Timber Design

Surname 1
Name
Professor
Course
Date
Q.1
Timber Design
Structural integrity in buildings is always vital aspect in ensuring a specific structure perform its
intended purpose for intended design life. Structures components must be able to withstand all
loads i.e. live and dead loads so as to avoid associated failures resulting from deformation,
fracture or fatigue. An element in a structure fails when its material is stressed past its
allowable limit. We had a visit to a nearby five multistorey building and observed the following
structural features;
Beams the multistorey building had several concrete beams which could be seen and
distinguished.
Beams are structural elements which are used to carry lateral loads. They achieve this by
resisting load imposed to them via beam axis. Their loading results from reactions from their
supports. They usually carry shear forces and imposed moments as a result of loading. Different
types of beams were distinguished not less than, simply supported beams these were very
prominent and involved beam supported on both sides, continuous beams i.e. buildings with
multiple supports or spans e.g. beam on grid A1-A7. Cantilever beams were also well
pronounced, especially at the end on the balconies.
Concrete columns were also well pronounced which are vertical members used to support
axial loads (compressive loads) from other members primarily slabs and beams. They were of
different sizes like 400 x200 mm and many other sizes dependent on the weight it carries.
Different column sections were also noted, including rectangular, circular and square.
Reinforced concrete slabs were also noted. Slabs refer to elements which carry loads from
occupants, furniture.
Walls
Surname 2
These enclose the building and can be loaded-bearing or non-load bearing depending on the
design.
Q2.
Length=2.4m
(i) 1.35*6=8.1KN/m from analysis Ndc=6.21 KN/M^2
(ii) (1.2*6.0) + (1.5*3) = 11.7KN/M Ndc=8.36 KN/m^2
(iii) 1.2*6+15=22.2 Ndc=19.4KN/M^2
Q3.
Solution
Live load= 3KN/M
Surname 3
Dead load =1.5KN/M Clear span=3.5 M
Design load =( 1.5 *4) +( 3*1.6)
5 .3 1.7 0.67 8800
Moment = WL /8 =(10.8* 2.6)/8 = 3.5N/
SINCE 3.51≤6,02 hence adequate.
(b)
Shear strength
Reaction =10.8*2.6/2 = 14.04KN/
TA =3FV÷2A = 1.5×14.04×10^3/23.6 x 10^3=0.892
Perms. =0.67N/mm^2 hence adequate.
© Bearing strength
dc= 1.7N/mm^2 F=14.04KN
Dca=F/blb= 14.04 *10^3 /23.6 = 0.8 <1.7 hence okay
(d) Deflection
Permi deflect= 0.0003 *2600=7.8 mm

Place new order. It's free, fast and safe

-+
550 words

Our customers say

Customer Avatar
Jeff Curtis
USA, Student

"I'm fully satisfied with the essay I've just received. When I read it, I felt like it was exactly what I wanted to say, but couldn’t find the necessary words. Thank you!"

Customer Avatar
Ian McGregor
UK, Student

"I don’t know what I would do without your assistance! With your help, I met my deadline just in time and the work was very professional. I will be back in several days with another assignment!"

Customer Avatar
Shannon Williams
Canada, Student

"It was the perfect experience! I enjoyed working with my writer, he delivered my work on time and followed all the guidelines about the referencing and contents."

  • 5-paragraph Essay
  • Admission Essay
  • Annotated Bibliography
  • Argumentative Essay
  • Article Review
  • Assignment
  • Biography
  • Book/Movie Review
  • Business Plan
  • Case Study
  • Cause and Effect Essay
  • Classification Essay
  • Comparison Essay
  • Coursework
  • Creative Writing
  • Critical Thinking/Review
  • Deductive Essay
  • Definition Essay
  • Essay (Any Type)
  • Exploratory Essay
  • Expository Essay
  • Informal Essay
  • Literature Essay
  • Multiple Choice Question
  • Narrative Essay
  • Personal Essay
  • Persuasive Essay
  • Powerpoint Presentation
  • Reflective Writing
  • Research Essay
  • Response Essay
  • Scholarship Essay
  • Term Paper
We use cookies to provide you with the best possible experience. By using this website you are accepting the use of cookies mentioned in our Privacy Policy.