Struc lec. no. 1

Theory of Structures(1) Lecture No. 1

Loads Reactions Supports Conditional Equations Stability and Determinancy Chapter 1 : Loads and Reactions

Chapter 2 : Statically Determinate Beams Types of Beams Normal Force, Shearing Force, and Bending Moment Relationship between Loads, Shearing Force and Bending Moment Standard Cases of S.F. and B.M.D(s) for Beams Principal Of Superposition Inclined Beams A Beam : is a structural member subjected to some external forces.

Chapter 3 : Statically Determinate Rigid Frames Internal Stability and Determinacy Method of Frames Analysis A Frame : is a structure composed of a number of members connected together by joints all or some are rigid.

Chapter 4 : Statically Determinate Arches Reactions , Thrust, Shearing Force and Bending Moment using Analytical Method Reactions Thrust, Shearing Force, and Bending Moment at any Point in the Arch using Graphical Method An Arch :is a curved beam or two curved beams connected together by intermediate hinge.

Chapter 5 : Statically Determinate Trusses Stability and Determinacy Analysis for Trusses: Methods of: joints, Sections, and Stress Diagram A truss : Consists of a number of straight members pin- connected together and subjected to concentrated loads at hinges.

Chapter 1 Loads and Reactions Loads classified according to: Cause of Loading : dead ,live, and lateral Shape of Load : concen. and distrib. Rate of application : static and dynamic Reactions: resistance by supports to counteract the action of loads Supports : Roller, Hinge, Fixed, and Link

Determination of the Reactions: ” External loads acting on the structure together with the reactions at the supports must constitute a system of non concurrent forces in equilibrium” ∑ x = 0 (sum of horizontal forces) = 0 ∑ y = 0 (sum of vertical forces) = 0 ∑ M = 0 (sum of moments ) = 0

Y B X A Y A Solution of Example 1 1- ∑ X = 0 XA – 5 = 0 XA = 5t 2- ∑MB = 0 YA × 6 – 2 × 6 × 3 + (4 × 3)/2 × 1 = 0 YA= 5t 3- ∑ Y = 0 YA + YB = 2 × 6+ (4 × 3)/2=18 YB = 13t

Solution of Example 2 1- ∑ X = 0 X A = 3t 2- ∑ M B = 0 Y A × 4 + 8 + 4 × 1 – 3 × 1= 0 Y A = -2.25t ↑ Y A =2.25t ↓ 3- ∑ y = 0 Y A + 2× 2 = Y B 2.25+ 4 = Y B Y B = 6.25 t ↑ Y A X A Y B

Solution of Example 3 1- ∑ X = 0 XB – 4 = 0 XB = 4t 2- ∑ MA = 0 10YB – 4 × 12 – 2 × 9 – 2 × 5× 2.5 – 4 × 3= 0 YB = 10.3 t 3- ∑ Y = 0 YA + YB = 4 + 2 + 10 = 16 YA = 5.7 t Y B Y A X B

Solution of Example 4 1- ∑ X = 0 XA – 4 = 0 XA = 4t 2- ∑ Y = 0 YA – 1 × 6 – 5 – 5 – 2.5 = 0 YA = 18.5 t 3- ∑ M / A = 0 MA – 5 × 2 – 5 × 4 – 2.5 ×6 - 1×6×3 = 0 M A = 63 t.m

Solution of Example 5 R 1 = 1/2 × 4 × 8 = 16 t R 2 = 0.5 × 4 = 2 t 1- ∑ X = 0 X A – R 2 = 0 X A = 2t 2- ∑ M A = 0 11 Y B + 2 × 13 – R 2 × 2 – 2 × 1 – R 1 × 7= 0 Y B = 8.36 t 3- ∑ Y = 0 Y A + Y B + 2 – 2 – 16 = 0 YA= 7.64t

Solution of Example 6 1- ∑ X = 0 X B – 20 = 0 X B = 20t 2- ∑ M A = 0 6 Y B + 20 × 3 – 10 × 6 – 20 × 6 = 0  Y B = 20t 3- ∑Y = 0 Y A + Y B = 20 + 30 + 10 = 60  Y A = 40t

Solution of Example 7 ∑ MO = 0 3 × 0 + 8 × 1 + 1 × 3 – Rd × 5 = 0 Rd = 2.2 t ↑ ∑ Mp = 0 1 × 2 + 8 × 4 + 3 × 5 – Rb × 5 = 0 Rb = 6.93t ∑ Mq = 0 1 × 2 + 8 × 4 + 3 × 5 – Rc × 5 = 0 Rc = 6.93t 8t

Conditional Equations Some structures are divided to several parts connected together by hinges, links or rollers. Forces are transmitted through these connections from one part to the others.

Solution of Example 8 Method I 1- Part FD 1) ∑ MD = 0 8 × 2 = 4 YF YF = 4t 2) ∑Y = 0 YD = 8 – 4 YD = 4t 2- Part ECF 1) ∑ ME = 0 10×2.5+4×5= 4YC YC= 11.25t 2) ∑Y = 0 YE=10+4-11.25 YE=2.75t 3- Part ABE 1) ∑MA=0 12×3+2.75×6=5YB YB=10.5t 2) ∑Y = 0 YA= 2.75 + 12-10.5 YA= 4.25t

Solution of Example 8 “ Continued ” Method II 1)∑MF(right)= 0 8 × 2 – 4 YD = 0 YD = 4t 2) ∑ME(right)= 0 18×4.5 – 4×9 – 4YC= 0 YC=11.25t 3) ∑MA = 0 30×7.5 – 4×15 – 11.25× 10 - 5YB = 0 YB=10.5t 4) ∑Y = 0 30 – 4 – 11.25 – 10.5 – YA=0 YA=4.25t

= 0 = X d × 8 – 8 × 2 X d = 2 t  = 8 ×12 + 2.5 × 12 × 6 Xd - 2 × 8 - 10 ×Y d =0 Y d = 26 t ∑ Y = 0 = Y a + 26 – 2.5 × 12 – 8 Y a = 30 + 8 – 26 = 12 t. ∑ X = 0 = X a – 0.5 × 8 – 2 Xa = 4 + 2 = 6 t. ← Solution of Example 9

= 0 = Ma + 0.5 × 8 × 4 – 6 × 8 M a = 48 – 16 = 32 m.t (anticlockwise) Check : ∑Mc for part ac Xa Ya 32 + 0.5 × 8 × 4 + 2.5 × 12 × 4 – 6 × 8 – 12 × 10 = 32 + 16 + 120 – 48 – 120 = 0 Solution of Example 9 “Continued”

1) ∑MF(right) = 0 RC cosα × 4.5 – 4 × 4.5 × 2.25 – 10 × 10.5 = 0 RC = 40.4 t 2) ∑MA = 0 YB×9 + XB × 2 + RC cosα × 18 + RC sinα × 8 –10×24 - 20×4.5 – 5 × 5 – 4 × 9 × 13.5 + 5 × 2 = 0 4.5 YB = 27.57 – XB (1) 3) ∑ME (right) = 0 4.5 YB - 6XB + RC cosα × 13.5 – 4 × 9 × 9 - 10 × 19.5 = 0 4.5 YB = 6XB + 82. 545 (2) Solution of Example 10

By solving equations (1) & (2) we get YB = 7.87 t. XB= - 7.85t 4) ∑Y = 0 YA = 36 + 10 + 20 + 5 – 32.33 – 7.87 YA = 30.8 t. 5) ∑ME(left) = 0 4.5 YA – 8 XA – 5 × 3 – 5 × 6.5 = 0 XA = 11.4 t. Solution of Example 10 “Continued”

Stability and Determinancy External concerned with the reactions. Internal concerned with the internal forces and moments. A Stable Structure is one that support : a system of loads. These loads together with the support reactions have to be in equilibrium and satisfy the three equilibrium equations in addition to conditional equations (if any). Stability of Structures Statically Unstable Structure is one in which: the number of the unknown reactions is less than the sum of the available equilibrium and conditional equations (if any).

Externally Statically Unstable Structures Geometrical Unstable Structures

Statically Determinate Structure is one in which t he reactions can be determined by application of the equations of equilibrium in addition to conditional equations(if any). Determinancy of Structures Statically Indeterminate Structure is one in which the number of unknown reactions is more than the number of available equations Stable-Once Statically Indeterminate Stable-Three Times Statically Indeterminate

Check the stability and determinacy for the following structures. Then, show how to modify it for stability and determinancy (a) (b) (c) (d) Solution of Example 11

Structure (a) : Is unstable because there are only two unknown reactions. The modification : Change either of the two roller supports to a hinge.

Structure (b) : Is unstable because the beam is rested on four link members connected at a point. The modification : Separate the link members Modification for structure (b)

Structure (C) : Is stable and three times statically indeterminate because there are six unknown support reactions. The modification : Add three intermediate hinges Modification for structure (C)

Structure (D) : Is stable and five times statically indeterminate because there are 8 unknown support reactions The modification : One of fixed supports is changed to a roller In addition to three intermediate hinges arranged as given in Struc. (c).

Struc lec. no. 1

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Struc lec. no. 1