conclusion - springer978-3-662-48044-1/1.pdf(2) technical speciﬁcation for steel structure of...

Conclusion

It is mentioned that the fuzzy machine learning and the neural network learning arethe fundamental approaches in the intelligent systems for structural design. Thepreliminary results presented in this chapter show that the previous one is devotedto the classification of samples, which form the sample base of the system, sorted byits attributes; the latter one aims to “cultivate” the system through NN learning inorder to accumulate the design expertise of human being in the system.

It is necessary to point out that since the number of the samples used in thischapter is not adequate, the characteristics of the building are also different fromeach other. It seems that further efforts must be paid for collecting more samples ofvarious kinds until the sample base can be improved for practical use. Nevertheless,the methodology of intelligent design introduced in this chapter is actually based oncomparative design philosophy. It has been much encouraged in this chapter thateven by using such an incomplete sample base, the numerical error estimation oftest examples show the mean error of the samples inferred by NN system can belimited within 6 %.

Nevertheless, for achieving benefits from intelligent comparative design ofstructure, the input fuzzy membership function is the key for success, hopefully thefuzzy machine learning method can effectively solve the problem by [62].However, for achieving our goals of benefit by intelligent comparative design ofstructure, there will be a lot of common efforts to be spent for researchers andpractitioners in the future.

© Shanghai Jiao Tong University Press, Shanghai and Springer-Verlag Berlin Heidelberg 2016S. Lin and Z. Huang, Comparative Design of Structures,DOI 10.1007/978-3-662-48044-1

369

Appendix A

A.1 Example of Conceptual Design for Single-StoryMulti-span Steel Factory Building

Previous chapters of this book have introduced in detail some indispensablemechanical concepts and methods of simplified calculation in modern structuraldesign. Now we’ll come to the application of these concepts and methods inpractical construction project, specifically, in the conceptual design for single-storymulti-span steel structural factory building.

1. Project Introduction

This is a construction project of an industrial factory building, taking up a con-struction area of about 20,000 m2. The factory building is comprised of twosingle-storied multi-span steel structural sections A and B (as shown in Fig. A.1).Each covers an area of 10,000 m2. In light of the length limitations of this book, wewill take factory building A as an example and talk about the conceptual design ofthis part of the factory only.

2. The purpose of conceptual design

(1) It is to select a better structural system and the way of arrangement, inwhich case the resultant structure should not only be up to the standard butalso minimize the cost and guarantee the safety.

Fig. A.1 Rendering of thisprospective factory building


371

(2) Through the estimation of steel consumption of the main structure, theappraisal of economic efficiency and applicability of the structural systemcould be better made along with the total cost of construction.

(3) Conceptual design could serve as a guide to the following preliminarydesign and detailed design. The control of the most critical section and thecombination of loads ensures the safety and rationality of structural design.

3. Conditions of design

(1) Proprietor-provided plane and elevation layouts,(2) Proprietor-provided live loads,(3) Preliminary survey report of foundation soil.

4. Design norms

(1) Load Code for the Design of Building Structures (GB50009-2001), 2001,Beijing,

(2) Technical Specification for Steel Structure of Lightweight Building withGabled Frames (CECS102), 2002, Beijing,

(3) Code for Design of Steel Structures (GB50017-2002), 2002, Beijing,(4) Code for Design of Building Foundation (GB50007-2002), 2002, Beijing,(5) Code for Design of Concrete Structures (GB50010-2002), 2002, Beijing.

5. The assumptions

(1) Assume the structural system works in an elastic state.(2) Live loads are simplified as equivalent uniformly or linearly distributed

loads, conducted with reference to “Load Code for the Design of BuildingStructures”.

(3) All supports or connecting joints are assumed to be rigid connections, inwhich case the bending moments are transferred.

(4) The slope of the roof is assumed to be 5 %.

6. The diagram of simplified calculation of the main steel structure

According to the design principles of optimization of steel structures, multiplespans of single-storied portal frames of steel structure are employed in this project.Since the load of crane shall be imposed on each span, identical-sectionedwide-flanged I-shaped steels are employed in each section of column of the mainrigid frame; as for the section of the beams, differential-sectioned wide-flangedI-shaped steels are employed to reduce the consumption of steel. Since the cranegoes over each span of the structure, beams and columns are rigidly connected andso it is with the columns and the foundation, which enables the transference of thebending moment.

With reference to the requirements of construction technology and the opti-mization principles of structure, the main structure of factory building A adopts theportal frame of four spans. The span crosses over a distance of 25 m and there’s aspacing of 6 m between main rigid frames. The side column is 12 m high, as shownin Fig. A.2.

372 Appendix A

7. Estimation of the loads in structural design

The estimation principles of loads introduced in Chap. 2 shall be applied. Takinginto consideration the requirements of construction technology for industrial pro-ject, major loads as listed in Table A.1 have to be considered in the design of thisfactory building.

8. The estimation of the internal forces of the cross section under the action ofthe most unfavorable combination of vertical loads

According to the Load Code for the Design of Building Structures (GB50009-2001), the most unfavorable vertical uniform load of beam is 1.2 × deadweight +1.4 × overhanging + 1.4 × load of snow or live load upon the roof

w ¼ 1:2� 0:45þ 1:4� 0:10þ 1:4� 0:30 ¼ 1:10 kN=m2

The deadweight of the roof structure (beam, purlin, and the panel included) isestimated to be 0.45 kN/m2.

The uniform load derived from the equation has to be transformed into equiv-alent uniform linear load q, being borne by each single-span portal frame.

Since the spacing between the frames is 6 m, we have

q ¼ 6m� 1:10 ¼ 6:6 kN=m

Table A.1 Summary of major loads for factory building A

Type Load

Uniform roof live load 0.3 kN/m2

Uniform roof snow load (Shanghai area) 0.2 kN/m2

Overhanging roof load 0.1 kN/m2

Wind load (Shanghai area) 0.55 kN/m2

Type of crane 12.5 t × 24 m

Total weight of the crane 19.316 t

Weight of the crane itself 1.879 t

Uniform floor live load 30 kN/m2

Load of the forklift 15 t

25m 25m 25m 25m

12m

5%5%

Fig. A.2 Diagram of simplified design and calculation of the main steel structure

Appendix A 373

http://dx.doi.org/10.1007/978-3-662-48044-1_2

The maximum bending moment and shear force borne by the primary beam ofthe portal frame are estimated as (the primary beam is assumed to be continuousbeam)

Mmax ¼ qL2=12 ¼ 6:6 kN=m� 25m� 25m=12 ¼ 343:75 kNm

Qmax ¼ qL=2 ¼ 6:6 kN=m� 25m=2 ¼ 82:5 kN

The maximum axial force of the central column is estimated as (the load of cranealong with possible combinations considered; crane weighs 19.316 t)

Nmax ¼ qLþ 4 Qcð Þ ¼ 6:6� 25þ 4� 1:1� 0:8 125þ 193:16=2ð Þ ¼ 944:96 kN

The maximum bending moment and axial force of the side column is estimatedas (the load of crane and possible combinations considered)

Mmax ¼ 2=3ð ÞqL2=12 ¼ 2=3ð Þ343:75 ¼ 229:17 kNm

Nmax ¼ 0:5 qLþ 4 Qcð Þð Þ ¼ 0:5� 944:96 ¼ 472:48 kN

Figure A.3 is the diagram of bending moment distribution of major rigid frameunder the action of most unfavorable vertical loads.

9. Estimation of internal force of the cross section under the action of mostunfavorable combination of horizontal loads

(1) The most unfavorable combination of horizontal loads that parallel with theaxis of the major rigid frame is 1.4 × wind load + horizontal braking forceof the crane. Specifically,

Fig. A.3 The distribution of bending moment of major rigid frame under vertical loads

374 Appendix A

Wind load : hw ¼ 1:4� 0:8þ 0:7ð Þ � 0:55� 6 ¼ 6:01 kN=m

0.8 and 0.7 in the above equation are the shape coefficients of the wind load

Horizontal braking force of the crane: hc ¼ 2� 0:7� 0:1 125þ 18:79ð Þ¼ 20:13 kN

125 kN in the equation above is the lifting weight of the crane and 18.79 kN thedeadweight of the crane itself.

Assume the action point of the crane is at the midpoint of the side column (themost unfavorable position), then the maximum bending moment and shear force ofthe central column (the height being 14.5 m) are estimated as

Mmax ¼ 6:01� 12� 2=8ð Þ � 14:5 m=2þ 20:13 kN� 14:5m=8 ¼ 165:95 kNm

According to the method of simplified calculation introduced in Chap. 4Sect. 4.7 concerning portal frame and frame structure, we assume that thewind-induced shear force distributed to the central column is twice that to the sidecolumn and the inflection point is the midpoint of the column.

Qmax ¼ 6:01� 12 2=8ð Þþ 20:13=2 ¼ 28:10 kN

The maximum bending moment and shear force of the side column are estimatedas

Mmax ¼ 6:01� 12� 1=8ð Þ � 12m=2þ 0:5� 20:13� 12m=8 ¼ 69:19 kNm

Qmax ¼ 6:01� 12� 1=8ð Þþ 0:5� 20:13 ¼ 19:08 kNm

Figure A.4 is the distribution diagram of the bending moment of major rigidframe under the action of the most unfavorable combination of loads that parallelwith the axis of the frame.

(2) The most unfavorable combination of horizontal loads that are perpendicularto the axis of major rigid frame

Fig. A.4 Distribution diagram of the bending moment of rigid frame under horizontal loads

Appendix A 375

http://dx.doi.org/10.1007/978-3-662-48044-1_4

Horizontal braking force of the crane:

hc0 ¼ 2� 0:7� 0:1 12:5 tþ 19:316 tð Þ=2 ¼ 22:27 kN

And the maximum bending moment and shear force of the central column areestimated as

Mmax ¼ 22:27 kN� 14:5m=8 ¼ 40:36 kNm

Qmax ¼ 22:27=2 ¼ 11:14 kN

The maximum bending moment and shear force of the side column are estimatedas

Mmax ¼ 0:5� 22:27� 12=8 ¼ 16:7 kNm

Qmax ¼ 0:5� 11:14 ¼ 5:57 kNm

10. Estimation and selection of cross section of beam and side column in majorrigid frame

The control internal forces within the section of beam and side column under theaction of load combinations are summarized as

Mxmax ¼ 343:75 kN m

Mymax ¼ 16:7 kN m

Qmax ¼ 82:5 kN

Nmax ¼ 472:48 kN

(1) Checking of the bending moment:Assume the major rigid frame adopts steel Q345, whose design strength is310 mpa, the sectional modulus of the selected section shall meet therequirement

W [Mmax=310 ¼ 343;750;000=310 ¼ 1;108;871 mm3

Symmetrical wide-flanged I-shaped section H400 × 280 × 8 × 16 could beadopted as the maximum cross section for both the side column and thebeam.The differential section of the beam is H500-300 × 280 × 8 × 16 and thecalculation of deformation and strength in conceptual design adopts theequivalent identical section H400 × 280 × 8 × 16. The area moment ofH400 × 280 × 8 × 16 is

376 Appendix A

Wx ¼ 2� 280� 16� 200þ 8� 400� 400=6 ¼ 1;952;000mm3 [ 1;108;871mm3 OK!

Wy ¼ 2 16� 280� 280=6ð Þ ¼ 418;133 mm3 [ 16;700;000=310 ¼ 53;871m3 OK!

(2) Checking of the shear force:The maximum shear stress of the cross section is

smax ¼ 1:5� 82:5 kN= 8� 400ð Þ ¼ 38:67 MPa\ s½ � ¼ 180 MPa OK!

(3) Checking of the axial compression:The maximum axial compressive stress within the section of the sidecolumn is

rmax ¼ 472;480= 8� 400þ 2� 16� 280ð Þ ¼ 38:86 MPa\ r½ �¼ 310 MPa

OK!

(4) Checking of vertical deformation of the beam:Sectional moment of inertia is first estimated:

Ix ¼ 2� 280� 16� 200� 200þ 8� 4003=12 ¼ 401;066;667 mm4

The combination of loads selected for the checking of deformation (dis-similar with the one chosen in the checking of strength is 1.0 deadweight +0.5 overhanging load + 0.5 live load upon the roof

w0 ¼ 0:45þ 0:5� 0:1þ 0:5� 0:3 ¼ 0:65 kN=m2

q0 ¼ 6 m� 0:65 kN=m2 ¼ 3:9 kN=m

The maximum vertical deformation of the beam is

Db ¼ 1=384ð Þ � 3:9 kN=m� 254 m4= 2:06� 105 N=mm2 � 401;066;667 mm4� �

¼ 0:048 m\L=400 ¼ 0:0625 m OK!

(5) Checking of horizontal deformation of the side column:Sectional moments of inertia are first estimated:

Ix ¼ 2� 280� 16� 200� 200þ 8� 4003=12 ¼ 401;066;667 mm4

Iy ¼ 2� 16� 2803=12 ¼ 58;538;666 mm4

The selected combination of loads is different from the one in the checkingof strength.The combination of horizontal loads that parallel with the axis of the majorrigid frame is as follows:

Appendix A 377

Wind load + horizontal braking force of the crane. Specifically,Wind load: hw = (0.8 + 0.7) × 0.55 × 6 = 4.95 kN/mHorizontal braking force of the crane: hc = 2 × 0.7 × 0.1(125 + 18.79) = 20.13 kNAssume the action point of the crane is at the midpoint of the side column(the most unfavorable position).The shear force imposed upon the side column is estimated as

Qx ¼ 4:95� 12� 1=8ð Þþ 0:5� 20:13 ¼ 17:49 kN m

The deformation of the side column under the action of axis-parallelinghorizontal loads is estimated as

Dc ¼ Qxð Þh3= 12EIð Þ¼ 17:49 kN� 113 m3= 12� 2:06� 105 N=mm2 � 401;066;667 mm4� �

¼ 0:0235 m\0:0275 m ¼ h=400 OK!

The deformation of the side column under the action of axis-perpendicularhorizontal loads is estimated as

Dc ¼ Qyð Þh3= 12EIð Þ¼ 5:57 kN� 123 m3= 12� 2:06� 105 N/mm2 � 58;538;666 mm4� �

¼ 0:007\0:03 ¼ h=400 OK!

11. Estimation and selection of cross section of the central column in majorrigid frame

(1) The control internal forces of the cross section of the central column underthe action of the most unfavorable combination of loads, derived fromprevious calculation, are

Mxmax ¼ 165:95 kN m

Mymax ¼ 40:36 kN m

Qmax ¼ 28:1 kN

Nmax ¼ 944:96 kN

(2) Symmetrical wide-flanged I-shaped section H300 × 260 × 8 × 16 isadopted and the according area moments are

Wx ¼ 2� 260� 16� 150þ 8� 300� 300=6 ¼ 1;368;000mm3

Wy ¼ 2 16� 260� 260=6ð Þ ¼ 3;605;300mm3

378 Appendix A

(3) Checking of the bending moment, shear, and axial force:

Mxmax ¼ Wxð Þ � 310 ¼ 1;368;000� 310 ¼ 424:32 kNm[ 165:95 kNm OK!

Mymax ¼ Wyð Þ � 310 ¼ 3;605;300� 310 ¼ 111:76 kNm[ 40:36 kNm OK!

smax ¼ 1:5� 28:1 kN = 8� 300ð Þ ¼ 17:56MPa\ s½ � ¼ 180MPa OK!

rmax ¼ 944;960= 8� 300þ 2� 16� 260ð Þ ¼ 88:15 MPa\ r½ � ¼ 310 MPa OK!

(4) Checking of deformation:The sectional moments of inertia are

Ix ¼ 2� 260� 16� 150� 150þ 8� 3003=12 ¼ 205;200;000mm4

Iy ¼ 2� 16� 2603=12 ¼ 46;869;000 mm4

The selected combination of loads in the checking of deformation is dif-ferent from the one in the checking of strength.

The combination of axis-paralleling horizontal loads is wind load + horizontalbraking force of the crane.

Wind load hw = (0.8 + 0.7) × 0.55 × 6 = 4.95 kN/mHorizontal braking force of the crane hc = 2 × 0.7 × 0.1(125 + 18.79) = 20.13 kNAssume the action point of the crane is at the midpoint of the side column (the

most unfavorable position).The shear force imposed upon the central column is estimated as

Qx ¼ 4:95� 12� 1=8ð Þþ 0:5� 20:13=2 ¼ 7:43þ 5:03ð Þ kN m

The deformation of the central column under the action of axis-parallelinghorizontal loads is estimated as follows:

Dc ¼ h3= 12EIð Þþ Qx00ð Þh3= 192EIð Þ ¼ Qx0=12þQx00=192ð Þh3=EI¼ 7:43 kN m=12þ 5:03 kN m=192ð Þ � 14:53 m3= 2:06� 105 N/mm2 � 205;200;000 mm4

� �

¼ 0:032 m\0:036 m ¼ h=400 OK!

The deformation of the central column under the action of axis-perpendicularhorizontal loads is estimated as follows:

Horizontal braking force of the crane is

hc0 ¼ 2� 0:7� 0:1 12:5 tþ 19:316 tð Þ=2 ¼ 22:27 kN

Dc ¼ Qyð Þh3= 192EIð Þ¼ 11:14 kN� 123 m3= 192� 2:06� 105 N=mm2 � 46;869;000 mm4

� �

¼ 0:003\0:036 ¼ h=400 OK!

Appendix A 379

12. Estimation of steel consumption of major rigid frame

The steel consumption of each major rigid frame in factory building A:

1:1 8� 400þ 2� 16� 280ð Þ � 10�6 � 7850� 25� 4þ 12� 2ð Þ�

þ 8� 300þ 2� 16� 260ð Þ10�6 � 7850� 14:5� 3ð Þ� ¼ 17:045 T

The coefficient 1.1 in the equation above has taken into consideration the extrasteel consumption entailed in the connecting plates and the indispensable structuralsupports.

Unit steel consumption is as follows:

17;045= 4� 25� 6ð Þ ¼ 28:41 kg/m2

The total steel consumption of major rigid frames in factory building A is asfollows:

1:05� 28:41� 10;000 m2 ¼ 298 T

The coefficient 1.05 in the equation above comes from the extra steel employedin wind-resisting columns.

13. Simplified calculation of the supporting ground and foundation

According to above static calculation, the loads in side column are

Mxmax ¼ 229:17 kNm

Mymax ¼ 26:7 kNm

Qmax ¼ 19:08 kN

Nmax ¼ 472:48 kN

According to above static calculation, the loads in middle column are

Mxmax ¼ 165:95 kNm

Mymax ¼ 40:36 kNm

Qmax ¼ 19:08 kN

Nmax ¼ 944:96 kN

With reference to the results of preliminary survey of the supporting ground andtaking into consideration the supporting capacity of the ground and a rational formof force bearing, independent below-column pile foundation is adopted, as shown inFig. A.5. The independent below-column pile foundation consists of atriangle-shaped pile cap and three prestressed-prefabricated supporting piles.Diameter of the pile is 400 mm and the spacing between piles is 1200 mm. The

380 Appendix A

length of the pile cap is no less than 2 m and the thickness no less than 700 mm.Based on the distribution of soil layers, the length of the pile is determined to be13 m.

The checking of bending and compressive resistance of the central pile cap:

Mx ¼ My ¼ Nmax=3ð Þ s� 31=2c=4� �

¼ 766 kN=3ð Þ 1200� 31=20:866� 400 mm=4� �

¼ 268 kNm[ 229:17 kNm [ 165:95 kNm OK!

Nmax ¼ 3� 766 kN ¼ 2298 kN[ 944:96 kN[ 472:48 kN OK!

14. Simplified design and calculation of floor

With reference to the condition of soil layers of the supporting ground and a strictrestriction upon the settlement of floor, piles are needed in the treatment of thesupporting ground under the floor. Prestressed-prefabricated 300-mm-diameterpiles, the length of and the spacing between which are respectively 8 m and2.5 m × 2.5 m, fit the job.

The checking of vertical loads:

30� 2:5� 2:5ð Þ ¼ 188 kN\Ra ¼ 259 kN OK!

300-mm-thick C30 reinforced concrete floor is preselected, as shown in Fig. A.6.Readers could manually design and calculate the two-way double-layered rein-forcements of the floor.

The design of the author goes with two-way double-layered Grade II rein-forcements, the diameter of and the spacing between which, respectively, are 16 and150 mm, for the plate strip above the column and 12-mm-diametered and200-mm-spaced two-way double-layered reinforcements for the mid-span platestrip. And the average reinforcement consumed is

Fig. A.5 Diagram of below-column pile foundation and the pile cap

Appendix A 381

42:13þ 17:8ð Þ=2 ¼ 30 kg=m2

15. Summary of design

The conceptual design of this project could be summarized as follows:

(1) The design of the major rigid frame is the key in this project. The esti-mation of loads, the simplification of calculation model and the consid-eration of the most unfavorable combination of loads are to be conductedwith careful attention. Afterward, practical construction experiences andsimplified methods of calculation introduced in previous chapters could bereferred to in the design and checking calculation. Figure A.7 is the dia-gram of selected sections for the major rigid frame.

(2) Unit steel consumption of major rigid frame of this factory building is28.41 kg/m2, on the basis of which the total steel consumption of majorsteel structure could be calculated as 298 t along with the probable cost ofconstruction.

(3) The design of the supporting ground and foundation could determine thetype of pile foundation as well as the length and type of the pile. And thisis quite significant for the estimation of cost of construction concerningthis part of the structure.

Fig. A.6 Diagram of floor slabs

H300 260 8 16

H400 280 8 16

H500-300 280 8 16

Fig. A.7 Diagram of selected sections for the major rigid frame

382 Appendix A

A simple design of floor could determine whether or not to adopt pile, techniqueentailed in the construction of floor and the estimation of reinforcements to beinstalled within the floor. At the completion of this process, the cost of constructionof this part could be estimated, too.

Appendix A 383

Appendix B

B.1 Example of Conceptual Design of Multi-storySteel Frame Factory Building

In previous chapters of this book, detailed introduction has been made of necessaryconcepts of mechanics and simplified methods of calculation to be applied inmodern structural design. And now we will come to the application of these con-cepts and methods in construction projects and to the analysis of example ofconceptual design of multi-storied multi-span steel frame factory building.

1. Project Introduction

The project of building this industrial factory building covers an area of about30,000 m2. And factory building C is a four-story steel frame structure (as shown inFig. B.1), taking up an area of 6400 m2 with a plane layout of 60 m × 60 m. As forthe others, they are all single-storied. Due to the limited length of this book, we willonly discuss the conceptual design of the representative four-storied steel framefactory building C in here.

Fig. B.1 Rendering of thisprospective factory building


385

2. The purpose of conceptual design

(1) It is to select a better structural system and the way of arrangement, inwhich case the resultant structure should not only be up to the standard butalso minimize the cost and guarantee the safety.

(2) Through the estimation of steel consumption of the main structure, theappraisal of economic efficiency and applicability of the structural systemcould be better made along with the total cost of construction.

(3) Conceptual design could serve as a guide to the following preliminarydesign and detailed design. The control of the most critical section and thecombination of loads ensures the safety and rationality of structural design.

3. Conditions of design

(1) Proprietor-provided plane and elevation layouts,(2) Proprietor-provided live loads,(3) Preliminary survey report of foundation soil.

4. Design Norms

(1) Load Code for the Design of Building Structures (GB50009-2001), 2001,Beijing,

(2) Technical Specification for Steel Structure of Lightweight Building withGabled Frames (CECS102), 2002, Beijing,

(3) Code for Design of Steel Structures (GB50017-2002), 2002, Beijing,(4) Code for Design of Building Foundation (GB50007-2002), 2002, Beijing,(5) Code for Design of Concrete Structures (GB50010-2002), 2002, Beijing.

5. The assumptions

(1) Assume the structural system works in an elastic state.(2) Live loads are simplified as equivalent uniformly or linearly distributed

loads, conducted with reference to “Load Code for the Design of BuildingStructures”.

(3) All supports or connecting joints are assumed to be rigid connections, inwhich case the bending moments are transferred.

(4) Factory building C is assumed to be a frame structure as shown in Fig. B.2.In both directions, there are five twelve-meter spans in the structure.

6. The diagram of simplified calculation of the main steel structure

According to the design principles of optimization of steel structures, multi-storiedframe structure is adopted in this project. Since all stories are involved in the forcebearing, box section is employed for the major columns; as for the primary beams,identical-sectioned wide-flanged I-shaped steel is adopted; all beam–column jointsand supports are rigidly connected to make a frame structure.

In light of the construction technology and the principles concerning the opti-mization of structures, the main structure of factory building C takes form of a

386 Appendix B

four-storied steel frame structure. The column grid is 12 m × 12 m. The groundfloor is 8 m high and the rest are 5 m high, as shown in Fig. B.2.

7. Estimation of the loads in structural design

The estimation principles of loads introduced in Chap. 2 shall be applied. Takinginto consideration the requirements of construction technology for industrial pro-ject, major loads as listed below have to be considered in the design of this factorybuilding (Table B.1):

8. Estimation of the internal forces of the cross section under the action of themost unfavorable combination of vertical loads

(1) The most unfavorable combination of vertical loads on the roof.The structural system of the roof is a primary–secondary beam system. Thesecondary beams are simply supported, spanning over a distance of 12 mwith an in-between spacing of 6 m. The primary beams are rigidly sup-ported, spanning over a distance of 12 m with an in-between spacing of12 m as well.

12m 12m 12m 12m 12m

8m

5m

5m

5m

Fig. B.2 Diagram of simplified calculation of the main steel structure

Table B.1 Summary of major design loads for factory building C

Type Load

Uniform roof live load 0.3 kN/m2

Uniform roof snow load (Shanghai area) 0.2 kN/m2

Overhanging roof load 0.1 kN/m2

Overhanging floor load 0.3 kN/m2

Wind load (Shanghai area) 0.55 kN/m2

Load of crane 0

Uniform live load on the 4th floor 5 kN/m2

Uniform live load on the 3th floor 7.5 kN/m2

Uniform live load on the 2nd floor 7.5 kN/m2

Uniform live load on the ground 30 kN/m2

Load of the forklift on the ground 5 t

Appendix B 387

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According to the Load Code for the Design of Building Structures(GB50009-2001), the most unfavorable combination of vertical uniformloads is 1.2 × deadweight + 1.4 × overhanging load + 1.4 × load of snowor live load on the roof.

w ¼ 1:2� 0:45þ 1:4� 0:10 þ 1:4� 0:3 ¼ 1:10 kN=m2

The equivalent uniform linear load shall then be (since the spacingbetween secondary beams is 6 m)

q0 ¼ 6m� 1:10 ¼ 6:6 kN=m

And the maximum bending moment and shear force of the simply sup-ported secondary beam are estimated as

M0max ¼ q0L2=8 ¼ 6:6 kN=m� 12 m� 12 m=8 ¼ 118:8 kNm

Q0max ¼ q0L=2 ¼ 6:6 kN=m� 12 m=2 ¼ 39:6 kN

Surface load is imposed upon 12-m-spaced, rigidly supported primarybeams and

q ¼ 12� 1:10 ¼ 13:2 kN=m

Mmax ¼ qL2=12 ¼ 13:2 kN=m� 12 m� 12 m=12 ¼ 158:4 kNm

Qmax ¼ qL=2 ¼ 13:2 kN=m� 12 m=2 ¼ 79:2 kN

The maximum axial compression of the middle column is estimated as

Nmax ¼ 1:1� 12� 12 ¼ 158:4 kN

The maximum bending moment and axial force of the side column areestimated as

Mmax ¼ 0:5� 158:4 ¼ 79:2 kNm

Nmax ¼ 0:5� 158:4 ¼ 79:2 kN

(2) The most unfavorable combination of vertical loads upon the 4th floorThe structural system of the floor is a primary–secondary beam system.The secondary beams are simply supported, spanning over a distance of12 m with an in-between spacing of 2 m. The primary beams are rigidlysupported, spanning over a distance of 12 m with an in-between spacing of12 m as well.The most unfavorable combination of vertical loads is 1.2 × dead-weight + 1.4 × overhanging load + 1.3 × live load on the roof.

388 Appendix B

w ¼ 1:2� 3:5þ 1:4� 0:30 þ 1:3� 5 ¼ 11:12 kN=m2

In the equation above, we assume the deadweight of the 4th floor to be3.5 kN/m2 and the equivalent uniform linear load (since the spacingbetween secondary beams is 2 m) would then be

q0 ¼ 2 m� 11:12 ¼ 22:24 kN=m

And the maximum bending moment and shear force of simply supportedsecondary beam are estimated as

M0max ¼ q0L2=8 ¼ 22:24 kN=m� 12 m� 12 m=8 ¼ 400:32 kNm

Q0max ¼ q0L=2 ¼ 22:24 kN=m� 12 m=2 ¼ 133:44 kN


q ¼ 12� 11:12 ¼ 133:44 kN=m

Mmax ¼ qL2=12 ¼ 133:44 kN=m� 12 m� 12 m=12 ¼ 1601:28 kNm

Qmax ¼ qL=2 ¼ 133:44 kN=m� 12 m=2 ¼ 800:64 kN

The maximum axial compression of the central column is estimated as

Nmax ¼ 11:12� 12� 12þ 158:4 ¼ 1759:68 kN


Mmax ¼ 0:5� 1601:28 ¼ 800:64 kNm

Nmax ¼ 0:5� 1759:68 ¼ 879:84 kN

(3) The most unfavorable combination of vertical loads upon the 3rd floorThe structural system of the floor is a primary–secondary beam system.The secondary beams are simply supported, spanning over a distance of12 m with an in-between spacing of 1.5 m. The primary beams are rigidlysupported, spanning over a distance of 12 m with an in-between spacing of12 m as well.The most unfavorable combination of vertical loads is 1.2 × dead-weight + 1.4 × overhanging load + 1.3 × live load on the roof.

w ¼ 1:2� 3:5þ 1:4� 0:30þ 1:3� 7:5 ¼ 14:37 kN=m2

The equivalent uniform linear load (since the spacing between secondarybeams is 1.5 m) would then be

Appendix B 389

q0 ¼ 1:5 m� 14:37 ¼ 21:56 kN=m

The maximum bending moment and shear force of simply supportedsecondary beam are estimated as

M0max ¼ q0L2=8 ¼ 21:56 kN=m� 12 m� 12 m=8 ¼ 388:08 kNm

Q0max ¼ q0L=2 ¼ 21:56 kN=m� 12 m=2 ¼ 129:364 kN


q ¼ 12� 14:37 ¼ 172:44 kN=m

Mmax ¼ qL2=12 ¼ 172:44 kN=m� 12 m� 12 m=12 ¼ 2069:28 kNm

Qmax ¼ qL=2 ¼ 172:44 kN=m� 12 m=2 ¼ 1034:64 kN

The maximum axial conpression of the central column is estimated as

Nmax ¼ 14:37� 12� 12þ 1759:68 ¼ 3824:64 kN


Mmax ¼ 0:5� 2069:28 ¼ 1034:64 kNm

Nmax ¼ 0:5� 3824:64 ¼ 1912:32 kN

(4) The most unfavorable combination of vertical loads upon the 2nd floorThe structural system of the floor is a primary–secondary beam system.The secondary beams are simply supported, spanning over a distance of12 m with an in-between spacing of 1.5 m. The primary beams are rigidlysupported, spanning over a distance of 12 m with an in-between spacing of12 m as well.The most unfavorable combination of vertical loads is 1.2 × dead-weight + 1.4 × overhanging load + 1.3 × live load on the roof.

w ¼ 1:2� 3:5þ 1:4� 0:30þ 1:3� 7:5 ¼ 14:37 kN=m2

The equivalent uniform linear load (since the spacing between secondarybeams is 1.5 m) would then be

q0 ¼ 1:5m� 14:37 ¼ 21:56 kN=m

The maximum bending moment and shear force of simply supportedsecondary beam are estimated as

390 Appendix B

M0max ¼ q0L2=8 ¼ 21:56 kN=m� 12m� 12 m=8 ¼ 388:08 kNm

Q0max ¼ q0L=2 ¼ 21:56 kN=m� 12m=2 ¼ 129:364 kN


q ¼ 12� 14:37 ¼ 172:44 kN=m

Mmax ¼ qL2=12 ¼ 172:44 kN=m� 12 m� 12 m=12 ¼ 2069:28 kNm

Qmax ¼ qL=2 ¼ 172:44 kN=m� 12 m=2 ¼ 1034:64 kN

The maximum axial compression of the central column is estimated as

Nmax ¼ 14:37� 12� 12þ 3824:64 ¼ 5893:92 kN


Mmax ¼ 0:5� 2069:28 ¼ 1034:64 kNm

Nmax ¼ 0:5� 5893:9 ¼ 2946:96 kN

Figure B.3 is a diagram of the distribution of bending moment in the main framewhen it is under the action of the most unfavorable combination of vertical loads.

9. The estimation of the internal forces of cross section under the action of themost unfavorable combination of horizontal loads

(1) The most unfavorable combination of horizontal loads on the 4th floorThe calculation of wind load:

hw ¼ 1:4� 0:55 kN=m2 � 12 m ¼ 9:24 kN=m

Fig. B.3 Diagram of vertical-load-induced distribution of the bending moment in the main frame

Appendix B 391

And the internal forces at the foot of the column are

Qmax ¼ 9:24� 5 1=6ð Þ ¼ 7:7 kN

Mmax ¼ 7:7� 2:5 m ¼ 19:25 kNm

The simplified calculation of seismic load: 7° seismic fortification intensityand total seismic-load-induced horizontal force on the 4th floor is

H4 ¼ 0:1� 60� 60� 1:1 ¼ 396 kN

Since there are 36 columns, the shear force shared by each column wouldbe

Qmax ¼ 396=36 ¼ 11 kN

Mmax ¼ 11� 2:5 m ¼ 27:5 kNm

(2) The most unfavorable combination of horizontal loads on the 3rd floorThe calculation of wind load:

hw ¼ 1:4� 0:55 kN=m2 � 12 m ¼ 9:24 kN=m


Qmax ¼ 9:24� 2� 5 1=6ð Þ ¼ 15:4 kN

Mmax ¼ 15:4� 2:5 m ¼ 38:5 kNm

The simplified calculation of seismic load: 7° seismic fortification intensityand total seismic-load-induced horizontal force on the 3rd floor is

H3 ¼ 0:1� 60� 60� 1:1þ 0:1� 60� 60� 11:12 ¼ 4399:2 kN

Since there are 36 columns, the shear force shared by each column wouldbe

Qmax ¼ 4399:2=36 ¼ 122:2 kN

Mmax ¼ 122:2� 2:5 m ¼ 305:5 kNm

(3) The most unfavorable combination of horizontal loads on the 2nd floorThe calculation of wind load:

hw ¼ 1:4� 0:55 kN=m2 � 12 m ¼ 9:24 kN=m

392 Appendix B


Qmax ¼ 9:24� 3� 5 1=6ð Þ ¼ 23:1 kN

Mmax ¼ 23:1� 2:5m ¼ 57:75 kNm

The simplified calculation of seismic load: 7° seismic fortification intensityand total seismic-load-induced horizontal force on the 2nd floor is

H2 ¼ 0:1� 60� 60� 1:1þ 0:1� 60� 60� 11:12þ 0:1� 60� 60� 14:37

¼ 9572:4 kN

Since there are 36 columns, the shear force shared by each column would be

Qmax ¼ 9572:4=36 ¼ 265:9 kN

Mmax ¼ 265:9� 2:5 m ¼ 664:75 kNm

(4) The most unfavorable combination of horizontal loads on the 1st floorThe calculation of wind load:

hw ¼ 1:4� 0:55 kN=m2 � 12 m ¼ 9:24 kN=m


Qmax ¼ 9:24 8þ 5þ 5þ 5ð Þ 1=6ð Þ ¼ 35:42 kN

Mmax ¼ 35:42� 4 m ¼ 141:68 kNm

The simplified calculation of seismic load: 7° seismic fortification intensityand total seismic-load-induced horizontal force on the 1st floor is

H1 ¼ 0:1� 60� 60� 1:1þ 0:1� 60� 60� 11:12þ 0:1� 60� 60� 14:37� 2

¼ 14;745:6 kN

Since there are 36 columns, the shear force shared by each column would be

Qmax ¼ 14; 745:6=36 ¼ 409:6 kN

Mmax ¼ 409:6� 4 m ¼ 1638:4 kNm

Figure B.4 is a diagram of the distribution of bending moment in the main frameunder the action of the most unfavorable combination of horizontal loads (seismicload is the control load).

Appendix B 393

10. Estimation and selection of sections for the beams and columns in the mainframe

From the calculation above, the combination of loads affecting first floor beams andcolumns should be the control forces.

As for the primary beams above the first floor,

Mmax ¼ 2069:28 kNm

Qmax ¼ 1034:64 kN

As for the secondary beams above the first floor,

M0max ¼ 388:08 kNm

Q0max ¼ 129:36 kN

As for the columns on the first floor,

Mmax ¼ 1638:4 kNm

Qmax ¼ 409:6 kN

Nmax ¼ 5893:92 kN

Symmetrical wide-flanged I-shaped section H800 × 400 × 12 × 20 is adopted forthe primary beams above the ground floor;

Symmetrical wide-flanged I-shaped section H380 × 280 × 8 × 16 is adopted forthe secondary beams;

As for the ground floor columns, box-shaped section 560 × 560 × 16 × 16 isadopted and the thickness of the box-shaped section is gradually decreased from thesecond floor to the top.

(1) The checking of the bending momentAssume steel Q345, strength of design being 310 MPa, is employed in themain frame.

Fig. B.4 Distribution of horizontal-load-induced bending moment in the main frame

394 Appendix B

Then, the sectional modulus of the selected section for the primary beamsshould meet the requirement of

W [Mmax=310 ¼ 2;069;280;000=310 ¼ 6;675;100 mm3

The moment of area of the primary beams shall be

W ¼ 2� 400� 20� 400þ 12� 800� 800=6¼ 7;680;000 mm3 [ 6;675;100 mm3

OK!

The moment of area of the secondary beams shall be

W 0 ¼ 2� 280� 16� 190þ 8� 380� 380=6 ¼ 1;895;000 mm3

W 0 [ 388;080;000=310 ¼ 1;252;000 mm3 OK!

And the moment of area of the columns shall be

W 00 ¼ 2� 560� 16� 280þ 2� 16� 560� 560=6 ¼ 6;690;000 mm3

W 00 [ 1;638;400;000=310 ¼ 5;290;000 mm3 OK!

(2) The checking of the shear forceThe maximum shear stress on the section of primary beams is

smax ¼ 15� 1034:64 kN= 12� 800ð Þ ¼ 162 MPa\½s� ¼ 180 MPa OK!

The maximum shear stress on the section of secondary beams is

smax ¼ 1:5� 129:36 kN= 8� 380ð Þ ¼ 67:4 MPa\½s� ¼ 180 MPa OK!

The maximum shear stress on the section of columns is

smax ¼ 1:5� 409:6 kN= 2� 16� 560ð Þ ¼ 34:3 MPa\½s� ¼ 180 MPa OK!

(3) The checking of axial compressionThe maximum axial stress on the section of columns is

rmax ¼ 5893:62 kN= 4� 16� 560ð Þ ¼ 164 MPa\ r½ � ¼ 310 MPa OK!

(4) The checking of vertical deformation of primary beamsSectional moment of inertia is first estimated.

Appendix B 395

Ix ¼ 2� 400� 20� 400� 400þ 12� 8003=12 ¼ 3;072;000;000 mm4

The combination of loads selected for the checking of deformation (dif-ferent from that for the checking of strength) is 1.0 deadweight + 0.5overhanging load + 0.5 live load on the floor

w ¼ 3:5þ 0:5� 0:3þ 0:5� 7:5 ¼ 7:4 kN=m2

q ¼ 12 m� 7:4 kN=m2 ¼ 88:8 kN=m

The maximum vertical deformation of primary beams is

Db ¼ 1=384ð Þ88:8 kN/m� 124 m4= 2:06� 105 N/mm2 � 3;072;000;000 mm4� �

¼ 0:008 m\L=400 ¼ 0:03 m OK!

The same method could be applied to the checking of the maximum verticaldeformation of secondary beams, which by the way meets the requirementof design, and readers could conduct the checking to see for yourselves.With reference to Appendix A and Chap. 3 of this book, lateral dis-placement of columns could also be checked.

11. The estimation of steel consumption in the main rigid frame

Steel employed in each frame in factory building C

1:10� 0:8 16� 560� 4ð Þ10�9 � 7850� 23� 6þ 12� 800þ 2� 20� 400ð Þ10�9 � 7850� 12� 20�

þ 8� 380þ 2� 16� 280ð Þ10�9 � 7850� 12� 6� 20Þ� ¼ 196 T

In the equation above, coefficient 1.1 is for the steel needed by the connectingplates and the coefficient 0.8 is for the gradually reduced load with the elevation ofstory.

The average unit-employed steel is

196;000= 60� 12� 4ð Þ ¼ 68:06 kg/m2

And the total steel employed in the main rigid frame in factory building C is

1:05� 68:06� 60� 60� 4 ¼ 1029 T

In the equation above, coefficient 1.05 has considered the extra steel employed inall diagonal braces in the main rigid frames.

12. The simplified design and calculation of the supporting ground andfoundation

From the calculation conducted with the most unfavorable combination of loads wecould have the most unfavorable internal forces of the central columns on theground floor.

396 Appendix B

http://dx.doi.org/10.1007/978-3-662-48044-1_3

Mmax ¼ 1638 kNm

Qmax ¼ 410 kN

Nmax ¼ 5894 kN

According to the results of proprietor-provided preliminary survey of the sup-porting ground and taking into consideration the bearing capacity of the supportingground and a rational form of force bearing, an independent below-column pilefoundation is adopted, as shown in Fig. B.5. The independent pile foundation belowthe central column consists of a square pile cap and nine prestressed prefabricatedsupporting piles. The diameter is 1200 mm; the length of each side is no less than3 m; and the thickness of the cap is no less than 1000 mm. According to thedistribution of soil layers of the supporting ground, the length of the piles shall be14 m (Ra = 810 kN).

The checking of the bending moment and compression of the cap below centralcolumn

Mx ¼ My ¼ 3Nmax1:6 ¼ 3� 810� 1:6 ¼ 3888 kNm[ 1638 kNm OK!

Nmax ¼ 9� 810 kN ¼ 7290[ 5894 kN OK!

13. Simplified design and calculation of floor

According to the soil conditions of the supporting ground and a strict restrictionproposed by the proprietor upon the settlement of the floor, piles are needed in thesupporting ground below the floor. Prestressed prefabricated piles with 300 mmdiameter, 8 m length and 2.5 m × 2.5 m spacing could meet the requirement.

Fig. B.5 Diagram of the below-column pile foundation and the cap

Appendix B 397

The checking of vertical loads

30� 2:5� 2:5ð Þ ¼ 188 kN\Ra ¼ 259 kN OK!

C30 reinforced concrete floor with a thickness of 300 mm is preselected, asshown in Fig. B.6.

The design yields two-way double-layered II reinforcements with 16 mmdiameter and 150 mm spacing for the plate strip above the column and those with12 mm diameter and 200 mm spacing for the mid-span plate strip. The average steelemployed could be calculated as (42.13 + 17.8)/2 = 30 kg/m2.

14. Summary of design

Conceptual concepts adopted in the design of this project are summarized asfollows:

(1) The design of the main rigid frame is the key in this project. The estimationof loads, the simplification of calculation model, and the consideration ofthe most unfavorable combination of loads are to be conducted with carefulattention. Afterward, practical construction experiences and simplifiedmethods of calculation introduced in previous chapters could be referred toin the design and checking calculation. Figure B.7 is the diagram of selectedsections for the main rigid frame. With the elevation of story and thedecrease of load, the steel employed could be gradually reduced.

Fig. B.6 Diagram of floor

560 10 10

560 12 12

560 14 14

560 16 16

H 0 400 12 20

Box

Box

Box

Box

Fig. B.7 Diagram of selected sections for the main rigid frame

398 Appendix B

(2) Unit steel employed in themain frame of this factory building is 68.06 kg/m2,on the basis ofwhich the total steel employed in themain steel structure couldbe derived as 1029 t along with the probable cost of construction.

(3) The design of the supporting ground and foundation could determine thetype of the pile foundation as well as the length and type of the pile. Andthis is of importance for the estimation of cost of construction concerningthis part of the structure.

(4) A simple design of floor could determine whether or not to adopt pile, thetechnique entailed in the construction of floor and the estimation of rein-forcements to be installed within. At the completion of this process, thecost of construction of this part could be estimated.

Appendix B 399

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