Application Sheets:

Crane Tension Cable

Context

Design of a tower crane

Objective

To assess the adequacy of the cable stay for the jib

Concepts used in this application sheet

Force: applied load, moments, force resolution

Engineering model

Jib Analysis Model

Pinned Connection

Jib Free Body Diagram

Structural Analysis

Analysis Model

• The self-weight of the jib is modelled as a uniformly distributed load along the length of the jib.
• As a simplifying assumption for this calculation, it is assumed that there is no moment continuity in the jib at the mast i.e. the end of the jib is pinned to the mast.

• All forces and dimensions are given in kN and m respectively, unless otherwise stated

Calculate Tension Force in the Stay

Calculate the resultant load due to the self weight of the jib: $3.5\times 50=175kN$ at 25m.

Take moments about A

$\sum {M}=(100\times 45)+(175\times 25)-(F_{y}\times 30)=0$

$F_{y}=296kN$

Resolve forces to find the total cable force

$\theta =\tan ^{-1}({\frac {20}{30}})=34^{\circ }$

$F_{t}={\frac {F_{y}}{\sin \theta }}={\frac {296}{\sin 45^{\circ }}}=529kN$

Assessment:

Criterion

Factor of Safety $={\frac {\sigma _{u}}{\sigma _{w}}}$

• $\sigma _{u}$ is the ultimate tensile stress
• $\sigma _{w}$ is the working stress

Data input

Force in stay: $F_{t}=529kN=5.29\times 10^{5}N$

Diameter of the stay: $D=25mm$

Cable stay ultimate tensile stress: $\sigma _{u}=345N/mm^{2}$

Minimum required: $FoS=3.0$

Calculations

Area of stay: $A={\frac {\pi {}D^{2}}{4}}={\frac {\pi \times 25^{2}}{4}}=491mm^{2}$

Stress in stay: $\sigma _{w}={\frac {F_{t}}{A}}={\frac {5.29\times 10^{5}}{491}}=1077N/mm^{2}$

Apply the criterion

$FoS={\frac {\sigma _{u}}{\sigma _{w}}}={\frac {345}{1077}}=0.32$ $\ll 3.0$

This is much less than the required FoS of 3.0 so the design is not acceptable

Decision: cable size must be increased

Validation

The force in the cable may be significantly less than the estimated value because of the assumption of the pin connection for the jib. Overestimating the force is a safe assumption. Crane collapses are relatively common and can result in deaths. Therefore, a higher than normal factor of safety is used to ensure the crane will not be overloaded.

Glasgow Central Station Truss

Context

Roof truss in the main concourse of the Glasgow Central Station

Objective

Estimating the size of members in an existing truss based on bending

Concepts used in this application sheet

• Force: applied load, point load, internal force, resolution into components
• Equilibrium: equilibrium equation, force equilibrium
• Free Body Diagram

Engineering Model

Structural Analysis

Truss model for analysis

Analysis Model

For this calculation only axial forces in the members are taken into account i.e. moment continuity is neglected.

Calculations

Two of the heavily loaded members are the end diagonal AB and the top chord DE at the centre of the beam.

Calculate the force in the end diagonal AB:

Caclulate the support reaction at A and H:

Total load on the truss: $500kN$

Vertical support reaction: ${\frac {500}{2}}=250kN$

FBD at end support

Use the free body diagram at the end support:

$\theta =\tan ^{-1}({\frac {BC}{AC}})=\tan ^{-1}({\frac {5}{5}})=\tan ^{-1}(1)=45^{\circ }$

$\sum {F_{y}}=250+F_{AB}\times \sin(\theta )=0$

$F_{AB}={\frac {-250}{\sin(\theta )}}={\frac {-250}{\sin(45)}}=-356kN$

$F_{AB}$ is negative, i.e. the force in AB is compressive

Truss with cut for further analysis

Calculate the force in the chords at the centre of the truss, use a cut just to the left of the central post.

Take moment equilibrium about joint G:

$-F_{DE}\times 5-(250\times 30)+(25\times 25)+50\times (20+15+10+5)=0$

$F_{DE}=-875kN$ (i.e. compressive)

Assessment

Criterion

Factor of Safety $={\frac {\sigma _{u}}{\sigma _{w}}}$

• $\sigma _{u}$ is the ultimate compressive stress
• $\sigma _{w}$ is the working stress

Data Input

Minimum required FoS: $2.0$ (given)

Ultimate compressive stress $\sigma _{u}=240N/mm^{2}$ (this value is an older steel classification that is no longer used. However it is appropriate here because the truss was built in the late 1800s)

Area of section, $A_{s}=3870mm^{2}$

End diagonal AB assessment:

Axial force $F_{AB}=356kN=3.56\times 10^{5}N$ (from analysis)

$\sigma _{w}={\frac {F}{A}}={\frac {356000}{3870}}=92.0N/mm^{2}$

$FoS={\frac {240}{92.0}}=2.6$

Decision: The chosen steel section is suitable for the compressive forces it has to withstand and meets the minimum FOS criterion.

Top chord DE assessment

Axial force $F_{AB}=875kN=8.75\times 10^{5}N$ (from analysis)

$\sigma _{w}={\frac {F}{A}}={\frac {875000}{3870}}=226.1N/mm^{2}$

$FoS={\frac {240}{226.1}}=1.06$

Decision: The FOS is less than 2 and hence a bigger section should be chosen to ensure that the structure is safe.

The next steps can either be:

• Choosing an arbitrary section and performing the same calculations as above or
• In this case it is possible to rearrange the equations to solve for the minimum required section area

Adjustment calulations for top chord DE

Substitute the equations for working and ultimate stress into the factor of safety equation and setting the required FoS to 2:

$FoS={\frac {\sigma _{u}}{\sigma _{w}}}={\frac {240}{\frac {F}{A_{s}}}}={\frac {240}{\frac {875000}{A_{s}}}}=2$

Solve for $A_{S}$:

${\frac {875000}{A_{s}}}={\frac {240}{2}}$

$A_{s}={\frac {875000}{120}}=7292mm^{2}$

Therefore, any section with an area greater than $7292mm^{2}$ will be acceptable.