BE Civil Engineering (IOE, TU) Applied Mechanics - Statics (IOE, CE 402) Question Paper 2080 Nepal

Concept. For a concurrent force system the resultant is obtained by adding the rectangular components: $R_x=\sum F_x$, $R_y=\sum F_y$, $R=\sqrt{R_x^2+R_y^2}$, and $\theta=\tan^{-1}(R_y/R_x)$.

Step 1 — Resolve each force.

Step 2 — Sum the components.

$$R_x=103.923-27.362-129.904+70.711 = +17.368\ \text{kN}$$ $$R_y=60.000+75.175-75.000-70.711 = -10.536\ \text{kN}$$

Step 3 — Magnitude.

$$R=\sqrt{17.368^2+(-10.536)^2}=\sqrt{301.65+111.01}=\sqrt{412.66}=20.31\ \text{kN}$$

Step 4 — Direction. Since $R_x>0$ and $R_y<0$, the resultant lies in the fourth quadrant.

$$\theta=\tan^{-1}\!\left(\frac{-10.536}{17.368}\right)=-31.25^\circ$$

(a) Result: $\boxed{R = 20.31\ \text{kN at } 31.25^\circ \text{ below the +x axis (i.e. } 328.75^\circ \text{ CCW)}}$.

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

$$\boxed{E = 20.31\ \text{kN at } 148.75^\circ \text{ CCW from +x axis.}}$$

Geometry note. $AF$ runs from $(0,0)$ to $(4,3)$: length $=5\ \text{m}$. $FC$ runs from $(4,3)$ to $(8,0)$: length $=5\ \text{m}$. $GE$ runs from $(8,3)$ to $(12,0)$: length $5\ \text{m}$.

Step 1 — Support reactions. Total downward load $=40+60=100\ \text{kN}$. Take moments about $A$ (CCW +). Roller at $E$ gives vertical reaction $R_E$:

$$\sum M_A=0:\; R_E(12)-40(4)-60(8)=0$$ $$R_E=\frac{160+480}{12}=\frac{640}{12}=53.333\ \text{kN}\ (\uparrow)$$ $$\sum F_y=0:\; R_A=100-53.333=46.667\ \text{kN}\ (\uparrow)$$

Pin at $A$ has no horizontal load, so $H_A=0$.

Step 2 — Section cut. Cut a vertical section through members $FG$ (top), $FC$ (diagonal) and $BC$ (bottom), just to the right of joint $B$/$F$. Consider the left portion (joints $A$, $B$, $F$). Acting on it: $R_A=46.667\ \text{kN}$ up at $A$, the $40\ \text{kN}$ load down at $F$, and the three cut member forces.

Force in $FG$. Take moments about $C(8,0)$ to eliminate $FC$ and $BC$ (both pass through or point at $C$). $FG$ is horizontal at $y=3$; its moment arm about $C$ is $3\ \text{m}$.

$$\sum M_C=0:\; R_A(8)-40(4)+F_{FG}(3)=0$$

Taking $F_{FG}$ positive in tension (acting away from the joint, i.e. toward $+x$ on the left segment, giving a CCW moment about $C$ of $F_{FG}\times3$ if compressive... let us be explicit). Sum of moments about $C$ of external forces on the left part:

$$46.667(8)-40(4)=373.33-160=213.33\ \text{kN·m (CCW)}$$

This must be balanced by the top chord force. $FG$ acts horizontally along the top; for the left segment the only member able to balance this CCW moment is a compressive $FG$ pushing in $-x$ direction at height $3$, giving a clockwise moment $F_{FG}(3)$:

$$F_{FG}(3)=213.33\Rightarrow F_{FG}=71.11\ \text{kN (Compression)}$$

Force in $BC$. Take moments about $F(4,3)$ to eliminate $FG$ and $FC$. $BC$ is horizontal at $y=0$, arm $=3\ \text{m}$ about $F$.

$$\sum M_F=0:\; R_A(4)+F_{BC}(3)=0\ ?$$

External moment about $F$ from left-segment loads: $R_A$ at $A(0,0)$, arm horizontal distance $=4$, gives $46.667(4)=186.67\ \text{kN·m}$ CCW (the $40$ kN at $F$ passes through $F$, zero moment). A tensile $BC$ pulls the segment toward $+x$ along $y=0$, producing a clockwise moment $F_{BC}(3)$ about $F$ that balances:

$$F_{BC}(3)=186.67\Rightarrow F_{BC}=62.22\ \text{kN (Tension)}$$

Force in $FC$. Use vertical equilibrium of the left segment. $FC$ goes from $(4,3)$ to $(8,0)$, direction cosines: $\Delta x=4,\ \Delta y=-3,\ L=5$, so vertical component $=\tfrac{3}{5}F_{FC}$.

$$\sum F_y=0:\; R_A-40+ (\text{vertical comp. of }FC)=0$$ $$46.667-40+F_{FC,y}=0\Rightarrow F_{FC,y}=-6.667\ \text{kN}$$

The member must pull downward on the left segment by $6.667\ \text{kN}$, meaning $FC$ is in tension with its line pulling the joint $F$ toward $C$ (down-right):

$$F_{FC}=\frac{6.667}{3/5}=\frac{6.667\times5}{3}=11.11\ \text{kN (Tension)}$$

Answers.

$$\boxed{R_A=46.67\ \text{kN}\uparrow,\quad R_E=53.33\ \text{kN}\uparrow}$$ $$\boxed{F_{FG}=71.11\ \text{kN (C)},\quad F_{BC}=62.22\ \text{kN (T)},\quad F_{FC}=11.11\ \text{kN (T)}}$$

Step 1 — Reactions. UDL resultant $=15\times4=60\ \text{kN}$ acting at $2\ \text{m}$ from $A$. Take moments about $A$ (CCW +). The applied couple is anticlockwise $=+20\ \text{kN·m}$.

$$\sum M_A=0:\; R_B(8)-60(2)-30(6)+20=0$$ $$R_B=\frac{120+180-20}{8}=\frac{280}{8}=35.0\ \text{kN}\ (\uparrow)$$ $$\sum F_y=0:\; R_A=60+30-35=55.0\ \text{kN}\ (\uparrow)$$

Step 2 — Shear force (from left).

• Just right of $A$: $V=+55.0\ \text{kN}$.
• At $x=4\ \text{m}$ (end of UDL): $V=55-15(4)=55-60=-5.0\ \text{kN}$.
• The shear crosses zero within the UDL. $55-15x=0\Rightarrow x=3.667\ \text{m}$.
• From $4\ \text{m}$ to $6\ \text{m}$ (no load): $V=-5.0\ \text{kN}$ constant.
• Just right of $6\ \text{m}$: $V=-5-30=-35.0\ \text{kN}$.
• From $6\ \text{m}$ to $8\ \text{m}$: $V=-35.0\ \text{kN}$ constant up to $B$, where $R_B=35$ closes it to $0$. (The couple at $7\ \text{m}$ does not change shear.)

Step 3 — Bending moment.

• $M_A=0$.
• For $0\le x\le4$: $M=55x-15\frac{x^2}{2}$. Maximum where $V=0$ at $x=3.667\ \text{m}$:

$$M_{max}=55(3.667)-7.5(3.667)^2=201.67-100.83=100.83\ \text{kN·m}$$

• At $x=4\ \text{m}$: $M=55(4)-7.5(16)=220-120=100.0\ \text{kN·m}$.
• At $x=6\ \text{m}$: $M=R_A(6)-60(6-2)=55(6)-60(4)=330-240=90.0\ \text{kN·m}$.
• At $x=7^-\ \text{m}$ (just left of couple): $M=55(7)-60(5)-30(1)=385-300-30=55.0\ \text{kN·m}$.
• At $x=7^+\ \text{m}$ (just right of couple): the anticlockwise couple causes a jump. Using right-side check $M_{7^+}=R_B(1)=35(1)=35.0\ \text{kN·m}$. The drop of $20\ \text{kN·m}$ matches the applied couple.
• $M_B=0$.

Salient diagram (values in kN, kN·m):

SFD:  +55 ......... drops linearly under UDL ......... -5 (at x=4)
      -5 constant to x=6, then -35 to B
BMD:  0 -> rises parabolically -> 100.83 (peak at x=3.667 m)
      -> 100 (x=4) -> 90 (x=6) -> 55 (x=7-) | jump | 35 (x=7+) -> 0 (B)

Answers.

$$\boxed{R_A=55.0\ \text{kN}\uparrow,\quad R_B=35.0\ \text{kN}\uparrow}$$ $$\boxed{M_{max}=100.83\ \text{kN·m at } x=3.667\ \text{m from }A.}$$

Step 1 — Areas and centroid heights (from bottom of web).

• Flange: $A_1=200\times40=8000\ \text{mm}^2$, located with its centroid at $y_1=160+\tfrac{40}{2}=180\ \text{mm}$.
• Web: $A_2=40\times160=6400\ \text{mm}^2$, centroid at $y_2=\tfrac{160}{2}=80\ \text{mm}$.

Total area $A=8000+6400=14400\ \text{mm}^2$.

Step 2 — Centroid.

$$\bar{y}=\frac{A_1 y_1+A_2 y_2}{A}=\frac{8000(180)+6400(80)}{14400}=\frac{1{,}440{,}000+512{,}000}{14400}=\frac{1{,}952{,}000}{14400}=135.56\ \text{mm}$$

(a) Centroid is $\boxed{135.56\ \text{mm}}$ above the bottom of the web.

Step 3 — Moment of inertia about centroidal axis (parallel-axis theorem).

For each rectangle: $I=\frac{bh^3}{12}+A d^2$, where $d=|y_i-\bar{y}|$.

Flange: $\frac{bh^3}{12}=\frac{200\times40^3}{12}=\frac{200\times64000}{12}=1{,}066{,}667\ \text{mm}^4$. $d_1=180-135.56=44.44\ \text{mm}$, $A_1 d_1^2=8000\times44.44^2=8000\times1974.9=15{,}799{,}200\ \text{mm}^4$. Flange total $=1{,}066{,}667+15{,}799{,}200=16{,}865{,}867\ \text{mm}^4$.

Web: $\frac{bh^3}{12}=\frac{40\times160^3}{12}=\frac{40\times4{,}096{,}000}{12}=13{,}653{,}333\ \text{mm}^4$. $d_2=135.56-80=55.56\ \text{mm}$, $A_2 d_2^2=6400\times55.56^2=6400\times3086.9=19{,}756{,}160\ \text{mm}^4$. Web total $=13{,}653{,}333+19{,}756{,}160=33{,}409{,}493\ \text{mm}^4$.

Step 4 — Sum.

$$I_{\bar{x}}=16{,}865{,}867+33{,}409{,}493=50{,}275{,}360\ \text{mm}^4$$

(b) $\boxed{I_{\bar{x}}\approx 50.28\times10^{6}\ \text{mm}^4}$.

Setup. Let $N_f$ = normal reaction at floor, $F=\mu N_f$ = friction at floor (acts toward the wall), $N_w$ = horizontal reaction at the smooth wall. Ladder inclined at $\theta=60^\circ$. Weight $W=250\ \text{N}$ at the midpoint ($3\ \text{m}$ along).

(a) Ladder alone.

Vertical: $N_f=W=250\ \text{N}$. Friction available $=\mu N_f=0.30\times250=75\ \text{N}$, so horizontal equilibrium gives $N_w=F=75\ \text{N}$ (if on the verge).

Check moment about the foot $O$ to find the $N_w$ actually required for equilibrium:

$$\sum M_O=0:\; N_w(L\sin\theta)-W\left(\tfrac{L}{2}\cos\theta\right)=0$$ $$N_w=\frac{W\cos\theta}{2\sin\theta}=\frac{250\cos60^\circ}{2\sin60^\circ}=\frac{250(0.5)}{2(0.8660)}=\frac{125}{1.7321}=72.17\ \text{N}$$

Required friction $=72.17\ \text{N} < 75\ \text{N}$ available, so the ladder is safe under its own weight.

$$\boxed{N_f=250\ \text{N},\quad N_w=72.17\ \text{N},\ F=72.17\ \text{N (ladder does not slip)}}$$

(b) Person of weight $P=800\ \text{N}$ at distance $s$ along the ladder.

Vertical: $N_f=W+P=250+800=1050\ \text{N}$. Maximum friction (slipping impends): $F_{max}=\mu N_f=0.30\times1050=315\ \text{N}$. Hence at impending slip $N_w=315\ \text{N}$.

Moment about foot $O$ (CCW +), with horizontal distances: weight at $\tfrac{L}{2}\cos\theta$, person at $s\cos\theta$, wall reaction arm $L\sin\theta$:

$$N_w(L\sin\theta)=W\left(\tfrac{L}{2}\cos\theta\right)+P\,(s\cos\theta)$$ $$315(6\sin60^\circ)=250(3\cos60^\circ)+800\,s\cos60^\circ$$ $$315(6\times0.8660)=250(3\times0.5)+800\,s(0.5)$$ $$315(5.196)=375+400s$$ $$1636.7=375+400s\Rightarrow 400s=1261.7\Rightarrow s=3.154\ \text{m}$$

(b) The person can climb $\boxed{s\approx3.15\ \text{m}}$ along the ladder (about $52.6\%$ of its length) before slipping begins.

(a) Varignon's theorem. The moment of a force about any point equals the sum of the moments of its components about the same point.

Proof. Let force $\vec{F}$ act at position $\vec{r}$ from point $O$, and let $\vec{F}=\vec{F_1}+\vec{F_2}$ (its components). The moment about $O$ is $\vec{M}_O=\vec{r}\times\vec{F}$. By distributivity of the cross product:

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

Thus the moment of the resultant equals the sum of the moments of the components. $\blacksquare$

(b) Application. Resolve the $50\ \text{N}$ force at $(3,1)$:

$$F_x=50\cos40^\circ=38.302\ \text{N},\qquad F_y=50\sin40^\circ=32.139\ \text{N}$$

Moment about origin (CCW positive) $M_O=x F_y - y F_x$:

$$M_O=(3)(32.139)-(1)(38.302)=96.418-38.302=58.12\ \text{N·m}$$ $$\boxed{M_O=58.12\ \text{N·m (anticlockwise).}}$$

Principle of virtual work. If a system in equilibrium is given an arbitrary, infinitesimal, geometrically compatible (virtual) displacement, the total virtual work done by all the active forces is zero: $\delta W=0$.

Step 1 — Release the redundant for $R_B$. To find $R_B$, remove the roller and replace it by the unknown $R_B$. Give the beam a virtual rotation $\delta\theta$ about the pin $A$. Then:

• Vertical virtual displacement at $B$ (distance $10\ \text{m}$ from $A$): $\delta_B=10\,\delta\theta$ (upward, in the sense of $R_B$).
• Vertical virtual displacement under the load (at $4\ \text{m}$): $\delta_L=4\,\delta\theta$ (in the same rotational sense, downward where the load acts).

Step 2 — Virtual work equation. $R_B$ acts upward and moves up; the $120\ \text{kN}$ load acts downward and the point moves down by $4\delta\theta$ (negative work as a resisting load, positive as an applied downward force moving down). Setting total virtual work to zero with consistent signs (upward reaction does $+$ work for upward $\delta_B$, downward load does $+$ work for downward $\delta_L$ — but they oppose for equilibrium):

$$R_B(10\,\delta\theta)-120(4\,\delta\theta)=0$$ $$\delta\theta\,[10R_B-480]=0$$

Since $\delta\theta\neq0$:

$$R_B=\frac{480}{10}=48.0\ \text{kN}$$

Check by statics: $\sum M_A=0\Rightarrow R_B(10)=120(4)\Rightarrow R_B=48\ \text{kN}$. ✓

$$\boxed{R_B=48.0\ \text{kN}\ (\uparrow).}$$

Part 1 — Quarter circle centroid by integration. Consider a quarter circle of radius $R$ in the first quadrant centred at the origin. Using polar elements, by symmetry $\bar{x}=\bar{y}$. The standard result derived from $\bar{x}=\dfrac{1}{A}\int x\,dA$ with $A=\tfrac{\pi R^2}{4}$ gives:

$$\bar{x}=\bar{y}=\frac{4R}{3\pi}\approx0.4244R$$

(Derivation: $\int_A x\,dA=\int_0^{\pi/2}\!\int_0^R (r\cos\phi)\,r\,dr\,d\phi=\frac{R^3}{3}[\sin\phi]_0^{\pi/2}=\frac{R^3}{3}$; dividing by $A=\frac{\pi R^2}{4}$ gives $\frac{4R}{3\pi}$.)

Part 2 — Composite area. Take the full rectangle as positive area and the removed quarter circle as negative.

Rectangle: $A_1=120\times80=9600\ \text{mm}^2$, centroid $(\bar{x}_1,\bar{y}_1)=(60,40)$.

Removed quarter circle (radius $40$, centred at $(120,80)$, occupying the region inside the rectangle near that corner — i.e. extending toward $-x$ and $-y$ from the corner): $A_2=\tfrac{\pi(40)^2}{4}=\tfrac{\pi\times1600}{4}=1256.64\ \text{mm}^2$. Its centroid lies at distance $\frac{4R}{3\pi}=\frac{4(40)}{3\pi}=16.977\ \text{mm}$ from the corner along each inward axis:

$$\bar{x}_2=120-16.977=103.023\ \text{mm},\qquad \bar{y}_2=80-16.977=63.023\ \text{mm}$$

Net area: $A=9600-1256.64=8343.36\ \text{mm}^2$.

Centroid:

$$\bar{x}=\frac{A_1\bar{x}_1-A_2\bar{x}_2}{A}=\frac{9600(60)-1256.64(103.023)}{8343.36}=\frac{576000-129463}{8343.36}=\frac{446537}{8343.36}=53.52\ \text{mm}$$ $$\bar{y}=\frac{A_1\bar{y}_1-A_2\bar{y}_2}{A}=\frac{9600(40)-1256.64(63.023)}{8343.36}=\frac{384000-79199}{8343.36}=\frac{304801}{8343.36}=36.53\ \text{mm}$$ $$\boxed{\bar{x}=53.52\ \text{mm},\quad \bar{y}=36.53\ \text{mm}\ \text{(from the origin).}}$$

Setup. $W=500\ \text{N}$, $\alpha=25^\circ$, $\mu=0.25$. Normal reaction (force normal to plane): $N=W\cos\alpha=500\cos25^\circ=500(0.9063)=453.15\ \text{N}$. Maximum friction $=\mu N=0.25\times453.15=113.29\ \text{N}$. Weight component along plane (down-slope) $=W\sin\alpha=500\sin25^\circ=500(0.4226)=211.31\ \text{N}$.

(a) Motion impending up the plane. Friction acts down the slope (opposing upward motion):

$$P=W\sin\alpha+\mu N=211.31+113.29=324.60\ \text{N}$$ $$\boxed{P_{up}=324.60\ \text{N (up the incline).}}$$

(b) Block about to slide down — minimum holding force. Friction now acts up the slope (opposing downward motion), aiding $P$:

$$P=W\sin\alpha-\mu N=211.31-113.29=98.02\ \text{N}$$

Since this is positive, a force is indeed needed to prevent sliding (the slope angle $25^\circ$ exceeds the friction angle $\phi=\tan^{-1}0.25=14.04^\circ$).

$$\boxed{P_{hold}=98.02\ \text{N (up the incline).}}$$

Free-body of joint $O$. Three forces meet: the $900\ \text{N}$ weight (down), $T_A$ along $OA$ (up-left at $30^\circ$ above horizontal), $T_B$ along $OB$ (up-right at $45^\circ$ above horizontal).

Equilibrium equations.

Horizontal ($\sum F_x=0$): $T_A\cos30^\circ=T_B\cos45^\circ$

$$T_A(0.8660)=T_B(0.7071)\Rightarrow T_A=0.8165\,T_B \quad (i)$$

Vertical ($\sum F_y=0$): $T_A\sin30^\circ+T_B\sin45^\circ=900$

$$0.5\,T_A+0.7071\,T_B=900 \quad (ii)$$

Solve. Substitute (i) into (ii):

$$0.5(0.8165\,T_B)+0.7071\,T_B=900$$ $$0.40825\,T_B+0.7071\,T_B=900$$ $$1.11535\,T_B=900\Rightarrow T_B=806.92\ \text{N}$$ $$T_A=0.8165\times806.92=658.85\ \text{N}$$ $$\boxed{T_A=658.85\ \text{N},\quad T_B=806.92\ \text{N}.}$$

(a) Definitions.

• Moment of inertia (second moment) of an area about an axis: $I_x=\int y^2\,dA$, $I_y=\int x^2\,dA$. It measures the distribution of area about that axis.
• Polar moment of inertia about an axis perpendicular to the plane through a point: $J=\int r^2\,dA=\int (x^2+y^2)\,dA$.
• Radius of gyration about an axis: $k=\sqrt{I/A}$; the distance at which the whole area may be concentrated to give the same $I$.
• Perpendicular-axis theorem: for a plane area, $J=I_x+I_y$ (the polar moment about an axis normal to the plane equals the sum of the two in-plane moments of inertia about perpendicular axes through the same point).

(b) Solid circle, diameter $d=100\ \text{mm}$, radius $r=50\ \text{mm}$.

Area $A=\dfrac{\pi d^2}{4}=\dfrac{\pi(100)^2}{4}=7853.98\ \text{mm}^2$.

Moment of inertia about a diameter:

$$I_x=I_y=\frac{\pi d^4}{64}=\frac{\pi (100)^4}{64}=\frac{\pi\times10^8}{64}=4{,}908{,}739\ \text{mm}^4\approx4.909\times10^{6}\ \text{mm}^4$$

Polar moment (perpendicular-axis theorem):

$$J=I_x+I_y=2(4.909\times10^{6})=9.817\times10^{6}\ \text{mm}^4=\frac{\pi d^4}{32}$$

Radius of gyration about a diameter:

$$k_x=\sqrt{\frac{I_x}{A}}=\sqrt{\frac{4{,}908{,}739}{7853.98}}=\sqrt{625.0}=25.0\ \text{mm}$$

(Equivalently $k_x=d/4=100/4=25\ \text{mm}$.)

$$\boxed{I_x=I_y=4.909\times10^{6}\ \text{mm}^4,\quad J=9.817\times10^{6}\ \text{mm}^4,\quad k=25.0\ \text{mm}.}$$