$***$How did they get the C$1$ values in the solution??$***$

A simply supported primary beam is required to span $10\mathrm{.}8m$ and to support the loads as shown in the figure below. Full lateral restraint to the beam may be assumed at the points of load application. Check the suitability of a UB $762267173$ in grade S$275$ steel. $($Adapted from $"$ Gardner L$,$ Nethercot DA $(2005).$ Designers' guide to EN $1993-1-1$ : eurocode $3$: design of steel structures : general rules and rules for buildings, Publisher. Thomas Telford, ISBN: ${9780727731630}^{\text{'}2})$

$($Ans: critical beam segment $CD$

$M_{B,RD}=1534>M_{Ed},$ beam is safe$)$

Answer:

$M_{c,Rd}=1704kNm$

By inspection either BC or CD is the critical segment of the beam for LTB $($CD is longer than $BC$ but has a more favourable BMD $($higher $c_1)$ so it is not clear in advance which is critical$)$

$BC$ :

$L_{cr}=3\mathrm{.}2m,C_1=1\mathrm{.}08,M_{cr,BC}=5851kNm$

$CD$:$L_{cr}=5\mathrm{.}1m,c_1=1\mathrm{.}77,M_{cr,CD}=4061\mathrm{.}3kNm$

$M_{Cr,CD}$ is the smallest hence $CD$ is the critical segment of the beam.

${}_{LT}=0\mathrm{.}65,{}_{LT}=0\mathrm{.}72,{}_{LT}=0\mathrm{.}86,f=0\mathrm{.}95,{}_{\text{L}\text{T}\text{m}\text{o}\text{d}}=0\mathrm{.}9,M_{B,Rd}=0\mathrm{.}9**1704\mathrm{.}45=1534\mathrm{.}06kNm>M_{Ed}$