Phase 5 Design Calculations Class of Concrete 3 months going to use the C40/50 concrete class during ruction. The first number, 40 in this case, is the class of concrete and it can be denoted as fck which represents the compressive strength of concrete. It is a very strong grade of concrete, making it especially effective for big construction projects such as a student accommodation that is 5 Storeys high. This class of concrete can be commonly found in foundations and beams for structural support in construction sites. Beam Calculations The first part of my design is going to the beam which is a horizontal element, but on the 2D sketch of a floor plan, it is the vertical component. The horizontal component will therefore be the slab. 4.33 433 BED ROOM BED ROOM 3.5m A The image above shows that the beam I have chosen to evaluate has a span of 3.5m and it is in the middle of 2 bedrooms. This beam is also going to be on the ground floor, so when I need to consider the loading conditions, the imposed loading will be that of the floor only which is 3KN//m². As the beam is going to take half the load from both the rooms, the characteristic imposed load would be the imposed loading multiplied by the total length across. Characteristic imposed load=3KN//m² x (2.15+2.15) m=12.9KN/m beff Open Word bw b This type of section is known as the Reinforced Concrete T section, as it is shaped like T. T sections play pivotal roles in construction as the web can resist shear stresses effectively and the flanges can resist bending stresses. The beams are vertical, and the slabs are horizontal. All my calculations have assumed that the slab is one-way • The value b(width) is 4.3m as it shows the length halfway between the 2 rooms. Characteristic Imposed 4.3 12.9KN/m load=3KN/m² X •The value hf is the thickness of the slab which I have chosen to be 150mm=0.15m •The value bw is the thickness of the beam which I have chosen to be 250mm-0.25m I have assumed the total vertical distance (depth) is 400mm-0.4m The span/depth is 3.5/0.4-8.75>2.5 so it is a shallow beam • Dimensions of beam are 0.4mx0.25m=0.1m² Reinforced Concrete beams contain steel bars, and this is because of the nature of both materials. Concrete works well in compression as it has a high compressive strength and steel works well under tension as it has a high tensile strength. The purpose of the steel bars in the concrete beams are to provide additional strength to the beam so it can withstand larger loads. • I have assumed the diameter of the bar in the reinforced concrete beam to be 20mm • The cross-sectional area of the bar is therefore πD²/4=π x 202/4=314mm² The distance between the bottom of the steel bar and the bottom of the beam is called the nominal cover which I have assumed to be 40mm. • Weight of concrete=25KN/m³ ⚫fyk-500N/mm2=500x10°N/m² Beam self-weight Weight of concrete x Area of T section Open Word Design au 1 Design load=135gb+1 5gb – (1 35-17 6875)+ d • Dimensions of beam are 0.4mx0.25m=0.1m² Reinforced Concrete beams contain steel bars, and this is because of the nature of both materials. Concrete works well in compression as it has a high compressive strength and steel works well under tension as it has a high tensile strength. The purpose of the steel bars in the concrete beams are to provide additional strength to the beam so it can withstand larger loads. • I have assumed the diameter of the bar in the reinforced concrete beam to be 20mm • The cross-sectional area of the bar is therefore лD²/4=π x 20²/4=314mm² The distance between the bottom of the steel bar and the bottom of the beam is called the nominal cover. which I have assumed to be 40mm. • Weight of concrete=25KN/m³ • fyk=500N/mm²-500x10°N/m² Beam self-weight Weight of concrete x Area of T section Design load =25[(4.3x0.15) + (0.25x0.25)] =17.6875KN/m=17.69KN/m(2dp) Design load-1.35gk+1.5qk = (1.35x17.6875) + (1.5x12.9) 43.228125KN/m = 43.23KN/m(2dp) The Design load is the uniformly distributed load across the beam's length, 3.5m. Maximum Bending moment The equation for the maximum bending moment is wL2/8=43.23x 3.52/8=66.20KNm(2dp) Below is the Bending Moment diagram for the simply supported beam with a Uniformly Distributed Load. 66.20 kNm Reaction forces on a simply supported beam Open Word 1 maximum shear force is the design load x half the length=43.23, 1.75 75.65KN(24p). Design load =25[(4.3x0.15) + (0.25x0.25)] =17.6875KN/m=17.69KN/m(2dp) Design load 1.35gk+1.5qk = (1.35x17.6875) + (1.5x12.9) 43.228125KN/m = 43.23KN/m(2dp) The Design load is the uniformly distributed load across the beam's length, 3.5m. Maximum Bending moment The equation for the maximum bending moment is wL2/8=43.23x 3.52/8=66.20KNm(2dp) Below is the Bending Moment diagram for the simply supported beam with a Uniformly Distributed Load. 66.20 KNM Reaction forces on a simply supported beam There will be 2 reaction forces on the beam (1 at each end) as it is a simply supported beam, so the maximum shear force is the design load x half the length=43.23x1.75=75.65KN(2dp). Below is the Shear Force diagram for the simply supported beam with a Uniformly Distributed Load. 7565 к -756541 Effective Depth The effective depth is the depth from the top of the beam to the halfway point of the steel bar. In reinforced concrete beams, there are vertical links that provide shear reinforce to help maintain the beam's stability under loading. We can assume the vertical link to be 10mm. Effective depth= Total depth of beam - Nominal Open Word beffective (Effective flange width) b1=b2=(h-bw)/2= (4 3-0 25№/2=2.025m. tic link 400-40-10-10-340mm=0.34m beffective (Effective flange width) bl=b2 (b-bw)/2= (4.3-0.25)/2=2.025m lo=3.5m (span) beffective, 1= beffective,2=0.2b1 +0.110= (0.2x2.025) + (0.1x3.5)=0.755m beffective beffective, 1+ beffective, 2 + bw=0.755 + 0.755+0.25=1.76m Area of Concrete == Ac (Area of concrete) = (1.76x0.15) + (0.4x(0.4- 0.15))=0.364m² Calculating K Assume the neutral axis lies in the flange The value of b we use is the effective width and for d we use the effective depth fck 40MPa 40x10°N/m²= K-MED/bd2fck= (66.2x103)/1.76x0.342x40x106-8.13x103<0.207 As our value of K is less than 0.207, there is no compression reinforcement required Calculating Z Z= Z= =0.338 0.95d 0.95x0.34=0.323 0.338 0.95d Therefore z=0.95d=0.323m=323mm Neutral Axis Calculation x=(d-z)/0.4 (0.34-0.323)/0.4-0.0425m hf/0.8 0.15/0.8=0.1875m X<hf so the neutral axis is in the flange Area of steel As-MED/0.87 x fyk x z As= (66.2x106)/ (0.87 x 500 x 323) Open Word .. Therefore 2 steel bars are needed 211 'm 10-6 2 =/n UNIVERSITY of BRADFORD Faculty of Engineering and Informatics BEng/MEng Civil and Structural Engineering CSE5014-B Structural Design Project Coursework brief Presentation Structural Design Project CSE5014-B 2023-24 Brief Ten years after completing the construction of the accommodation block, owing to an increase in the student population, the university proposes to modify their existing accommodation. You are asked to modify the existing block to accommodate a further 25 students. The previous height restriction of 14.0 m has been removed. Main Requirements 1. Outline how your group could alter the existing structure to accommodate the extra students 2. If additional floors are added, calculate the increase in load on supporting columns and foundation are they able to withstand the additional loads? If other changes are made to the existing structure, how will these affect the assumed load path. 3. If needed, propose any methods that could be employed to strength the existing structure 4. Outline any impact that this construction will have on the surrounding campus Assessment This presentation is worth 30% of the module mark, and the assessment marks will be broken down as follows: Presentation element Marks (% of component grade) Description of proposed scheme 20 Calculation of additional load 25 Strengthening methods/adaptations 20 Assessment of impact on surroundings Appearance and layout of slides 15 10 Confidence of delivery 10 Submission Your group must prepare a 15 minute Powerpoint presentation covering the requirements above. You do not need to prepare a separate written report. You must submit a copy of your Powerpoint slides on Canvas by 5 pm Monday 22nd April. Each member of the group should contribute to the presentation. Presentations will take place during the timetabled slots on Tuesday 23rd and Thursday 25th April Page 1 of 1

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Phase 5 Design Calculations Class of Concrete 3 months going to use the C40/50 concrete class during ruction. The first number, 40 in this case, is the class of concrete and it can be denoted as fck which represents the compressive strength of concrete. It is a very strong grade of concrete, making it especially effec