BE Civil Engineering (IOE, TU) Applied Mechanics - Statics (IOE, CE 402) Question Paper 2079 Nepal

Approach. Resolve every force into rectangular components, sum them to obtain $R_x$ and $R_y$, then recombine.

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

Component table (N):

Sums:

$$R_x = 173.21 - 75.00 - 86.60 + 176.78 = +188.39\ \text{N}$$ $$R_y = 100.00 + 129.90 - 50.00 - 176.78 = +3.12\ \text{N}$$

(a) Resultant magnitude:

$$R = \sqrt{R_x^{2}+R_y^{2}} = \sqrt{188.39^{2}+3.12^{2}} = \sqrt{35490.8 + 9.7} = 188.42\ \text{N}$$

Direction (from $+x$-axis, anticlockwise):

$$\theta_R = \tan^{-1}\!\left(\frac{R_y}{R_x}\right) = \tan^{-1}\!\left(\frac{3.12}{188.39}\right) = 0.95^{\circ}$$

Both components positive $\Rightarrow$ first quadrant.

Resultant $R \approx 188.4\ \text{N}$ acting at $0.95^{\circ}$ above the positive $x$-axis.

(b) Equilibrant. The equilibrant is equal in magnitude and opposite in direction to the resultant.

Equilibrant $E \approx 188.4\ \text{N}$ at $0.95^{\circ}+180^{\circ}=180.95^{\circ}$ (i.e. $\approx 0.95^{\circ}$ below the negative $x$-axis).

Geometry. Span $AE = 8$ m. Member $AB$ runs from $(0,0)$ to $(2,3)$: length $L_{AB}=\sqrt{2^2+3^2}=\sqrt{13}=3.606$ m, so $\cos\alpha = 2/3.606 = 0.5547$, $\sin\alpha = 3/3.606 = 0.8321$.

(a) Support reactions. Take pin at $A$ ($A_x, A_y$) and roller at $E$ ($E_y$).

$\sum M_A = 0$ (anticlockwise +): loads act at $x_B=2$, $x_D=6$.

$$E_y(8) - 20(2) - 30(6) = 0 \Rightarrow 8E_y = 40 + 180 = 220$$ $$\boxed{E_y = 27.5\ \text{kN}\ (\uparrow)}$$

$\sum F_y = 0$: $A_y + E_y - 20 - 30 = 0 \Rightarrow A_y = 50 - 27.5 = 22.5$ kN.

$$\boxed{A_y = 22.5\ \text{kN}\ (\uparrow)}$$

$\sum F_x = 0$: no horizontal loads $\Rightarrow A_x = 0$.

(b) Method of joints.

Joint $A$ — members $AB$ (up-right at angle $\alpha$) and $AC$ (horizontal). Reaction $A_y=22.5\uparrow$, $A_x=0$. Assume both members in tension (pulling away from joint).

$\sum F_y = 0:\quad A_y + F_{AB}\sin\alpha = 0$

$$F_{AB} = -\frac{22.5}{0.8321} = -27.04\ \text{kN}$$

Negative $\Rightarrow$ $F_{AB} = 27.04$ kN (Compression).

$\sum F_x = 0:\quad F_{AC} + F_{AB}\cos\alpha = 0$

$$F_{AC} = -F_{AB}\cos\alpha = -(-27.04)(0.5547) = +15.00\ \text{kN}$$

$F_{AC} = 15.0$ kN (Tension).

Joint $B$ — members $AB$, $BC$, $BD$ and the $20$ kN load. Geometry: $BA$ points down-left from $B$ (angle $\alpha$ below horizontal); $BC$ runs from $(2,3)$ to $(4,0)$, i.e. down-right, $L_{BC}=\sqrt{2^2+3^2}=3.606$ m, $\cos\beta=0.5547$, $\sin\beta=0.8321$; $BD$ is horizontal (to the right).

Known $F_{AB}=27.04$ kN compression (pushes joint $B$ up-right along $BA$ direction reversed, i.e. force on $B$ from member is along $B\to A$ reversed = up-right). Resolve carefully using assumed tension sign convention for $BC$ and $BD$.

$\sum F_y = 0$ at $B$: contributions — member $AB$ (compression $27.04$) pushes joint along direction from $A$ to $B$ = up-right, vertical component $+27.04\sin\alpha = +22.5$; load $-20$; member $BC$ tension pulls toward $C$ (down-right), vertical $-F_{BC}\sin\beta$:

$$22.5 - 20 - F_{BC}(0.8321) = 0 \Rightarrow F_{BC} = \frac{2.5}{0.8321} = 3.00\ \text{kN}$$

$F_{BC} = 3.0$ kN (Tension).

Summary: $A_x=0,\ A_y=22.5\ \text{kN}\uparrow,\ E_y=27.5\ \text{kN}\uparrow$; $F_{AB}=27.04$ kN (C), $F_{AC}=15.0$ kN (T), $F_{BC}=3.0$ kN (T).

Loads. UDL $10$ kN/m over $4$ m $= 40$ kN, acting at its centroid $x = 2$ m from $A$. Point load $24$ kN at $x = 5$ m.

(a) Reactions. $\sum M_A = 0:\ R_B(6) - 40(2) - 24(5) = 0$

$$6R_B = 80 + 120 = 200 \Rightarrow R_B = 33.33\ \text{kN}$$

$\sum F_y = 0:\ R_A = 40 + 24 - 33.33 = 30.67\ \text{kN}$

$$\boxed{R_A = 30.67\ \text{kN},\quad R_B = 33.33\ \text{kN}}$$

(b) Shear force (from left, just-after values):

• At $A$: $V = +30.67$ kN.
• Over $0\le x\le 4$ (UDL region): $V(x) = 30.67 - 10x$. At $x=4$: $V = 30.67 - 40 = -9.33$ kN.
• $4\le x\le 5$: constant $V = -9.33$ kN.
• Just left of $5$ m: $-9.33$ kN; just right of $5$ m (after $24$ kN): $-9.33 - 24 = -33.33$ kN.
• $5\le x\le 6$: constant $-33.33$ kN; at $B$ the reaction $+33.33$ brings $V$ to $0$. ✓

Shear is zero within UDL where $30.67 - 10x = 0 \Rightarrow x = 3.067$ m — location of maximum sagging moment.

Bending moment (sagging +):

• $M_A = 0$.
• For $0\le x\le4$: $M(x) = 30.67x - 10\dfrac{x^2}{2} = 30.67x - 5x^2$.
  • $M_{max}$ at $x=3.067$: $M = 30.67(3.067) - 5(3.067)^2 = 94.07 - 47.03 = 47.04$ kN·m.
  • At $x=4$: $M = 30.67(4) - 5(16) = 122.68 - 80 = 42.68$ kN·m.
• At $x=5$ (under point load): take moments of right side: $M = R_B(1) = 33.33$ kN·m.
• $M_B = 0$. ✓

Maximum bending moment $\approx 47.0$ kN·m at $x = 3.07$ m from $A$.

SFD (kN):  +30.67 ---\
                      \____  -9.33 (at x=4..5)
                            |drop 24|
                                    ---- -33.33 (x=5..6)
BMD (kN·m): 0 -> rises to 47.0 @3.07 -> 42.7 @4 -> 33.3 @5 -> 0 @6

Parts (areas and centroid heights $\bar{y}$ from bottom of web):

• Web: $A_1 = 20\times100 = 2000\ \text{mm}^2$, $\bar{y}_1 = 100/2 = 50$ mm.
• Flange: $A_2 = 120\times20 = 2400\ \text{mm}^2$, $\bar{y}_2 = 100 + 20/2 = 110$ mm.

(a) Centroid.

$$\bar{y} = \frac{A_1\bar{y}_1 + A_2\bar{y}_2}{A_1+A_2} = \frac{2000(50)+2400(110)}{2000+2400} = \frac{100000+264000}{4400} = \frac{364000}{4400}$$ $$\boxed{\bar{y} = 82.73\ \text{mm from the bottom of the web}}$$

(b) Moment of inertia about centroidal axis using $I = \sum(\bar{I}_i + A_i d_i^2)$, where $d_i = |\bar{y}_i - \bar{y}|$.

Web: $\bar{I}_1 = \dfrac{b h^3}{12} = \dfrac{20\times100^3}{12} = \dfrac{20\times10^6}{12} = 1.6667\times10^6\ \text{mm}^4$. $d_1 = 82.73 - 50 = 32.73$ mm; $A_1 d_1^2 = 2000(32.73)^2 = 2000(1071.2) = 2.1424\times10^6$.

Flange: $\bar{I}_2 = \dfrac{120\times20^3}{12} = \dfrac{120\times8000}{12} = 80000 = 0.08\times10^6\ \text{mm}^4$. $d_2 = 110 - 82.73 = 27.27$ mm; $A_2 d_2^2 = 2400(27.27)^2 = 2400(743.7) = 1.7849\times10^6$.

Total:

$$I_x = (1.6667 + 2.1424 + 0.08 + 1.7849)\times10^6 = 5.674\times10^6\ \text{mm}^4$$ $$\boxed{I_{x,\text{centroidal}} \approx 5.67\times10^{6}\ \text{mm}^{4}}$$

Setup. Wall smooth $\Rightarrow$ wall reaction $N_W$ is horizontal only. Floor gives normal $N_F$ (up) and friction $F$ (toward wall, horizontal). Ladder weight $W=300$ N at mid-length ($3$ m). Angle $\theta=60^{\circ}$.

Let foot be at origin, top at height $6\sin60^{\circ}=5.196$ m, horizontal reach $6\cos60^{\circ}=3.0$ m.

(a) Self-weight only. $\sum F_y=0:\ N_F = W = 300$ N. $\sum F_x=0:\ F = N_W$. $\sum M_{\text{foot}}=0:\ N_W(6\sin60^{\circ}) = W(3\cos60^{\circ})$

$$N_W = \frac{300(3)(0.5)}{6(0.8660)} = \frac{450}{5.196} = 86.6\ \text{N}$$

Required friction $F = 86.6$ N. Maximum available $F_{max} = \mu N_F = 0.30(300) = 90$ N.

Since $86.6 < 90$, the ladder is in equilibrium under its own weight (with a small margin).

(b) With person at distance $s$ along the ladder. Person weight $P=700$ N; its horizontal lever arm about the foot is $s\cos60^{\circ}=0.5s$.

$\sum F_y=0:\ N_F = W + P = 300 + 700 = 1000$ N. At impending slip $F = \mu N_F = 0.30(1000) = 300$ N, and $N_W = F = 300$ N.

$\sum M_{\text{foot}}=0:\ N_W(6\sin60^{\circ}) = W(3\cos60^{\circ}) + P(s\cos60^{\circ})$

$$300(5.196) = 300(3)(0.5) + 700(0.5)s$$ $$1558.8 = 450 + 350s$$ $$350s = 1108.8 \Rightarrow s = 3.168\ \text{m}$$

The person can climb about $3.17$ m up the ladder before it is on the verge of slipping (roughly $53\%$ of its length).

Varignon's theorem. The moment of a force about any point equals the algebraic sum of the moments of its components about the same point. (Equivalently, the moment of a resultant equals the sum of moments of the individual forces.)

Application. Moment about $O$ (anticlockwise positive) of a force with components $(F_x,F_y)$ applied at $(x,y)$:

$$M_O = x\,F_y - y\,F_x$$ $$M_O = (4)(240) - (3)(100) = 960 - 300 = 660\ \text{N\cdot m}$$

$M_O = +660\ \text{N\cdot m}$ (anticlockwise).

Check via perpendicular distance: the line of action has slope $F_y/F_x = 2.4$; the moment $660 = F\cdot d \Rightarrow d = 660/260 = 2.54$ m, the perpendicular distance from $O$ to the line of action.

Parts (the hole is subtracted, so its area is negative):

• Rectangle: $A_1 = 80\times60 = 4800\ \text{mm}^2$, centroid $(\bar{x}_1,\bar{y}_1) = (40, 30)$ mm.
• Hole: $r=15$ mm, $A_2 = -\pi r^2 = -\pi(15)^2 = -706.86\ \text{mm}^2$, centroid $(50, 30)$ mm.

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

$\bar{x}$:

$$\bar{x} = \frac{A_1\bar{x}_1 + A_2\bar{x}_2}{A} = \frac{4800(40) + (-706.86)(50)}{4093.14} = \frac{192000 - 35343}{4093.14} = \frac{156657}{4093.14} = 38.27\ \text{mm}$$

$\bar{y}$: Both centroids have $y=30$ mm, so by symmetry about $y=30$:

$$\bar{y} = 30\ \text{mm}$$

Centroid at $(\bar{x}, \bar{y}) = (38.27,\ 30.0)\ \text{mm}$. The hole, lying to the right of the rectangle's centre, shifts the centroid slightly left of $x=40$ mm.

Principle of virtual work. For a system in equilibrium, the total virtual work done by all the active (applied) forces during any small, kinematically admissible virtual displacement is zero: $\delta U = \sum F\,\delta s = 0$.

Setup for reaction at $B$. Remove the support at $B$ and replace it by its unknown reaction $R_B$. Allow the released beam to undergo a small virtual rotation $\delta\phi$ about $A$. Then a point at distance $x$ from $A$ rises by $\delta = x\,\delta\phi$ (taking upward virtual displacement positive).

• Point $B$ ($x=8$ m) moves up by $8\,\delta\phi$; $R_B$ acts upward $\Rightarrow$ work $= +R_B(8\,\delta\phi)$.
• Load point ($x=3$ m) moves up by $3\,\delta\phi$; load $W=60$ kN acts downward $\Rightarrow$ work $= -60(3\,\delta\phi)$.

Virtual work equation:

$$\delta U = R_B(8\,\delta\phi) - 60(3\,\delta\phi) = 0$$

Divide by $\delta\phi$ (non-zero):

$$8R_B - 180 = 0 \Rightarrow R_B = \frac{180}{8} = 22.5\ \text{kN}$$

$R_B = 22.5$ kN (upward). (Check by statics: $\sum M_A = R_B(8) - 60(3) = 0$ gives the same result.)

Resolve $W$ along/normal to the incline ($\theta=25^{\circ}$): along-plane component $W\sin\theta$, normal component $W\cos\theta$.

$$W\sin25^{\circ} = 500(0.4226) = 211.3\ \text{N},\quad W\cos25^{\circ} = 500(0.9063) = 453.2\ \text{N}$$

Normal reaction $N = W\cos\theta = 453.2$ N (no perpendicular applied force). Maximum friction $F_{max} = \mu N = 0.35(453.2) = 158.6$ N.

(a) Slide on its own? The driving (down-plane) component is $211.3$ N; maximum resisting friction is only $158.6$ N. Since $211.3 > 158.6$, friction cannot hold it. The block will slide down on its own. (Equivalently, $\tan25^{\circ}=0.466 > \mu=0.35$, so the angle exceeds the friction angle $\phi=\tan^{-1}0.35=19.3^{\circ}$.)

(b) Force to move up. When motion impends up the plane, friction acts down the plane at $F_{max}=\mu N$. With $P$ parallel to the incline:

$$P = W\sin\theta + \mu N = W\sin\theta + \mu W\cos\theta$$ $$P = 211.3 + 158.6 = 369.9\ \text{N}$$

Minimum $P \approx 369.9$ N (up the incline).

Reactions. By symmetry of loading (two equal $40$ kN loads symmetric about mid-span):

$$R_{L_0} = R_{L_3} = \frac{40+40}{2} = 40\ \text{kN (each, upward)}.$$

Section. Cut a vertical plane between panels $L_1$ and $L_2$ (i.e. cutting top chord $U_1U_2$, diagonal $U_1L_2$, and bottom chord $L_1L_2$). Consider the left free body (contains $L_0$, $L_1$, $U_1$).

To isolate the bottom chord, take moments about $U_1(4,3)$ — the intersection of the other two cut members ($U_1U_2$ and the diagonal both pass through or are eliminated at $U_1$).

Forces on the left segment about $U_1$ (anticlockwise +):

• $R_{L_0}=40$ kN up at $x=0$: lever arm (horizontal distance to $U_1$) $=4$ m $\Rightarrow$ moment $=+40(4)=+160$ kN·m.
• Load $40$ kN down at $L_1(4,0)$: it lies directly below $U_1$, horizontal lever arm $=0 \Rightarrow$ moment $=0$.
• $F_{L_1L_2}$ acts horizontally along the bottom chord at $y=0$, lever arm $=3$ m (vertical distance to $U_1$). Assume tension (pointing away, to the right): a rightward force at the bottom produces a clockwise moment about $U_1$, i.e. $-F_{L_1L_2}(3)$.

$$\sum M_{U_1} = 0:\quad 40(4) - F_{L_1L_2}(3) = 0$$ $$F_{L_1L_2} = \frac{160}{3} = 53.33\ \text{kN}$$

Positive $\Rightarrow$ tension. $F_{L_1L_2} = 53.33$ kN (Tension).

Free body of joint $C$. Three concurrent forces: weight $200$ N down, $T_{CA}$ up-left along $CA$ at $30^{\circ}$ above horizontal, $T_{CB}$ up-right along $CB$ at $45^{\circ}$ above horizontal.

Equilibrium equations.

$\sum F_x = 0:\quad -T_{CA}\cos30^{\circ} + T_{CB}\cos45^{\circ} = 0$

$$T_{CA}(0.8660) = T_{CB}(0.7071) \Rightarrow T_{CA} = 0.8165\,T_{CB} \quad (\ast)$$

$\sum F_y = 0:\quad T_{CA}\sin30^{\circ} + T_{CB}\sin45^{\circ} - 200 = 0$

$$T_{CA}(0.5) + T_{CB}(0.7071) = 200$$

Substitute $(\ast)$:

$$0.8165\,T_{CB}(0.5) + 0.7071\,T_{CB} = 200$$ $$0.40825\,T_{CB} + 0.7071\,T_{CB} = 200$$ $$1.11535\,T_{CB} = 200 \Rightarrow T_{CB} = 179.32\ \text{N}$$

Then $T_{CA} = 0.8165(179.32) = 146.42\ \text{N}$.

$T_{CA} \approx 146.4$ N, $T_{CB} \approx 179.3$ N.

Check (Lami's theorem): angle between cables at $C$ $=180^{\circ}-30^{\circ}-45^{\circ}=105^{\circ}$; $\dfrac{200}{\sin105^{\circ}} = \dfrac{T_{CA}}{\sin(45^{\circ}+90^{\circ})}=\dfrac{T_{CA}}{\sin135^{\circ}}$ gives $T_{CA}=200\sin135^{\circ}/\sin105^{\circ}=146.4$ N. ✓