En1915 Design Structures Assessment Answer

Question:

1.Design problems rarely have just one acceptable solution. There are usually many unacceptable solutions, and often quite a few acceptable solutions too, and the design exercise is to find a good (if not the “best”) acceptable solution. An acceptable solution would be one where the structure is safe and fit for purpose, but we would also look for solutions which have a “pleasing” form and appearance, economical use of the materials, and which have ease of construction and cost.

2. REPORT

Your report should illustrate and define the design solution which, in your opinion, is best suited to the locality, design brief, specification and identifiable constraints. The (electronic) submission includes:

● Design Report (in pdf format). An executive summary (150 words max) which defines what, where, why and how. A set of calculations which demonstrates that your proposal is feasible and safe.

● The design report should be accompanied with some drawings comprising an elevation, cross-section and plan of the proposed structure (dimensioned and with annotations as necessary) and some basic visuals.

3. CALCULATIONS

Calculations can be hand-written or typed but have to be contained within the Design Report and be submitted as a pdf document only. Ensure each page stipulates the group and project being designed. A4 paper format is to be used throughout.

4. SPECIFICATION

4.1 Geometry

Consider the span L of the element and the loaded-width (how much load is being carried by that element.

4.2 Loading

Characteristic permanent load (self weight - assumed) wkG = ??kN/m2 Ultimate load partial safety factor for permanent loads γG = 1.35 Design permanent load wdG = γG wkG

Characteristic variable imposed load (people, wind) wkQ = ?? kN/m2 Ultimate load partial safety factor for imposed loads γQ = 1.6Design variable imposed load wdQ = γQ wkQ

4.3 Material properties (for example steel)

Characteristic tensile and compressive strength of steel fk = σy (yield strength) = 275 N/mm2

Characteristic shear strength of steel τk = 0.6 fk = 165 N/mm2

Partial material safety factor γm = 1.05

Design tensile and compressive strength fd = fk / γm

Design shear strength τd = τk / γm

4.4 Flexural and lateral buckling

The design strengths specified in Section 4.3 do not take account of either the flexural buckling of compression members or the lateral buckling of beams, which depend upon representative slenderness ratios λ = Le / r where Le is the effective (unrestrained) length of the member and r is the minimum radius of gyration of the cross section. Pairs of beams, for example, are often cross braced at regular intervals to reduce their lateral buckling length. Codes of Practice provide tables of allowable (reduced) design stresses corresponding to λ values. Herein an approximate and much simplified allowance is made for flexural and lateral buckling by reducing the corresponding member resistance by a factor α = 0.6.

ANALYSIS

5.1 Background

It is essential that all equations are expressed in consistent units, so that they are dimensionally correct. Convenient units for determining design moments and shearing forces are kilo Newtons (kN) and metres (m). Convenient units for determining design resistances are Newtons (N) and millimetres (mm). 5.2 Design moment and shear force If n beams (or trusses) are used to support the applied loading the design load per unit length of span, per beam (or truss), wd is determined from the equation