E2027 Structural nalysis I TYPES OF STRUUTRES H in ivil Engineering Page 1-1
E2027 Structural nalysis I SUPPORTS Pin or Hinge Support pin or hinge support is represented by the symbol H or H V V Prevented: llowed: Horizontal translation and vertical translation Rotation Roller Support roller support is represented by the symbol or V V Prevented: llowed: Vertical translation Horizontal translation and Rotation H in ivil Engineering Page 1-2
E2027 Structural nalysis I Fixed Support fixed support is represented by the symbol M H or H V V M Prevented: llowed: Horizontal translation, Vertical translation and Rotation None H in ivil Engineering Page 1-3
E2027 Structural nalysis I EQUILIRIUM OF STRUTURES structure is considered to be in equilibrium if it remains at rest when subjected to a system of forces and moments. If a structure is in equilibrium, then all its members and parts are also in equilibrium. For a structure to be in equilibrium, all the forces and moments (including support reactions) acting on it must balance each other. For a plane structure subjected to forces in its own plane, the conditions for equilibrium can be expressed by the following equations of equilibrium: F x 0, Fy = 0, M z = = 0 The third equation above states that the sum of moments of all forces about any point in the plane of the structure is zero. Equations of ondition Sometimes internal hinges are present within a structure. n internal hinge cannot transmit moment. Therefore the bending moment at a hinge is zero. The condition that the moment is zero at a hinge provides an additional equation for analyzing the structure. Such equations are commonly called equations of condition. (a) (b) (c) H in ivil Engineering Page 1-4
E2027 Structural nalysis I eterminate and Indeterminate Structures structure is externally determinate if the support reactions can all be obtained by statics, i.e. No. of Support Reactions = No. of Equations (incl. Equilibrium & onditions) structure which is not determinate is called indeterminate. egree of Statical Indetermincy = No. of Support Reactions No. of Equations FREE-OY IGRMS 1. Free-body diagrams make use of the concept that if a whole structure is in equilibrium, any part of it is also in equilibrium. 2. Free-body diagrams can be constructed for various parts of a structure, and also for the entire structure. 3. When drawing a free-body, it is important to indicate on it all possible forces acting in the given structure at the cuts. 4. Internal forces common to two free-bodies (on opposite sides of the cut ) should be denoted as equal in magnitude but opposite in direction. 5. Free-body iagrams are very useful in finding the support reactions and determining the internal forces in structures. The use of free-body diagram is an important tool in structural analysis and stress analysis. H in ivil Engineering Page 1-5
E2027 Structural nalysis I Original Structure P1 P2 P3 1 1 H Free-body diagram of the structure to the left of 1-1 V P1 P2 P3 M H V Free-body diagrams of the individual elements P3 H V P1 P2 V H V H M H V Note the equal but opposite directed representation of the connecting forces at. H in ivil Engineering Page 1-6
E2027 Structural nalysis I Original Structure ( three-hinge arch) Free-body iagram for the whole structure P1 P2 P1 a b a b P2 H L L H H V L L V H 1 P1 a H H V b P2 Free-body iagrams of arch segments H H V L 1 V L V H V 1 M 1 H 1 P1 H 1 a V H Free-body iagrams to analyze internal forces at section 1-1 H V V 1 H in ivil Engineering Page 1-7
E2027 Structural nalysis I Example 1 eam has a pinned support at and a roller support at. It carries two concentrated loads of 20 kn each and a uniformly distributed load of 4 kn/m over the right hand half as shown. etermine the reactions. 20 kn 20 kn 4 kn/m 3m 1.5m 1.5m 3m Solution: 20 kn 20 kn 4 kn/m H V V 3m 1.5m 1.5m 3m X = 0, H = 0 Take moment about, 20*3 + 20*9 + 4*(4.5)*(4.5 + 4.5/2) V c *6 = 0 V c = 60.25 kn Y = 0, 20 + 20 + 4*4.5 = V + V V = -2.25 kn, (-ve sign indicates V acts in opposite direction) V = 2.25 kn ( ) H in ivil Engineering Page 1-8
E2027 Structural nalysis I Example 2 Find the support reactions for the simple beam shown. 5 40 kn 50 kn 3 4 5m 2.5m 2.5m Solution: 5 4 40 kn 40 kn 3 50 kn 30 kn H V V 5m 2.5m 2.5m Resolve the 50 kn inclined external load into horizontal and vertical components as shown. X = 0, H = 30 kn Take moment about, 40*5 + 40*7.5 V *10 = 0 V = 50 kn Y = 0, 40 + 40 = V + V V = 30 kn H in ivil Engineering Page 1-9
E2027 Structural nalysis I Example 3 etermine the truss reaction forces. 30 kn 30 kn 20 kn 5m 5m 5m 5m 5m Solution: 30 kn 30 kn 20 kn 5m H V V 5m 5m 5m 5m X = 0, H = 20 kn Take moment about, 30*10 + 30*15 20*2.5 V *20 = 0 V = 35 kn Y = 0, 30 + 30 = V + V V = 25 kn H in ivil Engineering Page 1-10
E2027 Structural nalysis I Example 4 etermine the support reactions for the frame shown. 20 kn 25 kn 5 3 4 4m 4m 10 kn E 12 m H in ivil Engineering Page 1-11
E2027 Structural nalysis I Solution: 20 kn 20 kn 5 3 25 kn 4 4m 4m 10 kn H V V E E 15 kn 12 m Resolve the 25 kn inclined external load into horizontal and vertical components as shown. X = 0, H + 10 = 15 kn H = 5 kn Take moment about, 10*4 + 20*12 15*8 V E *12 = 0 V E = 13.3 kn Y = 0, 20 + 20 = V + V E V = 26.7 kn H in ivil Engineering Page 1-12
E2027 Structural nalysis I Example 5 Find the reactions for the cantilever beam shown. 6 kn 4 knm 4 kn/m 1.5m 1.5m 1.5m 1.5m Solution: M H V 6 kn 4 kn/m 4 knm 1.5m 1.5m 1.5m 1.5m X = 0, H = 0 kn Y = 0, 6 + 4*1.5/2 = V V = 9 kn Take moment about, 6*1.5 + 4*(1.5/2)*(6-1.5*1/3) 4 M = 0 M = 21.5 knm H in ivil Engineering Page 1-13
E2027 Structural nalysis I Example 6 etermine the support reactions for the frame shown. 3 kn/m 8 kn/m 3 kn/m 12m 8m H in ivil Engineering Page 1-14
E2027 Structural nalysis I Solution: 3 kn/m 5 kn/m 3 kn/m 3 kn/m 12m H V M 8m X = 0, H = 3*12 kn = 36 kn Y = 0, 3*8 + 5*8/2 = V V = 44 kn Take moment about, 3*12*6 + 3*8*4 + 5*(8/2)*(8*2/3) M = 0 M = 418.7 knm H in ivil Engineering Page 1-15
E2027 Structural nalysis I Example 7 etermine the support reactions at the support, and F. Joints and E are internal hinges. 10 kn 4 kn/m E 2.5m 2.5m 2m 2m 4m F Solution: 10 kn 4 kn/m E V V 2.5m 2.5m 2m 2m 4m F M F H F V F H in ivil Engineering Page 1-16
E2027 Structural nalysis I reak the beam into three free-body diagrams, namely, E and EF. V 10 kn H V H V V E V E H E H E 4 kn/m M F E F H F V E V F onsider free-body diagram, X = 0, H = 0 kn y symmetry, V = V = 10/2 = 5 kn Remember the internal forces at hinge of and E are equal and opposite. This also applies to hinge E. onsider free-body diagram E, X = 0, H = H E = 0 kn Take moment about E, V *4 = V *2 V = 5*4/2 = 10 kn Y = 0, V + V E = V, V E = 10-5 = 5 kn H in ivil Engineering Page 1-17
E2027 Structural nalysis I onsider free-body diagram, EF. X = 0, H E = H F = 0 kn Y = 0, V E + V F = 4*4, V F = 16-5 = 11 kn Take moment about F, V E *4 + M F 4*4*2 = 0 M F = 4*4*2 5*4 = 12 knm H in ivil Engineering Page 1-18
E2027 Structural nalysis I Example 8 etermine the support reactions at,, G and H. Joints and F are internal hinges. 120 kn 150 kn E F G 10 kn/m 5m 5m 2m 3m 3m 2m 6m 5 3 4 H Solution: Resolve the inclined external load into vertical and horizontal components. 120 kn 120 kn 90 kn 10 kn/m H H E F G V V V G V H 5m 5m 2m 3m 3m 2m 6m H in ivil Engineering Page 1-19
E2027 Structural nalysis I reak the beam into three free-body diagrams, namely, EF and FGH. 120 kn 120 kn H V V 90 kn E F H V F F H V 10 kn/m F H F F G H V V V G V H H onsider the free-body FGH first, X = 0, H F = 0 onsider the free-body EF, y symmetry, V = V F = 120/2 = 60 kn X = 0, H = H F +90 = 0 + 90 = 90 kn. onsider the free-body, X = 0, H = H, H = 90 kn. Take moment about, 120*5 + 60*12 = V *10, V = 132 kn. Y = 0, V + V = 120 + 60, V = 48 kn. onsider the free-body FGH, Take moment about H, 60*8 + 10*8*4 = V G *6, V G = 133.3 kn. Y = 0, V G + V H = 10*8 + 60, V H = 6.7 kn. H in ivil Engineering Page 1-20
E2027 Structural nalysis I Tutorial 1 (Support Reactions) etermine the support reactions for the following structures. Q1. 10 kn 2 kn/m E 2m 2m 1m 3m is an internal hinge Q2. 30 kn/m is an internal hinge 200 kn 10.4m 1.6m 12 m 6m 6m H in ivil Engineering Page 1-21
E2027 Structural nalysis I Q3. 4 kn/m 2 kn 1m 3m 1m 2m Q4. 250 kn 40 kn/m 3m 3m 6m is an internal hinge Q5. 5 kn/m 5 kn/m 2m 2m 2m H in ivil Engineering Page 1-22
E2027 Structural nalysis I Q6. 30 kn 10 kn/m E F 3m 3m G 3m 3m, & F are internal hinges Q7. 20 kn/m 40 kn/m E F is an internal hinge 12 m G 3m 5m 3m 5m H in ivil Engineering Page 1-23
E2027 Structural nalysis I Q8. 2 kn/m is an internal hinge 4m 6m E 3 kn/m 3m 3m Q9. 60 kn 30 kn E 6 kn/m & E are internal hinges 5m 3m 3m 3m F H in ivil Engineering Page 1-24
E2027 Structural nalysis I Q10. 10 kn 20 kn E 6 knm is an internal hinge 2 kn/m 4m F 2m 1m 1m H in ivil Engineering Page 1-25