (from $A$), $b = L-a = 9\,\text{m}$, $L = 12\,\text{m}$. Peak ordinate (under $C$):

$$y_C = \frac{ab}{L} = \frac{3 \times 9}{12} = 2.25\,\text{m}$$

ILD is a triangle: $0$ at $A$, rising to $2.25$ at $C$ ($x=3$), falling to $0$ at $B$ ($x=12$).

Left slope (A→C): $2.25/3 = 0.75$ per metre. Right slope (C→B): $2.25/9 = 0.25$ per metre.

Positioning the two loads for maximum $M_C$

Place each load in turn over the peak; the heavier load over the peak usually governs.

Case 1 — heavier $80\,\text{kN}$ over $C$, the $50\,\text{kN}$ leading load $2\,\text{m}$ to the right of $C$ (on the C→B side):

• Ordinate under $80\,\text{kN}$ (at $C$) $= 2.25$.
• Ordinate under $50\,\text{kN}$ (at $x = 3+2 = 5\,\text{m}$, falling limb): $y = 2.25 - 0.25\times2 = 1.75\,\text{m}$.
• $M_C = 80(2.25) + 50(1.75) = 180 + 87.5 = 267.5\,\text{kN·m}$.

Case 2 — $50\,\text{kN}$ over $C$, the $80\,\text{kN}$ trailing load $2\,\text{m}$ to the left of $C$ (on the A→C side):

• Ordinate under $50\,\text{kN}$ (at $C$) $= 2.25$.
• Ordinate under $80\,\text{kN}$ (at $x = 3-2 = 1\,\text{m}$, rising limb): $y = 0.75\times1 = 0.75\,\text{m}$.
• $M_C = 50(2.25) + 80(0.75) = 112.5 + 60 = 172.5\,\text{kN·m}$.

Governing maximum

Case 1 governs:

$$\mathbf{M_{C,max} = 267.5\,kN·m}$$

with the $80\,\text{kN}$ load placed directly over the quarter point $C$.