CIV100 Mechanics. Module 5: Internal Forces and Design. by: Jinyue Zhang. By the end of this Module you should be able to:

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CIV100 Mechanics Module 5: Internal Forces and Design by: Jinyue Zhang Module Objective By the end of this Module you should be able to: Find internal forces of any structural members Understand how

1 CIV100 Mechanics Module 5: Internal Forces and Design by: Jinyue Zhang Module Objective By the end of this Module you should be able to: Find internal forces of any structural members Understand how Shear Force Diagrams (SFD) and Bending Moment Diagrams (BMD) relate to the internal forces in structures Know how to draw SFD for beams Know how to draw BMD for beams Understand how different load types (concentrated load, distributed load, couple) affect the shape of SFD and BMD Understand the design of structural members in tension Understand the design of structural members in bending 2010/10/26 1 1

2 Today s Objective Understand the concept of internal forces How to find internal forces of a structural member Understand the concept of SFD and BFD How to draw SFD and BFD of a beam with simple loading 2010/10/26 2 Internal Forces We have learnt this concept! 2010/10/26 3 2

3 Internal Forces Now let s look at multi-force member! Fx = 0 Fy = 0 Mo = /10/26 4 Internal Forces Naming those internal forces 2010/10/26 5 3

4 Finding Internal Forces Procedure for Analysis Solve the whole structure first, find all reaction forces if necessary. For frames and machines or their combinations, dismember the structure and obtain the reactions at each connection. Keep all distributed loadings, moments, and forces acting on the member in their exact location, then cut the member to analysis by an imaginary section at the point where the internal loading is to be determined. Draw FBD of the segment that has the least number of loads and apply EoE to find internal forces: Make assumption on unknown forces about their sense, if the EoE yields a negative scalar, your assumption was wrong. Moment should be summed at the cutting section where we eliminate the unknown normal and share forces. 2010/10/26 6 Example 1 The bar is fixed at its end and is loaded as shown. Determine the internal normal force at points B and C. 2010/10/26 7 4

5 Example 1 The bar is fixed at its end and is loaded as shown. Determine the internal normal force at points B and C. 2010/10/26 8 Example 2 Determine the internal normal force, shear force, and bending moment at point B. 2010/10/26 9 5

2010/10/26 10 Today s Objective What is Bending Moment Diagram (BMD)

6 Example 2 Determine the internal normal force, shear force, and bending moment at point B. 2010/10/26 10 Today s Objective What is Bending Moment Diagram (BMD) What is Shear Force Diagram (SFD) Understand the sign convention of BM and SF 2 examples of plotting SFD and BMD 2010/10/

Moments In North America, if the moment tends

positive; if the moment tends to cause it to

If the upper side is in compression, it is

7 SFD and BMD An Interesting Applet! 2010/10/26 12 Sign Convention for Bending Moments In North America, if the moment tends to cause the beam to curve upward it is positive; if the moment tends to cause it to curve downward it is negative. If the upper side is in compression, it is positive. If it holds water, it is positive. If it smiles, it is positive. 2010/10/

8 Sign Convention for Shear Forces The only rule: if the shear force causes clockwise rotation of the member on which it acts, it a positive shear force. 2010/10/26 14 Example 1 Draw the shear and moment diagrams for the cantilevered beam. 2010/10/

9 Example 1 Draw the shear and moment diagrams for the cantilevered beam. Check you lecture notes to see how we develop the SFD and BMD. 2010/10/26 16 Example 2 Sketch the shear and moment diagrams for the beam. 2010/10/

10 Example 2 Sketch the shear and moment diagrams for the beam. Check you lecture notes to see how we develop the SFD and BMD. 2010/10/26 18 Today s Objective Understand the relation between loads and SFD/BMD Two examples 2010/10/

Loading, Shear Force, and Bending Moment

area under the distributed loading curve

dm/dx=0 2010/10/26 20 Load V M 2010/10/26

11 Loading, Shear Force, and Bending Moment The slop is equal to the shear force. The change of the shear ΔV between two points is equal to the negative of the area under the distributed loading curve between the points. (if downward is +) The change of the moment ΔM between two points is equal to the negative of the area under the shear diagram between the points. Points of zero shear represents points of Max/Min moment as dm/dx=0 2010/10/26 20 Loading, Shear Force, and Bending Moment Load V M 2010/10/

12 Example 1 Find internal forces at point C and then draw the shear and moment diagrams for the beam. 2010/10/26 22 Example 1 Find internal forces at point C and then draw the shear and moment diagrams for the beam. To solve the reactions at A and B M B =0 -(Ay)(6) + (8)(4.5) + (4.5)(2) + (6)(1.5) = 0 Ay = 9 kn M A =0 -(8)(1.5) - (4.5)(4) - (6)(4.5) +(By)(6) = 0 By = 9.5 kn Fx=0 Bx = 6 kn Check answers Fy= =0 To solve internal forces at C M C =0 Mc - (1.125)(0.5) - (3)(0.75) + (9.5)(1.5) = 0 Mc = kn.m M B =0 - (Cy)(1.5) + (1.125)(1) + (3)(0.75) = 0 Cy = kn.m Fx=0 Cx = 6 kn Check answers Fy= =0 2010/10/

2010/10/26 24 Simply Supported Beam with UDL To Remember 1 M wl 2 Max = 8

13 Today s Objective Examples to review rules of drawing SFD and BMD Simple supported beam with uniformly distributed loading More complex examples 2010/10/26 24 Simply Supported Beam with UDL To Remember 1 M wl 2 Max = 8 The point where SF is zero means peak BM, because M/ x= /10/

14 Example 1 Draw SFD and BMD of the given beam. 2010/10/26 26 Example /10/

15 Example 2 Draw SFD, BMD, and NFD (normal force diagram) of the two members. 2010/10/26 28 Example 2 Draw SFD, BMD, and NFD (normal force diagram) of the two members. 2010/10/

16 Example 2 Draw SFD, BMD, and NFD (normal force diagram) of the two members /10/26 30 Today s Objective More examples on SFD and BMD 2010/10/

17 Example 1 A foundation beam supports four columns as shown. Knowing that the weight of each column has already been included in the load shown on the column, and the foundation beam has a weight of 30kN/m. Assuming the reaction of the ground is uniformly distributed, draw the SFD and BMD of the foundation beam. 2010/10/26 32 Example 1 A foundation beam supports four columns as shown. Knowing that the weight of each column has already been included in the load shown on the column, and the foundation beam has a weight of 30kN/m. Assuming the reaction of the ground is uniformly distributed, draw SFD/BMD of the foundation beam. 2010/10/

18 More Examples kN kN kN kN 2010/10/26 34 More Examples kN kN kN kN 2010/10/

19 Today s Objective Understand the following concepts Stress Strain Understand Hooke s Law The relationship between axial stress and axial strain Understand the stress-strain diagram The behaviour of ductile material The behaviour of brittle material Be able to design a structural member in tension Understand the concept of Load Factor (LF) 2010/10/26 36 Axial Stress Fcosɵ Fsinɵ ɵ F 2010/10/

Axial Stress Fcosɵ Fsinɵ ɵ F Axial Stress 2010/10/26

σ=p/a Units are the same units of pressure (kpa,

20 Axial Stress Fcosɵ Fsinɵ ɵ F Axial Stress 2010/10/26 38 Axial Stress Caused by an axial force. Can be tensile stress or compressive stress. Symbol used σ (sigma) Axial stress = Force / Area σ=p/a Units are the same units of pressure (kpa, MPa, etc ) To Remember P σ = A 1N 1Pa = 2 1m 1N 1MPa = 2 1mm 2010/10/

21 Axial Stress This equation tells us that: The greater the applied force the greater the stress. The smaller the area on which the force is applied the greater the stress. To Remember P σ = A 1N 1Pa = 2 1m 1N 1MPa = 2 1mm 2010/10/26 40 Axial Strain Stresses cause the material to deform: Tension will cause elongation Compression will cause shrinkage Symbol used ε (epsilon) Axial Strain = Change in length / Original length ε = ΔL / L (no units) To Remember ΔL ε = L 2010/10/

22 Hooke s Law Robert Hooke s fundamental discovery (in 1676): for a spring, Axial Force Axial Deformation Mathematically it is A P ΔL P ΔL P ΔL σ ε σ = E ε L A L To Remember σ = E ε modulus of elasticity Young s modulus of elasticity Young s modulus Unit of Young s modulus: same as stress, Pa or MPa Relates axial stress to axial strain The modulus of elasticity is a property of material (each material has a different E ). The modulus of elasticity tells you how stiff a material is. The larger E means a stiffer material. 2010/10/26 42 Stress-Strain Curve: Ductile Material Typical ductile material in construction: low-carbon steel Hooke s Law is only good from point A to point B Slop of line AB is the Young s modulus In design practice, yield stress is a key parameter we frequently use We normally put a load factor to make it much safer 2010/10/

23 Stress-Strain Curve: Brittle Material Typical brittle material in construction: cast iron, concrete, rock Hooke s Law is only good at very beginning of the curve when the stress is small Suddenly rupture without plastic region (yielding stage) Capacity of resisting compression >>>>>>>>>>>> Capacity of resisting tension 2010/10/26 44 What is Design? Design is to select a system (or a part of a system like a structural member) to safely resist the anticipated loads. In order to do a design, you need to analyze a given structure: Understand the structure and its overall geometry Know the external loads Know the material used to build the structure Steps to design: Analyze the structure and find the internal forces Look at the available material and find its strength Select a solution to make sure the strength is greater than the load effects (the internal forces caused by external loads) There maybe many feasible solutions, you need to think other criteria like cost or aesthetics. 2010/10/

24 Load Factor To make a structure safe, we need: Maximum Load Stress < Yield Stress for the Material Is This Good Enough? Why? Loading might be greater then your estimate Material might not be as strong as specified Structure analysis might be an approximation or simplification Errors and mistakes To provide reasonable level of safety, we introduce a concept Load Factor The load factor (LF) enlarge the effect of the anticipated or service loads, and thus requires the structure to have greater capacity. For example, if a member will have a tensile stress of 10 MPa, after applying a LF of 1.7, we design the member as it would have a tensile stress of 10 MPa x 1.7 = 17 MPa 2010/10/26 46 Example 1 A Standard channel (C200x21) of 2m length is subject to a tension force of 500kN. Given that the modulus of elasticity for steel E= 200x10 3 MPa. Calculate the elongation in the steel member. 2010/10/

25 Example 1 A Standard channel (C200x21) of 2m length is subject to a tension force of 500kN. Given that the modulus of elasticity for steel E= 200x10 3 MPa. Calculate the elongation in the steel member. From tables: Cross Sectional Area = 2600mm 2 σ = (500 x 1000) / 2600 = MPa (or N/mm 2 ) ε = σ / E = / = x 10-4 L = L x ε = 2000 x x 10-4 = mm 2010/10/26 48 Example 2 Design member AC of the truss. Structural Steel is to be used in the design. The yield stress for the steel is 310 MPa, E=2x10 5 MPa, and the load factor is 2.0. The rectangular cross-section (a=3b) will be cut from plates whose thickness, b, is only available in 5 mm increments. 2010/10/

26 Example 2 Design member AC of the truss. Structural Steel is to be used in the design. The yield stress for the steel is 310 MPa, E=2x10 5 MPa, and the load factor is 2.0. The rectangular cross-section (a=3b) will be cut from plates whose thickness, b, is only available in 5 mm increments. 2010/10/26 50 Today s Objective No buckling, ignore buckling issue in complementary notes! The tensile strength (yielding stress due to tension) may be different to its compressive strength (yielding stress due to compression). When design against axial forces, use the right yielding stress to calculate. Understand the relationship Between load, pressure, and stress block Between stress block and internal bending moment Calculate internal BM according to a given stress block Examples 2010/10/

27 Axial Stress in Bars We have studied stress in a two force member which is in either tension or compression. σ = P A We call it stress block 2010/10/26 52 Load, Pressure, and Stress External Load vs. Pressure Internal Forces vs. Stress A distributed load acting over an area can be visualized as a volume: (a) The volume is equal to the magnitude of the equivalent single force, and (b) The line of action of the equivalent single force passes through the centroid off the volume. H/2 H 2010/10/

28 Calculating Equivalent Force 2010/10/26 54 Calculating Equivalent Force 2010/10/

29 Stress Block and Internal BM When we say there is a internal BM at a certain section of a beam, what does it mean? How can the section have a BM? Who provides the moment? 2010/10/26 56 Example The stress blocks due to the internal bending moment at Q on QA of the shown beam is given in figure (b). Determine: 1. The total compression force due to the bending moment. 2. The internal bending moment at Q. 3. The value of the concentrated load P. 2010/10/

30 Example The stress blocks due to the internal bending moment at Q on QA of the shown beam is given in figure (b). Determine: 1. The total compression force due to the bending moment. 2. The internal bending moment at Q. 3. The value of the concentrated load P. 2010/10/26 58 Today s Objective More examples on stress blocks 2010/10/

Example 1 This section is used to fabricate a simply supported steel beam. The stress distribution due to the bending moment M in a certain section is shown in the diagram.

31 Example 1 This section is used to fabricate a simply supported steel beam. The stress distribution due to the bending moment M in a certain section is shown in the diagram. Determine the total tension force and compression force, as well as the bending moment M. 2010/10/26 60 Example 1 This section is used to fabricate a simply supported steel beam. The stress distribution due to the bending moment M in a certain section is shown in the diagram. Determine the total tension force and compression force, as well as the bending moment M. Step-1: Divide the stress distribution into 3 distinct shapes (2 triangles and one rectangle) as shown. Step-2: Find the values of the forces: F1= (1/2) (66.5)(2x9.5)=118848N = kN F2 = (102)(9.5)=182293N = kn F3 = (1/2)( )(9.5)(102)=13021N = kn The volume of the compression stress block is the total compress force which is F1 + F2 + F3 = 314 kn The total tension force is same as compression force because of the symmetry of the section. Step-3: Find the moment caused by these forces (note these forces are equal and opposite thus causing a couple): M = F1(2*44.33) + F2(2*71.25) +F3(2* ) = kn.mm = 38.4 kn.m 2010/10/

32 Example 2 The cross section of a reinforced concrete beam is shown below along with the assumed distribution of the compression stress due to a bending moment M (the beam is subject to bending moment only). Determine: (a) The value of the compression force acting on the cross section (b) The tension stress in the steel bars (c) The bending moment M 2010/10/26 62 Example 2 The cross section of a reinforced concrete beam is shown below along with the assumed distribution of the compression stress due to a bending moment M (the beam is subject to bending moment only). Determine: (a) The value of the compression force acting on the cross section (b) The tension stress in the steel bars (c) The bending moment M (a) The value of compression force is the volume of stress block measured 110mm from the top over the entire width of the beam. The volume is further divided into two components: one cube plus one wedge. For the cube: F1=(22MPa)(110mm)(250mm)=605kN For the wedge: F2=(30MPa-22MPa)(110mm)(0.5)(250mm)=110kN Total compression force is F1+F2=715kN (b) The tension force is equal to the compression force because the beam is subject to bending moment only, i.e. no normal force, so the total tension force is 715kN. The total area of the four reinforce bars A=4πr 2 =4(3.14)(15) 2 =2827mm 2 Tension stress σ=p/a=715000n/2827mm 2 =252 MPa (c) When we calculate the centroidal axis on the composite section, we ignore the impact of reinforce bars, i.e. we still trade the entire section is made by homogeneous material because the area of the steel bars is too small compared to the concrete area and thus is not significant. Then the centroidal axis is at the half height. M due to F1 = (605kN)(250mm-110mm/2)=117.98kNm M due to F2 = (110kN)(250mm-110mm/3)=23.467kNm 250mm 195mm 213mm M due to tension = (715kN)(250mm-50mm)=143kNm Total Moment = =284kNm 200mm 2010/10/

33 Today s Objective Understand the stress due to bending moment Three fundamental assumptions Bending stress formula, i.e. how to calculate bending stress Design a beam subject to bending moment 2010/10/26 64 Application We have leant How to calculate the bending moment of a section (the internal force) What is the bending stress How they are related In many circumstances we need To design a structural member (for example a beam) to hold the loads As such, we need Analyze the internal forces, i.e. draw SFD, BMD, and NFD Find the critical section Design against NF, BM, and SF: σ factored loads < σ yield 2010/10/

Bending stresses are independent of the stresses caused by internal axial and shear forces in the beam.

34 Three Fundamental Assumptions The material is linearly elastic, i.e. stress is proportional to strain. σ ε σ = E ε Plane sections before bending remain plane during bending. Bending stresses are independent of the stresses caused by internal axial and shear forces in the beam. 2010/10/26 66 Bending Stress Formula Let s look at a bending element isolated from a beam y z X σ = To Remember M I Z Z y σ ε ε y z σ y σ = K y M σ = I 2010/10/26 67 Z Z y 34

35 Design against Bending Moment Where is the maximum stress due to bending, if we can assume the beam is uniform (prismatic)? At the location of the maximum bending moment. This can be determined by checking BMD. At the location in the cross section where y is maximum, i.e. in the extreme top and bottom fibres. σ M LF yield I Section Modulus S Unit is mm 3 S I Z Z = y max max y max Z S σ = S To Remember M I Z Z I y Z Z = y max required M = σ max yield LF S required = I y Z max M = σ max yield LF 2010/10/26 68 Example /10/

36 Example /10/26 70 Example 2 The wooden beam has a cross section (a x 3a). The tensile strength of wood is 12MPa and compressive strength of wood is 15MPa. Using a LF of 2 to design the beam section. 2010/10/

Example 2 The wooden beam has a cross section (a x 3a, increments of a = 10mm). The tensile strength of wood is 12MPa and compressive strength of wood is 15MPa.

37 Example 2 The wooden beam has a cross section (a x 3a, increments of a = 10mm). The tensile strength of wood is 12MPa and compressive strength of wood is 15MPa. Using a LF of 2 to design the beam section. 2010/10/26 72 Be Prepared Final Exam: Monday, Dec. 8, 2008, 9:30am Previous final papers fall term and solution: posted! Earlier exams and spring term exams are available at: then browse by code, select CIV, then CIV101 Quiz #2 and Tutorial #10 Quiz #2, this week, the first hour of tutorial class Office hour for Quiz #2: Wednesday GB224 Get them back on Wednesday, Dec. 3 rd, 3-5pm, out of GB224 Bring your questions about marking to the TA during office hours No resubmission for Tutorial #10, you will get 2 marks as long as your submit and complete Office hours before your final: Dec. 4-7, GB224 (i2c lab) Luai (Thursday, Dec. 4, who is marking your Quiz 2 and Tutorial 10) Ahmed (Saturday, Dec. 6) I will be available all 4 days 2010/10/

Example 1 For the shown beam, find the maximum bending moment. Select the least square section (a by a sq. mm) that can be used if the factor of safety is 1.3 and maximum stress is 280 MPa.

38 Example 1 For the shown beam, find the maximum bending moment. Select the least square section (a by a sq. mm) that can be used if the factor of safety is 1.3 and maximum stress is 280 MPa. (Neglect beam weight, don t check buckling, increments of a = 5mm) 30º 2010/10/26 76 Example 1 For the shown beam, find the maximum bending moment. Select the least square section (a by a sq. mm) that can be used if the factor of safety is 1.3 and maximum stress is 280 MPa. (Neglect beam weight, don t check buckling) 30º 2010/10/

39 Example 2 Choose a steel wideflange beam to take the shown loads. The yield stress is 290MPa and the load factor is 1.8. Neglecting the selfweight of the beam, and design criterion is to minimize weight. (a) Assuming no depth limits. (b) The depth is not to exceed 400mm 2010/10/26 78 Example /10/

40 Example 2 Choose a steel wide-flange beam to take the shown loads. The yield stress is 290MPa and the load factor is 1.8. Neglecting the self-weight of the beam, and design criterion is to minimize weight. (a) Assuming no depth limits. (b) The depth is not to exceed 400mm Step 1: Solving for reactions Taking moment about A, solve By=85.714kN Taking moment about B, solve Ay=54.286kN Step 2: Draw BMD 1 2 M = ( x + 2) 20( x )( x ) = 10x x M = 0 x = m x M = kN m Step 3: calculate section modulus S S 6 M maxlf (10 )(1.8) 6 required = = = (10 ) σ yield 290 Step 4: choose a section (a) W457x89 (b) W305x97 mm /10/26 80 More Examples 2010/10/

41 More Examples 2010/10/

Mechanics of Structure

S.Y. Diploma : Sem. III [CE/CS/CR/CV] Mechanics of Structure Time: Hrs.] Prelim Question Paper Solution [Marks : 70 Q.1(a) Attempt any SIX of the following. [1] Q.1(a) Define moment of Inertia. State MI