Practice exam

Question 1Not yet answeredMarked out of 3.00Shear from moment

The expression of the bending moment is

5x^2

. What is the expression of the shear force?

10x

10

-10x

5x

Question 2Not yet answeredMarked out of 3.00Supports

How does the displacement of a node supported by a pin behave?

Zero displacement in all directionsZero displacement along the vertical onlyZero forceZero displacement along the horizontal only

Question 3Not yet answeredMarked out of 3.00Zero-force members

Which are the zero-force members in a structure consisting only of two bars AB and AC connected at a node A and supported by a roller?

Both AB and ACAC onlyAB onlyNeither AB nor AC

Question 4Not yet answeredMarked out of 3.00Polar moment of inertia

The polar moment of inertia of a rectangle of side lengths

3\ \mathrm{cm} \times 6\ \mathrm{cm}

about its centroid measures, in

\mathrm{cm}^4

, ...

72.5

60.5

62.5

67.5

Question 5Not yet answeredMarked out of 3.00Maximum bending moment

For the beam ACB (figure in the source): a triangular distributed load of peak

20\ \mathrm{kN/m}

runs over

AC = 1.5\ \mathrm{m}

, and a cable inclined at

60^\circ

holds

C

B

over

0.5\ \mathrm{m}

. What is the abscissa of the location of the maximum bending moment?

0.694\ \mathrm{m}

0.864\ \mathrm{m}

0.754\ \mathrm{m}

1.224\ \mathrm{m}

Question 6Not yet answeredMarked out of 3.00Equilibrium on an incline

A linear spring of stiffness

5000\ \mathrm{N/m}

and initial length

15\ \mathrm{cm}

holds a particle of weight

100\ \mathrm{N}

on a smooth incline. If the spring may not deflect by more than

10\%

of its length, the maximum inclination angle of the plane would be ...

55.3^\circ

41.5^\circ

48.6^\circ

45.1^\circ

Question 7Not yet answeredMarked out of 6.00Method of joints

Calculate the forces in the members EF, EG, and ED of the truss shown in the source (loads

40\ \mathrm{kN}

20\ \mathrm{kN}

10\ \mathrm{kN}

) using the method of joints.

Question 8Not yet answeredMarked out of 8.00Method of sections

For the truss in the source, calculate the forces in members JK, JD, and DK using the method of sections. Provide all intermediate results.

Question 9Not yet answeredMarked out of 10.00Shear and loading from the moment diagram

The bending-moment diagram of a simply supported beam is parabolic between

x = 0

and

x = 1.5\ \mathrm{m}

(peak

15\ \mathrm{kN\,m}

x = 1

) and linear from

x = 1.5

x = 2\ \mathrm{m}

(down to

0

). Determine the shear-force expressions and the original loading on the beam.

You are allowed to use that.

Unit 1Basic Definitions and Vectors

A vector perpendicular to both, with magnitude $V_1 V_2 \sin\theta$ .

A scalar; zero when the vectors are perpendicular.

Force is mass times acceleration; in statics the net force is zero, so $\vec{a} = 0$ .

A downward force; $g \approx 9.81\ \mathrm{m/s^2}$ .

Unit 2Geometry and Trigonometry

The cosines satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ .

For resolving force triangles that are not right-angled.

Unit 3Forces, Moments and Centroids

Replace $dA$ with $dw$ , $dm$ , or $dV$ for weight, mass, or volume centroids.

Split a shape into known pieces; subtract holes with a negative area.

Centroid sits one quarter of the height up from the base.

$a$ semi-major, $b$ semi-minor axis.

Build the unit vector from $A$ to $B$ , then scale it by the force magnitude.

$\bar{y}$ measured from the flat circular face.

Stiffer (larger $EA$ ) or shorter members stretch less under the same force.

The expansion of $\vec{r} \times \vec{F}$ .

$d$ is the perpendicular lever arm. Summing moments about a smart point removes unknowns.

Seen in Practice exam 3483.

$\vec{r}$ runs from the point to any point on the force line.

Apply a unit virtual load to find a single displacement.

Centroid distances measured from the right-angle corner.

$\bar{y}$ measured from the flat edge.

$b$ members, $n$ joints. Satisfied means statically determinate.

A member is zero-force when a joint has two non-collinear members and no load.

Seen in Practice exam 3483.

$F_j$ are the real member forces, $f_j$ the forces from the unit virtual load.

Unit 4Internal Forces

A vertical strip has its own centroid at half its height.

Differentiate the moment to get shear, again to get the distributed load.

Each step raises the order of the curve: a constant load gives linear shear and a parabolic moment. See the shear and moment diagrams note.

Seen in Practice exam 3483.

The second moment of area; measures resistance to bending.

Shift an inertia from the centroidal axis to a parallel axis a distance $d$ away.

Always add when moving away from the centroid.

For a rectangle, $J_O = \tfrac{bh(h^2 + b^2)}{12}$ .

Seen in Practice exam 3483.

Zero when either axis is an axis of symmetry.

The centroidal value is smaller because of the parallel axis shift.

Centroidal value about an axis through the centroid, parallel to the base.

Stack disks along the axis of revolution and integrate.

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Practice exam

Practice exam

Formulary

Unit 1Basic Definitions and Vectors

Unit 2Geometry and Trigonometry

Unit 3Forces, Moments and Centroids

Unit 4Internal Forces

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