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I Background and significance of the topic The title of this graduation project is under the guidance of Mr. Wu Jing and Mr. Qian Dejun. The construction design of the engineering structure largely...

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I      Background and significance of the topic
The title of this graduation project is under the guidance of Mr. Wu Jing and Mr. Qian Dejun. The construction design of the engineering structure largely integrates the knowledge we have learned in the past four years in our university. It is of great practical significance for us to integrate the knowledge we have learned and apply it to engineering practice.
This graduation project is part of the office building of Zhida R & D center, including some structures between 4 ~ 8 axes and a ~ C axes, which is a reinforced concrete frame structure; Plane 16.7 × 33.6m. The house has five floors and six floors in some parts. The bottom floor is 4.2m high and the above floors are 3.6m high. The live load of storage building is 4kn / m ², The project is located in Nanjing.
Graduation project is one of the most important links in the stage of undergraduate education. It is also an important means to test the quality of undergraduate education. It aims to cultivate students' comprehensive ability to use their knowledge to solve practical problems. Undergraduate education courses can be divided into three categories: basic courses, professional basic courses and professional courses. They have the dual characteristics of wide coverage and strong professionalism, which is particularly prominent in civil engineering. The knowledge we have learned is essentially inte
elated, but we artificially cut them apart in the process of learning, which makes our knowledge fragmented and one-sided, lack of overall grasp of the knowledge system, and can not be integrated in the process of application.
In the undergraduate stage, we have studied structural mechanics, engineering structure design principle and basic engineering, and also understood the specific methods of scheme design, structural analysis, component design and construction drawing in structural design. Although we have mastered some professional knowledge of civil engineering, we still lack a comprehensive understanding of the whole process of structural design. The graduation design makes up for this defect to a great extent. The graduation project will take the actual project as the background to investigate our knowledge from basic engineering to main engineering, from structural layout to component design, from vertical load to wind load and even seismic load, which can systematically combine the knowledge learned by undergraduates and deepen our understanding of the knowledge learned. In addition to using the knowledge learned, the graduation project also exercises our ability to communicate with people, consult literature norms and use all kinds of professional software, which is very helpful for our future growth and development. Therefore, graduation design is of great significance to the cultivation of professional talents.
In many structural systems, the concept of frame structure mechanics is simple and the design idea is clear. Taking frame structure design as the title can better reflect the goal of cultivating students' comprehensive ability in graduation design. The frame structure is composed of beams and columns. The bent structure with hinged beams and columns is mostly used in the concrete single-storey plant, while the rigid frame structure with rigid beam column connection is mainly used in the multi-storey house. In recent years, frame structure has been widely used in practical engineering because of its flexible space separation and low cost.
During design calculation, we should consult relevant national specifications; When drawing construction drawings, we need to learn the flat method atlas; At the same time, we will also use many computer software, such as word, Excel, AutoCAD and PKPM for computer verification. In the graduation project, the study of these standard atlas and software are combined with actual engineering cases, so as to exercise our ability to learn and apply in practice.
II      Key issues and difficulties of the project:
There are three key problems in the structural design:
one      It is the structural scheme design and component size selection. This is not only related to whether the stress analysis is reasonable, based on the consideration of the actual project, but also affects the project cost. In the structural layout, the layout of column grid, primary and secondary beams directly affects the selection of one-way slab and two-way slab; Foundation selection and stair design should compare various types and choose the best; The rationality of member section size estimation directly determines the co
ectness of stress analysis. How to choose a better scheme is the key in design.
two      It is to select the appropriate calculation model and simplify it reasonably. How to select and simplify the calculation model directly affects the stress analysis of engineering components and the rationality of the final result.
three      How to ca
y out load analysis and internal force combination. Ca
y out co
ect load stress analysis and select the layout of the most unfavorable load to obtain the maximum combined internal force of the most unfavorable section to ensure the safety of the design. At the same time, the practical significance should also be considered in the internal force combination. The seismic action should be fully considered, and the design principles of "strong column and weak beam", "strong shear and weak bending" and "strong node and weak member" should be observed to make it have a good yield mechanism, so as to ensure that the frame beam will yield first under the seismic action and form a beam hinge mechanism to abso
and dissipate seismic energy.
There are four difficulties:
one      Reasonable layout and size estimation of components. It is necessary to ca
y out co
ect load stress analysis on the premise of co
esponding load specifications, and then make reasonable selection of dimensions with reference to structural requirements and economy, so as to meet the subsequent safety requirements.
two      Appropriate simplification and stress analysis of the model. As the design project is a high-rise building, considering the seismic demand, it is designed as a cast-in-situ reinforced concrete frame structure. How to simplify the reasonable model directly affects the rationality and simplicity of the design. During the stress process, it needs to be analyzed in turn according to the co
ect stress transmission path, otherwise it will directly affect the accuracy of subsequent design.
three      Calculation and analysis of frame structure. As the building is a high-rise building, the stress analysis is more complex than the general building structure. We need to skillfully master the stress analysis method of high-rise buildings, reasonably simplify the conditions, and compare the e
or with the computer calculation results and analyze the causes.
four      Skilled use of CAD software and PKPM software. CAD drawing technology has greatly improved our drawing efficiency and accuracy. Mastering drawing technology can greatly reduce our workload. In addition, because it is the first time to come into contact with PKPM computer software to check the manual calculation, it is also a great challenge to our software learning ability. Moreover, PKPM software is not available in China. Since I'm not in China, it's a little difficult for me to download this software.
III      Literature review (or Research Report):
The structural system of modern high-rise buildings is constantly developing and changing with the development of building height and function. The key to the structural design of high-rise buildings is to select a reasonable lateral force resisting system, in which the basic lateral force resisting units are frame, solid web tube, frame tube and support. In the structural system of high-rise buildings, the most commonly used and mature systems are mainly frame structure, frame shear wall structure and tube structure. The mechanical characteristics of each building structure system, the ability and effect of resisting horizontal load, the magnitude of lateral stiffness and the sensitivity of seismic performance are different. Under different fortification intensities and site categories, that is, when bearing different wind loads and earthquake actions, each structural system has different advantages and disadvantages. Therefore, reasonable structural schemes should be selected for different buildings.
In reinforced concrete high-rise and multi-storey public buildings, when the stiffness and strength of the frame structure can not meet the requirements of earthquake resistance or wind resistance, it is easier to meet the relevant requirements of lateral stiffness by using the frame with large strength and stiffness. At the same time, most of the frames in the structure can provide convenient and flexible plane layout for the building, so as to meet the architects' full utilization of the building. It is also a very reasonable structural form in theory and design. It is also the most commonly used structural form in China's u
an housing and comprehensive building construction.
eference:
[1] Cao Shuangyin, chief editor, principles of engineering structure design [M], Southeast University Press, 2004.
[2] Qiu Hongxing, chief editor, architectural structure design [M], Southeast University Press, 2002.
[3] Cheng Wenzhen, et al., concrete structure design (Volume II) [M], China Construction Industry Press, 2005.
[4] Ding Dajun, chief editor, masonry structure [M], China Construction Industry Press, 2004.
[5] Li Aiqun, Gao Zhenshi, et al., earthquake resistance and disaster prevention of engineering structures [M], Southeast University Press, 2003
[6] Kim S E, Colin k J. Multi0storey Precast Concrete Framed Structures[M]. Wiley Blackwell, 2013 (P70-112)
7] National standard of the people's Republic of China. Code for design of concrete, GB XXXXXXXXXX.
[8] National standard of the people's Republic of China. Load code for building structures, GB XXXXXXXXXX.
[9] National standard of the people's Republic of China. Code for seismic design of buildings, GB XXXXXXXXXX.
[10] National standard of the people's Republic of China. Code for design of building foundation, GB XXXXXXXXXX.
[11] Industrial standards of the people's Republic of China. Technical specification for prefa
icated concrete structures JGJ 1-2014
IV      Scheme (design scheme, or research scheme, development scheme) demonstration:
According to the use function of the project and the height of the building, the main structure adopts cast-in-situ reinforced concrete frame structure.
one      Structural layout
The plate thickness is determined according to the stress, load and span of the plate. The preliminarily selected residential slab thickness is 120mm. The column size is determined by the axial compression ratio. Mainly 600x650, 600x500 and other sizes. The beam section size is estimated by the span. Selection of concrete strength: C40 for column, C40 for beam and C40 for slab. Reinforcement grade selection: HRB400 is selected for main reinforcement and sti
up of beam, slab and column.
two      Reinforcement design of slab and secondary beam
According to the elastic theory, the floor slab is divided into one-way slab and two-way slab according to whether L02 / L01 is greater than 3, and calculated according to their respective methods. The secondary beam is designed according to the elastic theory according to the design requirements of the beam.
three      Internal force calculation of frame under vertical load
The internal force calculation under vertical load includes dead load
Answered 62 days After Jan 21, 2022

Solution

Sonam answered on Mar 24 2022
115 Votes
( Dead Load + Imposed Load )
The various forces acting on the beam are the dead load and live load transfe
ed from the slab, the dead weight of the outer wall and the partition wall and the dead weight of the parapet wall. The dead weight of the outer and partition wall will be acting on the intermediate beams only and the dead weight of the parapet walls will be acting on the roof beam only.
Dead weight of outer wall
Thickness of wail = 10" = 0.25m    height of wall = 3.5m
Unit weight of wall material = 22KN/mJ
for beam 1 of span 2.29m, the total weight of outer wall = 0.25*3.5*22*2.29 = 44.08 KN which
will be uniformly distributed over a length of 2.29m, so outer wall load will be = 19.56 KN/m
Dead weight of inner wall
Thickness of wall = 6" = 0.15m    height of the wall = 3.5m
Unit weight of wall material = 22KN/m
For beam 2 of span 5.03m, the total weight of inner wall= 59.03 KN which will be uniformly distributed over a length of 5.03m . so inner wall load will be = 11.73 KN/m
Dead weight of parapet wall
Thickness of wall = 6" = 0.15m    height of the wall = 1 m
Unit weight of wall material = 22 KN/m
For beam 2 of span 5.03m, the total weight of parapet wall= 16.85 KN which will be uniformly distributed over a length of 5.03 m , so parapet wall load will be = 3.35 KN/m
Dead weight and live weight of slab dead weight of slab is 4.5KN/m2 and live load on slabs for commercial building is 2KN/m2 , beam 1 by assuming 45 degree load distribution of slab, the area of slab under beam is 7m2. therefore total load transfe
ed to the beam will be = 8.5KN distributed over a length of . 29m, so total load will be = 3.71KN/m in which 1.14 KN/m is the live load and 2.57KN/m is the dead load.
Similarly for beam 2, by assuming 45 degree load distribution of slab, the area of slab under beam2 is 11.04m2. Therefore total load transfe
ed to the beam will be = 71.78 KN distributed over a length of 5.03 m. so total load will be = 14.27 KN/m in which 4.39KN/m is the live load 2nd 9.88 KN/m is the dead load.
Design of Slab
Slab 1
· Internal dimensions 11 foot 7 inch * 11 foot 4 inch
· The two adjacent edges of slab is discontinuous
· M25 Grade concrete is used, Fe415 Steel is used
Aspect Ratio:
Aspect RaLio = long dimension/short dimension = 3.53/3.45 = 1.02. As the aspect ratio is less than 2, It is a two-way slab therefore both the major and minor axis will take load and designed.
Assume trial depth (d) of 120 mm, assuming 20 mm cover to steel, overall depth = 140 mm
c/c distance between support = 3.45 + 0.20 = 3.65m Effective Short Span
Lx + d = 3.45-J .12 = 3.57m c/c distance between support = 3.53 + 0.20 = 3.73m Effective Long Span
Ly ^ d = 3.53 + 0.11 = 3.64m
Minimum of the two values will be considered, Bar diameter has been assumed as 10 mm)
True aspect ratio = 3.64/3.57
Check: span/depth < 35, here (3.57/.12) - 29.79
Moment Coefficients: Ly/Lx=1.02    ax    ay
. -.ative moment at continuous edge    0.047    0.047
I : sitive moment at mid span    0.035    0.035
Load Calculation: for 140 mm thick sla
dead Load 0.14*1*1*25 = 3.5 KN/m2
live Load for commercial buildings = 2 KN/m" 7
finish Load = 1 KN/m2
Total Load = 6.5 KN/m2
Bending Moment Calculation: For short span Mx = ax*w*Lx
Hx = 3.9 KN-m at continuous edge, Mx = 2.9 KN-m at mid-span
Mx= 3.9 KN-m at continuous edge, My = 2.9 KN-m at mid-span ir Force Calculation:
Shear force on short span = force transfe
ed to the long edge =1/2*(3.73+0.08)*(3.65/2)*[l/(0.75)(3.65)]*6.5 = 8.09 KN for lm
Shear force on long span = force transfe
ed to the short edge np3.65*(3.65/2)*[l/(0.75)(3.65)]*6.5 = 7.92 KN for lm
Check for Depth:
Mmax. = 0.36fckbxm(d-0.42xm)
1.5*3.9*106 = 0.36*25* 1000*0.48d*(d-0.42*0.48d)
dy = 41mm, provided dy= 110mm. Hence, okay. Similarly, dx will also be okay of
Area of Steel
Mx = 0.87*fv*A,; *(dx-0.42xm)
1.3*.9*10° - 0.87*415*Ast.*(0.7984)*120
Astx170mm" (at continuous edge), Astx = 126mm" (at mid span)
My= 0.87*fy*Ast.*(dy-0.42xm) 3.9*10° =
1.5*3.9*1066=0.87*415*Ast.*(0.7984)*l10
Asty =185mm2 (at continuous edge), Astv = 137mm2 (at mid span)
Minimum reinforcement = 0.12% of the gross sectional area = (0.12/100)* 1000* 140 = 168mm2
At mid span in both long and short span 168mm2 reinforcement should be provided.
Spacing Sx and Sy (in mid span) = 1000*Ast/Astx = [1000*(-)*82]/168 = 300mm
8mm diameter bars are used. Taking 50% of the reinforcement to be bent up at the support section. Extra reinforcement required at continuous support = 170-( 168/2) = 86mm'
Provide 6mm diameter bars, Sx = [1000*(-)*6 ]/86 = 328mm, Provide Sx = 320mm
for long span extra reinforcement required = 185-( 168/2) = 101mm
[1000*Q)*62]/101 =280 mm
Reinforcement Detailing:
· Slabs are considered as divided in each direction into middle strips and edge strips. The middle strip being three-quarters of the width and edge strip one-eigth of the width.
· Tension reinforcement provided at mid-span in the middle strip shall extend in the lower part of the slab to within 0.25L of a continuous edge, or 0.15L of a discontinuous edge.
· Over the continuous edge of the middle strip, the tension reinfoi cement shall extend in the upper part of the slab at a distance of 0.15L from the supports, and at least 50' extend a distance of 0.3L
· At a discontinuous edge, negative moments may arise. They depend on the degree of the fixidity at the edge of the slab but. in general, tension reinforcement equal to 50 provided at mid span extending 0.1L into the span will be sufficient.
Torsion Reinforcement:
· Torsion Reinforcement shall be provided at any comer where the slab is simply supported on both edges meeting at that comer. It shall consist of top and bottom reinforcement each with layers of bars placed parallel to the sides of the slab and extending from the edges a minimum distance of one-fifth of the shorter span. The area of reinforcement in each of these four layers shall be three-quarters of the area required for the maximum mid-span moment in the slab.
· Torsion Reinforcement equal to half as described above shall be provided at a comer contained by edges over only one of which the slab is continuous. It should not be provided at any comer contained by edges over both of which the slab is continuous.
• Torsion Reinforcement AT = 3/4Astx = 127.5mm2 provided at both top and bottom
and extending from edge a minimum distance of one-fifth of short span = 3.57/5 = 715mm.
Check for shear and development length:
shear force in short span = shear force in long edges = 8.09 KN/m
nominal shear stress = (1.5*8.09*103) - (1000*120) = 0.101 N/mm2
Minimum shear strength of concrete for M25 is 0.29 N/mm2. Hence, safe.
Development Length = (0.87*fy*diameter) - (4*bond strength) = (0.87*415*8) - (4*1.4*1.6) = 322mm.
L0 + M/V > Ld
At support section, M = 0.87*fy*Astx(d-0.42xm) = 5.88 KN-m and V = i 5*8.09 =12.13 KN.
MV = 484mm.
Hence, safe. Similarly for long span it will be safe as shear force is less there.
Slab 4
Dimension 26 foot 10...
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