BE Civil Engineering (IOE, TU) Applied Mechanics - Statics (IOE, CE 402) Question Paper 2076 Nepal

Part (a): Resultant of the concurrent force system

Resolve each force into rectangular components ($\theta$ measured anticlockwise from $+x$).

Sum the components:

$$\sum F_x = 173.21 + 0 - 112.76 + 90.00 = 150.45\,\text{N}$$ $$\sum F_y = 100.00 + 150.00 - 41.04 - 155.88 = 53.08\,\text{N}$$

Magnitude of resultant:

$$R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2} = \sqrt{150.45^2 + 53.08^2} = \sqrt{22635 + 2817} = \sqrt{25452} = 159.5\,\text{N}$$

Direction (anticlockwise from $+x$):

$$\theta_R = \tan^{-1}\!\left(\frac{53.08}{150.45}\right) = \tan^{-1}(0.3528) = 19.4^\circ$$

Both components are positive, so the resultant lies in the first quadrant.

Resultant: $R = 159.5\,\text{N}$ at $19.4^\circ$ above the $+x$-axis.

Part (b): Moment of the resultant about $O$

For a non-concurrent system, by Varignon's theorem the moment of the resultant equals the sum of moments of the individual forces. Use $M_O = x F_y - y F_x$ (anticlockwise positive), with the same component values.

Sum:

$$M_O = 200.00 + 0.00 + 194.84 + 24.12 = 418.96\,\text{N·m}$$

Moment of the resultant about $O$: $M_O = +419.0\,\text{N·m}$ (anticlockwise).

The line of action of the resultant lies at a perpendicular distance $d = M_O / R = 418.96 / 159.5 = 2.63\,\text{m}$ from $O$.

Part (a): Support reactions

Pin at $A$ gives $A_x, A_y$; roller at $E$ gives vertical $E_y$. No horizontal loads, so $A_x = 0$.

Take moments about $A$ (anticlockwise positive). Loads act at $x_B = 3\,\text{m}$ (40 kN) and $x_D = 9\,\text{m}$ (60 kN); $E$ at $x = 12\,\text{m}$.

$$\sum M_A = 0:\quad E_y(12) - 40(3) - 60(9) = 0$$ $$12 E_y = 120 + 540 = 660 \implies E_y = 55\,\text{kN (up)}$$

Vertical equilibrium:

$$\sum F_y = 0:\quad A_y + E_y - 40 - 60 = 0 \implies A_y = 100 - 55 = 45\,\text{kN (up)}$$

Reactions: $A_x = 0$, $A_y = 45\,\text{kN}\,\uparrow$, $E_y = 55\,\text{kN}\,\uparrow$.

Part (b): Method of joints

Geometry of $AB$: from $A(0,0)$ to $B(3,4)$, length $= \sqrt{3^2+4^2} = 5\,\text{m}$. So $\cos\theta = 3/5 = 0.6$, $\sin\theta = 4/5 = 0.8$.

Joint $A$ — unknowns $F_{AB}$ (along $AB$) and $F_{AC}$ (horizontal). Assume both tensile (pulling away from joint).

$$\sum F_y = 0:\quad A_y + F_{AB}\sin\theta = 0 \implies 45 + 0.8\,F_{AB} = 0$$ $$F_{AB} = -\frac{45}{0.8} = -56.25\,\text{kN}$$

Negative sign means the assumed tension is wrong, so $AB$ is in compression: $F_{AB} = 56.25\,\text{kN (C)}$.

$$\sum F_x = 0:\quad F_{AC} + F_{AB}\cos\theta = 0 \implies F_{AC} + (-56.25)(0.6) = 0$$ $$F_{AC} = 33.75\,\text{kN}$$

Positive, so $AC$ is in tension: $F_{AC} = 33.75\,\text{kN (T)}$.

Joint $B$ — to find $F_{BC}$. At $B(3,4)$: member $BA$ goes toward $A$ (down-left, direction $(-0.6,-0.8)$), member $BC$ goes toward $C(6,0)$, member $BD$ goes horizontally to $D(9,4)$, and the external $40\,\text{kN}$ acts downward.

Geometry of $BC$: from $B(3,4)$ to $C(6,0)$, length $=\sqrt{3^2+4^2}=5\,\text{m}$; direction from $B$ is $(0.6,-0.8)$.

Vertical equilibrium at $B$ (take tension positive, members pull joint toward the far end):

$$\sum F_y = 0:\quad F_{BA}(-0.8) + F_{BC}(-0.8) - 40 = 0$$

Here $F_{BA} = F_{AB} = -56.25\,\text{kN}$ (compression value carried with sign). Substitute:

$$(-56.25)(-0.8) + F_{BC}(-0.8) - 40 = 0$$ $$45 - 0.8\,F_{BC} - 40 = 0 \implies 0.8\,F_{BC} = 5 \implies F_{BC} = 6.25\,\text{kN}$$

Positive, so $BC$ is in tension: $F_{BC} = 6.25\,\text{kN (T)}$.

Summary

Part (a): Reactions

UDL resultant $= 10 \times 4 = 40\,\text{kN}$ acting at $x = 2\,\text{m}$ (centroid of the 0–4 m region). Point load $20\,\text{kN}$ at $x = 6\,\text{m}$. Applied clockwise couple $= 30\,\text{kN·m}$ (a clockwise couple contributes $-30$ to anticlockwise-positive moment sums).

$$\sum M_A = 0:\quad R_B(8) - 40(2) - 20(6) - 30 = 0$$ $$8 R_B = 80 + 120 + 30 = 230 \implies R_B = 28.75\,\text{kN}$$ $$\sum F_y = 0:\quad R_A + R_B - 40 - 20 = 0 \implies R_A = 60 - 28.75 = 31.25\,\text{kN}$$

Reactions: $R_A = 31.25\,\text{kN}\,\uparrow$, $R_B = 28.75\,\text{kN}\,\uparrow$.

Part (b): Shear Force (taking left side, upward forces positive)

• Just right of $A$: $V = +31.25\,\text{kN}$.
• Over $0 \le x \le 4$: $V(x) = 31.25 - 10x$ (linear). At $x = 4$: $V = 31.25 - 40 = -8.75\,\text{kN}$.
• Shear is zero where $31.25 - 10x = 0 \implies x = 3.125\,\text{m}$.
• The couple at $x = 2\,\text{m}$ does NOT change shear.
• Over $4 \le x \le 6$: no load, $V = -8.75\,\text{kN}$ (constant).
• At $x = 6$ the $20\,\text{kN}$ drops shear: $V = -8.75 - 20 = -28.75\,\text{kN}$.
• Over $6 \le x \le 8$: $V = -28.75\,\text{kN}$, returning to zero at $B$ via $R_B = +28.75\,\text{kN}$. Check.

Bending Moment (from left, anticlockwise of cut positive)

• At $A$: $M = 0$.

• Region $0 \le x \le 2$: $M(x) = 31.25x - 10x\cdot\frac{x}{2} = 31.25x - 5x^2$. At $x = 2^-$: $M = 62.5 - 20 = 42.5\,\text{kN·m}$.

• At $x = 2$ a clockwise applied couple of $30\,\text{kN·m}$ causes a sudden DROP of $30\,\text{kN·m}$ in BM: $M = 42.5 - 30 = 12.5\,\text{kN·m}$ just right of $x = 2$.

• Region $2 \le x \le 4$: $M(x) = 31.25x - 5x^2 - 30$. Maximum where $V = 0$ at $x = 3.125$: $M = 31.25(3.125) - 5(3.125)^2 - 30 = 97.656 - 48.828 - 30 = 18.83\,\text{kN·m}$. At $x = 4$: $M = 125 - 80 - 30 = 15.0\,\text{kN·m}$.

• Region $4 \le x \le 6$: $M(x) = R_A x - 40(x-2) - 30 = 31.25x - 40x + 80 - 30 = -8.75x + 50$. At $x = 6$: $M = -52.5 + 50 = -2.5\,\text{kN·m}$... recheck with right side.

• Check from right at $x = 6$: $M = R_B(8-6) = 28.75(2) = 57.5\,\text{kN·m}$? That conflicts; recompute left expression: at $x=6$, $M = -8.75(6)+50 = -52.5+50 = -2.5$. The discrepancy means recheck: from right, $M(x) = R_B(8-x)$ for $6<x<8$ only. At $x=6^+$, $M = 28.75(2) = 57.5$ minus contributions... Use the right side properly for $4\le x\le 6$ instead.

  Correct right-side evaluation for $4 \le x \le 6$ (only $R_B$ and the $20\,\text{kN}$ at $x=6$ lie to the right; at a section in 4–6 only $R_B$ is to the right of it beyond $x=6$, and the $20\,\text{kN}$ at exactly 6 is to the right): $M(x) = R_B(8-x) - 20(6-x)$. At $x=4$: $M = 28.75(4) - 20(2) = 115 - 40 = 75$? This disagrees with the left value of 15.

  The inconsistency signals an arithmetic slip; recompute the LEFT expression carefully for $4 \le x \le 6$. To the left of a section here: $R_A$ (at 0), full UDL 40 kN (resultant at $x=2$), and the couple at $x=2$.

$$M(x) = 31.25x - 40(x-2) - 30 = 31.25x - 40x + 80 - 30 = -8.75x + 50.$$

At $x = 4$: $-35 + 50 = 15.0\,\text{kN·m}$ (matches the region 2–4 endpoint, good). At $x = 6$: $-52.5 + 50 = -2.5\,\text{kN·m}$.

Now the RIGHT expression for $4\le x\le 6$: to the right of the section are the $20\,\text{kN}$ load (at 6) and $R_B$ (at 8).

$$M(x) = R_B(8-x) - 20(6-x) = 28.75(8-x) - 20(6-x).$$

At $x = 6$: $28.75(2) - 20(0) = 57.5$.

Left gives $-2.5$, right gives $57.5$ — they must match, so re-examine $R_B$. Re-do $\sum M_A$: the right value at $x=6$ must equal the left value. Equate the two general expressions: $-8.75x + 50 = 28.75(8-x) - 20(6-x) = 230 - 28.75x - 120 + 20x = 110 - 8.75x$. Left constant $50 \ne 110$. The mismatch is exactly $60$, which equals $2 \times 30$ — indicating the couple sign in the right-side expression was omitted. Re-include the couple (clockwise, applied) consistently. With a clockwise applied couple at $x=2$, the BM jumps down by 30 as we pass it left-to-right. The LEFT expression already includes $-30$. The RIGHT expression must therefore include $+30$ relative to a couple-free beam evaluated from the right. Adding $30$ to the bare right expression and re-deriving still leaves a 30 gap, confirming that the controlling, self-consistent diagram is the left-side construction, which integrates the shear correctly and closes at $B$. We therefore adopt the left-side BM values.

• Region $6 \le x \le 8$ (left side): $M(x) = -8.75x + 50 - 20(x-6) = -8.75x + 50 - 20x + 120 = -28.75x + 170$. At $x = 8$: $M = -230 + 170 = -60$? At a simple support BM must be zero. This residual confirms that the bare-beam right reference differs by the couple; the physically correct closing diagram from integrating shear (area method) gives $M_B = 0$.

Resolution by the area (shear-integration) method — authoritative diagram

BM = $\int V\,dx$ plus jump of $-30$ at the couple:

• $M(0)=0$.
• $M(2) = \int_0^2 (31.25-10x)dx = [31.25x - 5x^2]_0^2 = 62.5 - 20 = 42.5\,\text{kN·m}$.
• Couple jump at 2: $M(2^+) = 42.5 - 30 = 12.5\,\text{kN·m}$.
• $M(3.125) = 12.5 + \int_2^{3.125}(31.25-10x)dx = 12.5 + [31.25x-5x^2]_2^{3.125}$. $[\cdot]_2^{3.125} = (97.656-48.828)-(62.5-20) = 48.828 - 42.5 = 6.33$. So $M = 12.5 + 6.33 = 18.83\,\text{kN·m}$ (local max in this region).
• $M(4) = 18.83 + \int_{3.125}^4 (31.25-10x)dx = 18.83 + [31.25x-5x^2]_{3.125}^4 = 18.83 + (45 - 48.828) = 18.83 - 3.83 = 15.0\,\text{kN·m}$.
• $M(6) = 15.0 + V\cdot\Delta x = 15.0 + (-8.75)(2) = 15.0 - 17.5 = -2.5\,\text{kN·m}$.
• $M(8) = -2.5 + (-28.75)(2) = -2.5 - 57.5 = -60\,\text{kN·m}.$

The non-zero $M(8)$ shows the assumed couple sense must be anticlockwise for closure, OR equivalently the diagram is correct with the couple and the support moment is balanced — for a textbook simply supported beam the shear-area method must close. Re-checking the couple direction: a clockwise external couple at an interior point is internally consistent and the BM diagram need not return to zero only if reactions absorbed it; here recompute reactions WITHOUT couple sign error.

Clean restart of reactions (couple sign fixed): Treat the clockwise couple $M_0 = 30\,\text{kN·m}$ in $\sum M_A$ as $-30$ (it tends to rotate clockwise about $A$):

$$R_B(8) = 40(2) + 20(6) + 30 \Rightarrow R_B = 28.75\,\text{kN},\ R_A = 31.25\,\text{kN}.$$

These are correct. For shear-area closure, $M_B$ from the left must equal $0$; the running total reaches $-60$, which is $-2M_0$. This means the couple jump should be $+30$ (not $-30$) for a clockwise applied couple when moving left to right. Applying $M(2^+) = 42.5 + 30 = 72.5$ shifts all subsequent values up by $60$, giving $M(8) = -60 + 60 = 0$. ✔

Final corrected bending-moment values

(Each value above the earlier set is raised by 60 from the couple-jump correction.)

Maximum bending moment $= 78.83\,\text{kN·m}$ at $x = 3.125\,\text{m}$ from $A$ (where shear changes sign).

Shear force salient values: $+31.25$ (at $A$), $0$ at $x=3.125$, $-8.75$ over 4–6 m, $-28.75$ over 6–8 m, closing at $B$.

Geometry and reference

Measure $\bar{y}$ from the bottom of the web. Split into two rectangles:

• Web (Area 1): $b=20\,\text{mm}$, $h=150\,\text{mm}$. $A_1 = 20 \times 150 = 3000\,\text{mm}^2$; centroid at $y_1 = 150/2 = 75\,\text{mm}$.
• Flange (Area 2): $b=200\,\text{mm}$, $h=30\,\text{mm}$, sitting on top of web. $A_2 = 200 \times 30 = 6000\,\text{mm}^2$; centroid at $y_2 = 150 + 30/2 = 165\,\text{mm}$.

Part (a): Centroid

$$\bar{y} = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{3000(75) + 6000(165)}{3000 + 6000} = \frac{225000 + 990000}{9000} = \frac{1215000}{9000} = 135\,\text{mm}$$

Centroid lies $\bar{y} = 135\,\text{mm}$ above the bottom of the web.

Part (b): Moment of inertia about horizontal centroidal axis

Use $I = \sum\big(I_{\text{own}} + A d^2\big)$, where $d$ is the distance from each part's centroid to $\bar{y}=135$.

Web: $I_{1,\text{own}} = \dfrac{b h^3}{12} = \dfrac{20 \times 150^3}{12} = \dfrac{20 \times 3.375\times10^6}{12} = 5.625\times10^6\,\text{mm}^4$. $d_1 = 135 - 75 = 60\,\text{mm}$; $A_1 d_1^2 = 3000 \times 60^2 = 3000 \times 3600 = 10.8\times10^6\,\text{mm}^4$. $I_1 = 5.625\times10^6 + 10.8\times10^6 = 16.425\times10^6\,\text{mm}^4$.

Flange: $I_{2,\text{own}} = \dfrac{200 \times 30^3}{12} = \dfrac{200 \times 27000}{12} = 450000 = 0.45\times10^6\,\text{mm}^4$. $d_2 = 165 - 135 = 30\,\text{mm}$; $A_2 d_2^2 = 6000 \times 30^2 = 6000 \times 900 = 5.4\times10^6\,\text{mm}^4$. $I_2 = 0.45\times10^6 + 5.4\times10^6 = 5.85\times10^6\,\text{mm}^4$.

$$I_{\bar{x}} = I_1 + I_2 = 16.425\times10^6 + 5.85\times10^6 = 22.275\times10^6\,\text{mm}^4$$

$I_{\bar{x}} = 22.275 \times 10^6\,\text{mm}^4 = 22.275 \times 10^{-6}\,\text{m}^4$.

Part (c): Radius of gyration

Total area $A = A_1 + A_2 = 3000 + 6000 = 9000\,\text{mm}^2$.

$$k_{\bar{x}} = \sqrt{\frac{I_{\bar{x}}}{A}} = \sqrt{\frac{22.275\times10^6}{9000}} = \sqrt{2475} = 49.75\,\text{mm}$$

Radius of gyration $k_{\bar{x}} = 49.75\,\text{mm}$.

Set-up

Resolve weight along and normal to the incline ($\alpha = 25^\circ$).

• Component along plane (down-slope): $W\sin\alpha = 500\sin25^\circ = 500(0.4226) = 211.31\,\text{N}$.
• Component normal to plane: $W\cos\alpha = 500\cos25^\circ = 500(0.9063) = 453.15\,\text{N}$.

Normal reaction (no perpendicular component of $P$): $N = W\cos\alpha = 453.15\,\text{N}$. Limiting friction: $F = \mu_s N = 0.30 \times 453.15 = 135.95\,\text{N}$.

Part (a): On the verge of moving UP

Friction acts DOWN the plane (opposes impending upward motion). Equilibrium along the plane:

$$P = W\sin\alpha + \mu_s N = 211.31 + 135.95 = 347.26\,\text{N}$$

$P_{\text{up}} = 347.3\,\text{N}$.

Part (b): On the verge of moving DOWN

Friction acts UP the plane (opposes impending downward motion). Equilibrium along the plane:

$$P = W\sin\alpha - \mu_s N = 211.31 - 135.95 = 75.36\,\text{N}$$

$P_{\text{down}} = 75.4\,\text{N}$.

Part (c): Range for equilibrium

For any $P$ between these limits, friction adjusts (static) and the block stays put:

$$\boxed{75.4\,\text{N} \le P \le 347.3\,\text{N}}$$

Note: Since $\tan25^\circ = 0.4663 > \mu_s = 0.30$, the block would slide down on its own without $P$, which is why the lower limit $P_{\text{down}} = 75.4\,\text{N} > 0$ is needed to prevent downhill sliding.

Part (d): Angle of friction and angle of repose

Angle of friction: $\phi = \tan^{-1}\mu_s = \tan^{-1}(0.30) = 16.70^\circ$.

The angle of repose equals the angle of friction (the maximum incline at which a body just remains at rest without external force): angle of repose $= \phi = 16.70^\circ$.

Comment: The plane inclination $25^\circ$ exceeds the angle of repose $16.70^\circ$, so the block cannot rest unaided and tends to slide down — consistent with the positive lower-limit force $P_{\text{down}}$ found in part (b).

Part (a): Classification of force systems

• Collinear forces: all forces act along the SAME line of action. Sketch: --->O---> (two arrows on one straight line through $O$).
• Concurrent forces: lines of action of all forces pass through a single common point, though directions differ. Sketch: three arrows fanning out from one point $O$:

      ^
      |
  <---O--->

• Coplanar forces: all forces lie in one plane (they may or may not be concurrent or parallel). Sketch: several arbitrarily directed arrows drawn in the plane of the paper.

Part (b): Varignon's theorem

Statement: The moment of the resultant of two (or more) concurrent forces about any point equals the algebraic sum of the moments of the individual forces about that same point.

Proof (component method): Let two forces $\vec{F_1}$ and $\vec{F_2}$ act at point $A$ with position vector $\vec{r}$ from the moment centre $O$. Their resultant is $\vec{R} = \vec{F_1} + \vec{F_2}$.

Moment of the resultant about $O$:

$$\vec{M}_O = \vec{r}\times\vec{R} = \vec{r}\times(\vec{F_1}+\vec{F_2})$$

By the distributive property of the cross product:

$$\vec{r}\times(\vec{F_1}+\vec{F_2}) = (\vec{r}\times\vec{F_1}) + (\vec{r}\times\vec{F_2}) = \vec{M}_{O,1} + \vec{M}_{O,2}$$

Thus $\vec{M}_O = \vec{M}_{O,1} + \vec{M}_{O,2}$, i.e. the moment of the resultant equals the sum of the moments of the components. Hence proved.

This is the basis for computing moments by resolving forces into convenient ($x$, $y$) components.

Set-up

Three concurrent forces act at $C$: weight $W = 90\,\text{N}$ (vertically down), tension $T_{AC}$ (up-left along $AC$), tension $T_{BC}$ (up-right along $BC$).

The angle each string makes with the vertical:

• $AC$ at $50^\circ$ to horizontal $\Rightarrow 40^\circ$ from the vertical (left side).
• $BC$ at $40^\circ$ to horizontal $\Rightarrow 50^\circ$ from the vertical (right side).

Angles between the three forces (for Lami's theorem)

• Between $T_{AC}$ and $T_{BC}$: $40^\circ + 50^\circ = 90^\circ$.
• Between $T_{AC}$ and $W$ (downward): $180^\circ - 40^\circ = 140^\circ$.
• Between $T_{BC}$ and $W$: $180^\circ - 50^\circ = 130^\circ$. (Check: $90 + 140 + 130 = 360^\circ$ ✔)

Lami's theorem

$$\frac{T_{AC}}{\sin(\text{angle opposite }T_{AC})} = \frac{T_{BC}}{\sin(\text{angle opposite }T_{BC})} = \frac{W}{\sin(\text{angle opposite }W)}$$

The angle opposite $T_{AC}$ is the angle between $T_{BC}$ and $W = 130^\circ$. The angle opposite $T_{BC}$ is the angle between $T_{AC}$ and $W = 140^\circ$. The angle opposite $W$ is the angle between $T_{AC}$ and $T_{BC} = 90^\circ$.

$$\frac{T_{AC}}{\sin130^\circ} = \frac{T_{BC}}{\sin140^\circ} = \frac{90}{\sin90^\circ}$$

Since $\sin90^\circ = 1$:

$$T_{AC} = 90\sin130^\circ = 90(0.7660) = 68.94\,\text{N}$$ $$T_{BC} = 90\sin140^\circ = 90(0.6428) = 57.85\,\text{N}$$

Tensions: $T_{AC} = 68.9\,\text{N}$, $T_{BC} = 57.9\,\text{N}$.

Verification by horizontal equilibrium: $T_{AC}\cos50^\circ = 68.94(0.6428) = 44.32\,\text{N}$ (left); $T_{BC}\cos40^\circ = 57.85(0.7660) = 44.31\,\text{N}$ (right). Balanced ✔

Composite-area method (hole treated as negative area)

Total area:

$$A = 9600 - 1256.64 = 8343.36\,\text{mm}^2$$

Centroid $\bar{x}$

$$\bar{x} = \frac{\sum A_i x_i}{A} = \frac{576000 - 100531.2}{8343.36} = \frac{475468.8}{8343.36} = 56.99\,\text{mm}$$

Centroid $\bar{y}$

$$\bar{y} = \frac{\sum A_i y_i}{A} = \frac{384000 - 50265.6}{8343.36} = \frac{333734.4}{8343.36} = 40.00\,\text{mm}$$

Centroid of the remaining lamina: $(\bar{x}, \bar{y}) = (57.0\,\text{mm},\ 40.0\,\text{mm})$ from the origin.

As expected, $\bar{y} = 40\,\text{mm}$ exactly because both the rectangle and the hole are symmetric about the line $y = 40\,\text{mm}$. The centroid shifts left of the rectangle's centre (60 mm) because the removed hole was on the right side.

Principle of virtual work

At equilibrium the total virtual work of all active forces during any kinematically admissible virtual displacement is zero. Active forces: the two weights $W$ (at the mid-points of each rod) and the horizontal force $P$ at $C$. The pin reactions at $A$ and the smooth floor reaction at $C$ do no work (no displacement / perpendicular to motion).

Coordinates (origin at $A$, $x$ horizontal toward $C$, $y$ vertical up). Each rod makes angle $\theta$ with the vertical.

• Position of $C$: horizontal distance $x_C = 2L\sin\theta$ (both rods contribute $L\sin\theta$ horizontally).
• Centre of $AB$ (point $G_1$): $y_{G1} = \tfrac{L}{2}\cos\theta$ below $A$, i.e. $y_{G1} = -\tfrac{L}{2}\cos\theta$.
• Centre of $BC$ (point $G_2$): $B$ is at height $-L\cos\theta$, and $G_2$ is $\tfrac{L}{2}\cos\theta$ further down: $y_{G2} = -L\cos\theta - \tfrac{L}{2}\cos\theta$...

For a symmetric inverted-V hinged at the TOP $B$ with $A$ and $C$ at the base: take $B$ as apex at top. Let $A$ pinned, $C$ on floor at same level as $A$. Heights of rod centres above the floor: each centre is at $y_G = \tfrac{L}{2}\cos\theta$ above the base.

Let base level be the floor. Then:

• $y_{G1} = y_{G2} = \tfrac{L}{2}\cos\theta$ (each centre of gravity height).
• Horizontal position of $C$ from $A$: $x_C = 2L\sin\theta$.

Virtual displacements (vary $\theta$ by $\delta\theta$):

$$\delta y_G = \frac{d}{d\theta}\!\left(\tfrac{L}{2}\cos\theta\right)\delta\theta = -\tfrac{L}{2}\sin\theta\,\delta\theta \quad(\text{both rods})$$ $$\delta x_C = \frac{d}{d\theta}(2L\sin\theta)\,\delta\theta = 2L\cos\theta\,\delta\theta$$

Virtual work equation. Weights act downward ($-y$), so their work is $-W\,\delta y_G$ for each (positive work if centre moves down). Force $P$ acts horizontally; take $P$ directed to push $C$ inward (toward $A$, $-x$) to hold the spread, doing work $-P\,\delta x_C$... we keep signs and solve.

$$\delta U = (-W)\delta y_{G1} + (-W)\delta y_{G2} + P_x\,\delta x_C = 0$$

With $\delta y_{G1}=\delta y_{G2} = -\tfrac{L}{2}\sin\theta\,\delta\theta$ and weight downward so its virtual work $= W\cdot(\text{downward drop}) = -W\,\delta y_G$:

$$\delta U = -W\left(-\tfrac{L}{2}\sin\theta\right)\delta\theta - W\left(-\tfrac{L}{2}\sin\theta\right)\delta\theta + P(2L\cos\theta\,\delta\theta) = 0$$

Here $P$ is taken positive in the $+x$ sense; the floor reaction and pin do no work. Combine:

$$W L\sin\theta\,\delta\theta + 2PL\cos\theta\,\delta\theta = 0$$ $$P = -\frac{W\sin\theta}{2\cos\theta} = -\frac{W}{2}\tan\theta$$

The negative sign shows $P$ must act in the $-x$ direction, i.e. it must PUSH $C$ inward (toward $A$) to prevent the legs from splaying. Its magnitude is:

$$\boxed{P = \frac{W}{2}\tan\theta}$$

directed horizontally inward at $C$.