Statics — Quiz 5
Questions in this quiz (6)
A shaft is subjected to loading $$\left(kN\right)$$ as shown in the figure. Determine the equivalent loading at point $$A$$ $$\left(kN,\ kN\cdot m\right)$$.
- $$R=260\ kN$$ downward $$M_A=4850\ kN\cdot m$$ counterclockwise
- $$R=230\ kN$$ upward $$M_A=1010\ kN\cdot m$$ counterclockwise
- $$R=260\ kN$$ upward $$M_A=1010\ kN\cdot m$$ clockwise
- $$R=260\ kN$$ upward $$M_A=1550\ kN\cdot m$$ counterclockwise
A signboard is $$3\ m$$ wide and $$1.5\ m$$ tall, mounted on top of a $$4\ m$$ pole. The total wind force on the signboard is $$240\ N$$. Determine the resulting moment at the base $$\left(N\cdot m\right)$$.
- $$840$$
- $$1140$$
- $$1320$$
- $$1560$$
A $$12$$-m long cantilever beam is subjected to a uniformly distributed load of $$3\ kN/m$$ over its entire length. The resulting moment $$\left(kN\cdot m\right)$$ at the fixed support is:
- $$216$$
- $$144$$
- $$108$$
- $$72$$
A traffic sign is hung from two points off a fixed pole as shown in Fig. $$6.23$$. If each point load is $$1\ kip$$, the resulting moment $$\left(kip\cdot ft\right)$$ at the base is most nearly:
- $$5$$
- $$8$$
- $$15$$
- $$35$$
A steel girder is fabricated using three welded plates as shown in the figure. Determine the moment of inertia $$\left(m^4\right)$$ about the major axis. The section is symmetric about the major axis, so the neutral axis is at mid-depth. Web height: $$h_w=0.9-2\left(0.04\right)$$ $$h_w=0.82\ m$$ Use the sum of the web inertia and the two flange inertias. For the web: $$I_{web}=\frac{bh^3}{12}$$ $$I_{web}=\frac{\left(0.03\right)\left(0.82\right)^3}{12}$$ $$I_{web}=\frac{\left(0.03\right)\left(0.551368\right)}{12}$$ $$I_{web}=0.00137842\ m^4$$ For one flange: $$A_f=\left(0.25\right)\left(0.04\right)$$ $$A_f=0.01\ m^2$$ Centroid distance from the neutral axis: $$d=\frac{0.9}{2}-\frac{0.04}{2}$$ $$d=0.45-0.02$$ $$d=0.43\ m$$ Centroidal inertia of one flange: $$I_{f,c}=\frac{bh^3}{12}$$ $$I_{f,c}=\frac{\left(0.25\right)\left(0.04\right)^3}{12}$$ $$I_{f,c}=\frac{\left(0.25\right)\left(0.000064\right)}{12}$$ $$I_{f,c}=0.0000013333\ m^4$$ Use the parallel axis theorem for one flange: $$I_f=I_{f,c}+A_fd^2$$ $$I_f=0.0000013333+\left(0.01\right)\left(0.43\right)^2$$ $$I_f=0.0000013333+0.001849$$ $$I_f=0.0018503333\ m^4$$ For two flanges: $$2I_f=2\left(0.0018503333\right)$$ $$2I_f=0.0037006666\ m^4$$ Total moment of inertia: $$I=I_{web}+2I_f$$ $$I=0.00137842+0.0037006666$$ $$I=0.0050790866\ m^4$$ $$I\approx0.00508\ m^4$$ Final answer: $$0.0051$$
- $$0.0062$$
- $$0.0041$$
- $$0.0028$$
- $$0.0051$$
Consider the ladder shown in the figure. A $$14$$-ft ladder leans against a smooth vertical wall making an angle of $$60^\circ$$ with the floor. A $$180$$-lb person stands at the midpoint of the ladder. Assume zero friction between the ladder and the wall. The minimum coefficient of friction at the floor required to prevent slipping is most nearly:
- $$0.20$$
- $$0.29$$
- $$0.35$$
- $$0.45$$
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