Learning Goal:

To know how to use shear and bending$-$moment diagrams to determine the design requirements of a steel beam used to support a loading.

When designing a steel beam to support a given loading, engineers often use bending characteristics as the primary criterion; shear characteristics are then taken into account, and, if multiple available designs are still viable, weight and cost are then considered to complete the specification.

The bending design consideration requires the calculation of the beam's section modulus. This is a property based solely on the geometry of the beam, specifically the moment of inertia and the centroid:

S$=$I$/$c

$.$

Once a loading is specified, we can use the flexure formula with the maximum bending moment, Mmax

$,$ to specify a lower bound on the section modulus:

Srequired$=$Mmax$\backslash$sigma allow

$.$

Where the maximum bending stress, $\backslash$sigma allow

$,$ is a parameter generally given in the design specifications.

The shear specification,

$\backslash$tau allow$>=$VmaxQIt

is generally less specific than the bending$-$moment specification, so it is used as the secondary design consideration.

Figure$1$ of $3$

The figure shows a horizontal shaft A E$.$ The shaft is supported by a horizontal surface with journal bearings at the left end A and at point D$,$ located $30$ millimeters to the left from the right end E$.$ A wheel is attached, and the force of $20$ newtons acts vertically downward to the shaft at the E$.$ Another wheel is attached and the force of $80$ newtons acts vertically downward to the shaft at point B $50$ millimeters to the right from the A$.$ A third wheel is attached, and the force of $125$ newtons acts vertically downward to the shaft at point C $90$ millimeters to the right from the B and $60$ millimeters from the D$.$

Part A $-$ Design a Circular Shaft

Part complete

Determine the minimum allowable diameter of the circular shaft to the nearest millimeter if the design parameters require $\backslash$sigma allow$=4\mathrm{.}5$ MPa

and $\backslash$tau allow$=500$ kPa

$.$ The bearings at A

and D

exert only vertical reactions on the shaft.$($Figure $1)$

Express your answer to the nearest tenth of a millimeter.

View Available Hint$($s$)$for Part A

d$=$

$23\mathrm{.}9$

mm

Previous Answers

Correct

Part B $-$ Determine Necessary Beam Width

Part complete

Determine the minimum width, w

$,$ of the beam that will safely support the loading P$=6$ kip

$.$ The beam is $20$ ft

long from the pin support to the point where P

is applied, $6$ in

tall, has a rectangular cross$-$section, and made of a material that has a maximum allowable bending stress of $\backslash$sigma allow$=14$ ksi

and a maximum allowable shear stress of $\backslash$tau allow$=0\mathrm{.}9$ ksi

$.($Figure $2)$

Express your answer to three significant figures.

View Available Hint$($s$)$for Part B

w$=$

$8\mathrm{.}57$

in

Previous Answers

All attempts used; correct answer displayed

Part C $-$ Maximum Distributed Load

Determine the maximum uniform distributed load w

that can be applied to the W$12\backslash$times $14$

beam shown below if the maximum allowable bending stress is $\backslash$sigma allow$=26$ ksi

and the maximum allowable shear is $\backslash$tau allow$=8\mathrm{.}8$ ksi

$.$ The distance between the supports is $32$ ft

$.$

$($Figure $3)$

The geometric properties of the beam are listed in the table below.

Designation Area Depth Web Thickness Flange x$$x

axis y$$y

axis

A $($in$2)$

d $($in$)$

t $($in$)$

width thickness I $($in$4)$

S $($in$3)$

r $($in$)$

I $($in$4)$

S $($in$3)$

r $($in$)$

bf $($in$)$

tf $($in$)$

W$12\backslash$times $14$

$4\mathrm{.}1611\mathrm{.}910\mathrm{.}2003\mathrm{.}9700\mathrm{.}22588\mathrm{.}614\mathrm{.}94\mathrm{.}622\mathrm{.}361\mathrm{.}190\mathrm{.}753$

Express your answer to three significant figures.

View Available Hint$($s$)$for Part C

Activate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

w$=$

nothing

lb$/$ft