Structure One One

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<p>Prof. Dr. Raddi M. Al Zubaidi PhD Civil Engineering United Kingdom 1984</p> <p>1</p> <p>Subjects to be Covered * Structure of building * Code of Practice *Reactions and Supports *Stability and determinacy * Trusses *Shear and moment concepts * Shear and moment diagrams</p> <p>2</p> <p>Assessment</p> <p>Exam 1 &amp; Assignments Midterm Exam Final Exam</p> <p>30% 20% 50%</p> <p>3</p> <p>Dependent Lecture -Course Text book - J.B. Gauld, Bryan. Structures for Architects.Longman.1995 - Reference Books - Ambrose,James, Simplified Mechanics and Strenght of Materials, John Wiley 2000 - -Schodek,L.Daniel ,Structures, Pearon,20044</p> <p>Structure of BuildingSlab</p> <p>Beam Tie beam G.L Foundation Road base PCC Column</p> <p>5</p> <p>Types of Footing 1- Wall Footing 2- Speard Footing 3- Combined Footing 4- Mat or Raft Foundation 5- Piles Foundation</p> <p>6</p> <p>Wall Footing</p> <p>7</p> <p>Load from column L</p> <p>B</p> <p>Spread Footing</p> <p>8</p> <p>COMBINED FOOTING</p> <p>P1</p> <p>P2</p> <p>9</p> <p>Mat or Raft Foundation</p> <p>10</p> <p>Piles Foundation</p> <p>11</p> <p>One Way Slab</p> <p>12</p> <p>Ribs Slab</p> <p>13</p> <p>Flat Slab</p> <p>14</p> <p>Flat Slab</p> <p>15</p> <p>Structure ____________________ The World structure describes much of what is seen in nature . Living plant from the frailest of fern to the most rugged of trees ,posses a structure ,other than aircraft ,ships and floating structures is to transfer applied loads to the ground16</p> <p>Types of Structure1- Concrete 2- Steel 3- Wood</p> <p>17</p> <p>Structure</p> <p>18</p> <p>Structure</p> <p>19</p> <p>Structure</p> <p>20</p> <p>Type of Structure</p> <p>21</p> <p>Type of Structure</p> <p>22</p> <p>Type of Structure</p> <p>23</p> <p>Type of Structure</p> <p>24</p> <p>Stress-Strain Curve</p> <p>25</p> <p>Stress-Strain CurveWhen external force is applied to abeam as shown in the previous slide the beam will exert stress and strain</p> <p>26</p> <p>Stress-Strain Curve</p> <p>27</p> <p>Stress- Strain CurveWhen a load is applied the beam will deflect ,but if the beam is behaving elastically it will return to its original position when the load is removed When the load is increased then the beam enter the plastic zone and the plastic deformation increases until the beam will fail. The point of breaking is called ultimate strength of the material28</p> <p>Stress-Strain Curve</p> <p>29</p> <p>Stress Strain CurveYield point =The point at which the permanent deformation occur and the material begins to yield Permissible safety=Yield point of material/safety factor Permissible safety=(Ultimate strength of material/safety factor) +(Applied loads x Safety factor)30</p> <p>Approximate Sizes of Structural Members During the planning stages of any project ,the architect requires to understand how the project is to be structured .Where at the detailed planning stage it is needed to know the sizes of the structural members.</p> <p>31</p> <p>Approximate Sizes of Structural Members</p> <p>32</p> <p>33</p> <p>34</p> <p>35</p> <p>36</p> <p>37</p> <p>38</p> <p>39</p> <p>40</p> <p>41</p> <p>42</p> <p>43</p> <p>44</p> <p>45</p> <p>46</p> <p>47</p> <p>48</p> <p>49</p> <p>Example No.1 Apiece of wire hanging from a stairwell with a weight on the end. The weight will cause the wire to stretch as shown in Fig.1 .If the length and the diameter of the wire and the magnitude of the weight are known. Determine the extension of the wire ?If the following information are available? .</p> <p>Fig 150</p> <p>Length of wire = 12 m Diameter of wire = 2 mm Weight = 31.5 kg Assume 1 kg = 10 N E = 200 KN /mm2</p> <p>51</p> <p>52</p> <p>53</p> <p>Example 2</p> <p>For the same information in Fig.1 but the weight is increased to 72 kg .Determine the extension</p> <p>54</p> <p>Force = 72 x 10 = 720 N Area = 3.14 x 12 = 3.14 mm2 Stress = Force / Area = 720/3.14 =229.3 N/mm2 Strain = Extension / Original length =Extension/12000 Young s modulus for mild steel=200 KN/mm2 Stress =E x Strain 229.3 =200000x strain Strain =229.3 /200000 =0.00114 Extension =12000 x 0.00114 = 13.758 mm</p> <p>55</p> <p>Example 3For the hard steel road shown in Fig 2 .There is a tension force equal to 250000 N acting on both sides . Determine the extension if the following information's are available ? Diameter = 50 mm ,E= 300 KN/ mm2L=3m 250000N Fig.2 250000 N56</p> <p>Length = 3m Diameter = 50 mm E =300 KN /mm2 Area of rod = 3.14 x(25)2 = 1962.5 mm2 Stress = Force /Area =250000/1962.5 =127.38 N/mm2 Stress = E x Strain 127.38 =300000 x strain Strain =127.38 /300000 =0.000424657</p> <p>Solution</p> <p>Solution Example 3Strain = Extension /Original Extension= 0.0004246 x 3000 =1.272mm</p> <p>58</p> <p>Example 4 A reinforcement bar shown in Fig.3 .The bar undergoes a tension force of 300000 N at both ends .Determine the extension with the following information's available : E = 300 KN/mm2 ,Diameter= 24 mmL = 15 m 300000 N Fig.3 300000 N</p> <p>59</p> <p>Solution Example 4</p> <p>Length = 15 m Diameter =24 mm E = 300 KN/mm2</p> <p>60</p> <p>Solution Area = 3.14 (12)2 = 452.16 mm2 Stress =300000 / 452.12 = 663.48 N /mm2 Strain =Stress / E =663.48 /300000 =0.00221 Strain =Extension / Original length Extension = 0.00221 x15000 = 33 mm</p> <p>61</p> <p>Effect of Loads on MembersA B C D E F Girder A,B,C,D,E,F h h G Girder G,H,I,J,K,L62</p> <p>qh h h I H J K L</p> <p>Beam AL</p> <p>qh /2 </p> <p>A</p> <p>L</p> <p>63</p> <p>Beam BK</p> <p>qh</p> <p>B</p> <p>K</p> <p>64</p> <p>Beam EH</p> <p>qh E H</p> <p>65</p> <p>Beam FG</p> <p>q h/2</p> <p>66</p> <p>Girder GHIJKL(qh/4) (qh/2) (qh/2) (qh/2) (qh/2) (qh/4)</p> <p>5(qh/4)</p> <p>G</p> <p>H</p> <p>I 5h</p> <p>J</p> <p>K</p> <p>L</p> <p>5(qh/4)</p> <p>column</p> <p>column</p> <p>67</p> <p>Girder FEDCBA(qh/4) (qh/2) (qh/2) (qh/2) (qh/2) (qh/4)</p> <p>F 5(qh/4)</p> <p>E</p> <p>D 5h</p> <p>C</p> <p>B</p> <p>A 5(qh/4)</p> <p>column</p> <p>column</p> <p>68</p> <p>For the concrete slab shown in Fig.1 .The following information's are available : Dead load = 4 KN /m2 Live load =1.5 KN/m2 L=20 m h= 4 m The slab rests on a number of beams and the beams are rests on two main girders. Determine the resulting effect on beams: AL, BK ,EH,FG The Girders GHIJKL and FEDCBA and the69</p> <p>Example 2</p> <p>Solution</p> <p>Factored load q= 1.2 D + 1.6 L q = 1.2 x 4 +1.6 x 1.5 = 7.2 KN /m2 Beams AL and FG Tributary area =(h/2) l= 4/2 x 20 =40 m2 Total load = q x 40 =7.2 x40 = 288 KN Uniform load = w = Total load/l = 288/20 =14.4 KN /m End support load =288/2 =144 KN Beams BK,CG ,DI and Eh Tributary area = 2(h/2) x20 =2x4/2x20 =80 70 m2</p> <p>Total load = 80 x7.2 = 576 KN Uniform load = 576/ 20=28.8 KN/m End load = 576 / 2= 288 KN</p> <p>71</p> <p>Beams AL and FG</p> <p>14.4 KN/m</p> <p>20 m 144 kN 144 kN</p> <p>72</p> <p>Beams BK and CG</p> <p>28.8 KN/m</p> <p>288 KN</p> <p>288 KN</p> <p>73</p> <p>Girders FEDCBA and GHIJKL144 KN 288 KN 288 KN 288 KN 288 KN 144 KN</p> <p>720 KN</p> <p>720 KN</p> <p>74</p> <p>Example 2For the floor system shown in Fig.1.25.Assum that the</p> <p>surface load q is applied on panel</p> <p>abef if q = 100 lb/ft2 determine the resulting effect on beams ab, cd, ef, girder ae and the columns at a and e Assume = 20 ft and h= 10 ft</p> <p>75</p> <p>76</p> <p>77</p> <p>78</p> <p>79</p> <p>80</p> <p>81</p> <p>82</p> <p>83</p> <p>Example 3 For the slab shown in Fig.3 .The following information's are available: Dead load = 4.5 KN /m2 Live load= 1.5 KN/m2 Find the total and uniform load on beam CD and BE and the loads on girder DEF and columns D and F ?</p> <p>84</p> <p>Example 3C 6m B 6m A 30 m Fig.385</p> <p>D</p> <p>E</p> <p>F</p> <p>Solution q = 1.2 x 4.5 + 1.6 x 1.5 = 7.8 KN/ m2 Load on beam CD Total load = 3 x30 x 7.8 = 702 KN Uniform load = 702 / 30 = 23.4 KN /m</p> <p>Beam CDTotal load = 702 KN uniform load = 3.4 KN/m</p> <p>351 KN</p> <p>351 KN86</p> <p>Beam BF</p> <p>Total load =6 x30 x 7.8 = 1404 KN Uniform load =1404 /30= 46.8 KN/m</p> <p>Total load = 1404 KN Uniform load = 46.8 KN/m</p> <p>702 KN</p> <p>702 KN87</p> <p>Girder FED351 KN 702 KN 351 KN</p> <p>F 702 KN</p> <p>E</p> <p>D 702 KN</p> <p>88</p> <p>For the reinforced concrete slab shown 4 the following information are available Dead load = 5 KN/ m2 Live load = 1.5 KN /m2 Find the total and uniform load on beams : EF, DG, CH, BI and AJ and the loads on girder FGHIJ and the column F . Determine also the size of the column ?89</p> <p>Example 4</p> <p>E D C 8m 3m 20 m Fig.490</p> <p>4m 6m H G</p> <p>F</p> <p>B A</p> <p>I J</p> <p>q =Factored load= 1.2 x5 + 1.6 x 1.5 = 8.4 KN/ m2 Beam EF Total load = 2 x 20 x 8.4 = 336 KN Uniform load =336 /20 = 16.8 KN/m</p> <p>Solution</p> <p>91</p> <p>Beam EFTotal load = 336 KN Uniform load = 16.8 KN/M</p> <p>168 KN</p> <p>168 KN</p> <p>92</p> <p>Beam DG Total load =5 x 20 x 8.4 =840 KN Uniform load = 840 /20 =42 KN /m</p> <p>420 KN</p> <p>420 KN</p> <p>93</p> <p>Beam CH Total load = 7 x 20 x 8.4 = 1176 KN Uniform load = 1176 /20 =58.8 KN/ m</p> <p>588 KN</p> <p>588 KN</p> <p>94</p> <p>Beam BI Total load = 5.5 x 20 x 8.4 = 924 KN Uniform load =924 /20 = 46.2 KN/m</p> <p>462 KN</p> <p>462 KN</p> <p>95</p> <p>Beam AG Total load =1.5 x 20 x 8.4 = 252 KN Uniform load = 252 / 20 = 12.6 KN</p> <p>126 KN</p> <p>126 KN</p> <p>96</p> <p>126 KN 462 KN 3m J I 8m H 588 KN 6m G 420 KN</p> <p>168 KN</p> <p>4m F</p> <p>GIRDER FGHIJ</p> <p>97</p> <p>Load on column F can be found from bending moment as 168x 21 + 420 x 17 + 588 x 11 + 462 x3 F x 21 =0 18525 F = 21 = 882 KN The size of the column should be 30 x 30 cm</p> <p>98</p> <p>Approximate size of column</p> <p>99</p> <p>Example 5 For the reinforced concrete structure shown in Fig.5 The following information are available Dead load =5.5 KN/m2 Live load =1.8 KN/m2 Find the total and uniform load on beams ABCD,EFGH,IJKL and MOQR .The total load on girder CGKQ and the columns C and Q are required. Find also the depth of the slab, beam BC, the girder CGKQ and the cantilever CD ,also find the size of the columns?</p> <p>100</p> <p>A E F</p> <p>B 6m G 8m</p> <p>C</p> <p>D</p> <p>H</p> <p>I 1m M O</p> <p>J 7m 8m Fig.5</p> <p>K L Q 1m</p> <p>R</p> <p>101</p> <p>Solution Q = 1.2 x 5.5 + 1.6 x 1.8 = 9.48 KN / m2 Beam ABCD Total load = 9.48 x 3x 10 =284.4 KN Uniform load = 284.4 / 10 = 28.44 KN /m</p> <p>142.2 KN</p> <p>142 KN102</p> <p>Beam EFGH Total load = 9.48 x 7 x10 = 663.6 KN Uniform load = 663.6 / 10 = 66.36 KN /m</p> <p>331.8 KN</p> <p>331.8 KN</p> <p>103</p> <p>Beam IJKL Total load = 9.48 x 7.5 x 10 = 711 KN Uniform load = 711 /10 = 71.1 KN /m</p> <p>355.5 KN</p> <p>355.5 KN</p> <p>104</p> <p>Beam MOQR Total load = 9.48 x 3.5 x 10 =331.8 KN Uniform load = 331.8 / 10 = KN /m</p> <p>165.9 KN</p> <p>165.9 KN</p> <p>105</p> <p>Girder CGKQ</p> <p>165.9 KN 7m Q</p> <p>355.5 KN 8m K</p> <p>331.8 KN 6m</p> <p>142.2 KN</p> <p>G</p> <p>C</p> <p>106</p> <p>Load on columns C &amp; Q Take the bending moment at C Q x 21 = 165.9 x 21 + 355.5 x 14 + 331.8 x 6 Q = 497.7 KN C = 497.7 KN</p> <p>107</p> <p>Depth of the slab</p> <p>Slab depth = 8/ 30 = 27 cm</p> <p>108</p> <p>Depth of beam BC</p> <p>Depth = 8 /18 = 45 cm</p> <p>109</p> <p>Depth of girder CGKQDepth = 21 /18 = 117 cm</p> <p>110</p> <p>Depth of cantileverDepth = 1 /7 = 15 cm</p> <p>111</p> <p>Size of column cSize of column = 24 x 24 cm</p> <p>112</p> <p>For the reinforced concrete structure building shown in Fig.6 .The following information are available:Dead Load = 5.6 KN/ m2 Live Load = 1.6 KN / m2 Find the total and uniform loads on beams AD, EH, IL, MN, OP, QR, ST, the load on girder NT and the load on column N and C . Find also the depth of :the first and ground slabs Beam BC and QR Girder NT Cantilever CD The size of column at N and C113</p> <p>Example 6</p> <p>2m</p> <p>8m</p> <p>2m</p> <p>Fig.6114</p> <p>A</p> <p>B</p> <p>E F 10 m</p> <p>C D</p> <p>I J 2m 8m Ground Floor K 8m</p> <p>G H</p> <p>2m</p> <p>L115</p> <p>M 5m O 7m Q 6m S 8m First Floor T R P</p> <p>N</p> <p>116</p> <p>Solution q = 1.2 x5.6 +1.6 x1.6 = 9.28 KN /m2 First Floor Load on beam MN Total load = 9.28 x8 x 2.5 = 185.6 KN Uniform load = 23.2 KN /m</p> <p>M 92.8 KN</p> <p>N 92.8 KN117</p> <p>Beam OP Total load = 9.28 x8 x6 = 445.44 KN Uniform load = 55.68 KN/m</p> <p>O</p> <p>P 227.7 KN</p> <p>227.7 KN</p> <p>118</p> <p>Beam QRTotal load = 60.32 KN /m Uniform load= 9.28 x 8 x 6.5 = 482.56 KN</p> <p>Q</p> <p>R</p> <p>241.3 KN</p> <p>241.3 KN</p> <p>119</p> <p>Beam STTotal load = 9.28 x 8 x 3 = 222.7 KN Uniform load = 27.84 KN/m</p> <p>S</p> <p>T</p> <p>111.36 KN</p> <p>111.36 KN120</p> <p>Load on Girder NT ,total load on column N =(111.3+241.3+222.7+92.8)/2= 334 KN</p> <p>111.36 KN 6m T</p> <p>241.3 KN 7m R</p> <p>222.7 KN 5m P</p> <p>92.8 KN N</p> <p>121</p> <p>Ground Floor</p> <p>Beam AD Total load = 9.28 x 12 x5 = 556.8 KN Uniform =46.4 KN/m</p> <p>A</p> <p>B 278.4 KN</p> <p>C 278.4 KN</p> <p>D</p> <p>122</p> <p>Beam EH Total load = 9.28 x12 x9 = 1002 KN Uniform load = 83.52 KN /m</p> <p>E F 501 KN</p> <p>G 501 KN</p> <p>H</p> <p>123</p> <p>Beam IL Total load = 9.28 x 12 x4 = 445.44 KN Uniform load = 37.12 KN /m</p> <p>I</p> <p>J 222.7 KN</p> <p>K 222.7 KN</p> <p>L</p> <p>124</p> <p>Girder CK Load on column C C x 18 = 278.4 x 18 + 501 x 8 C = 501 KN222.7 KN 8m K G 501 KN 10 m C 278.4 KN</p> <p>125</p> <p>First Floor</p> <p>Depth of Slab = 8/ 30 = 27 cm Depth of beams = 8 / 18 = 45 cm Depth of girder = 18 /18 = 100 cm Total load on column N =334 KN Size of column at N =23 x23 cm</p> <p>126</p> <p>Depth of slab =1000/30 = 34 cm Depth of beams = 800 / 18 = 45 cm Depth of girders = 1800 / 18 =100 c m Depth of cantilever = 200 / 7 = 29 cm Total load on column C = 501 + 334 = 835 KN Size of column = 30 x30 cm</p> <p>Ground floor</p> <p>127</p> <p>Reactions Forces</p> <p>Reactions forces develop at support points of structure to equilibrate the effects of applied forces. It is recalled that a force is completely defined by its magnitude ,direction and point of application since applied forces are fully specified ,these three characteristics are necessarily known for each. Reaction forces are not specified as priori however but develop as a consequences of applied forces and in conformance with the support mechanisms.</p> <p>128</p> <p>129</p> <p>130</p> <p>131</p> <p>132</p> <p>133</p> <p>134</p> <p>135</p> <p>136</p> <p>137</p> <p>RbY = Rb cos = Rb = RbY 5 /2 Rbx = Rb sin Rbx = Rb / 5 = Rb= Rbx 5 Rbx= RbY / 2 Or tan = 1 = Rbx Rby 2 Rbx = Rby 2 2</p> <p>Rbx 51</p> <p>Rb</p> <p>RbY</p> <p>138</p> <p>139</p> <p>Determine the reactions at points A&amp;B ?50 KN 75 KN2 1</p> <p>60 KN1</p> <p>30 KN A</p> <p>2</p> <p>B 3m 4m 4m 5m</p> <p>140</p> <p>Free Body diagram</p> <p>26.83 KN 30 KN 53.6 KN</p> <p>50 KN</p> <p>67 KN 33.54 KN</p> <p>Rax</p> <p>A Ray 3m 4m 4m 5m</p> <p>B Rby</p> <p>141</p> <p>Solution Fx =0 Rax + 53.6 + 30 33.54 = 0 Rax = - 50 KN MA =0 Taking the moment at point A 26.83x 3 + 50 x7 + 67x 11 Rby x 16 = 0 Rby = 73 KN Ray 26.83 50 67 + 73 =0 Ray = 71 KN142</p> <p>143</p> <p>144</p> <p>145</p> <p>146</p> <p>147</p> <p>148</p> <p>tan =Rby =</p> <p>Rby Rbx Rbx 3</p> <p>=</p> <p>1 3</p> <p>2 30 3 =Rbx 1 =Rby</p> <p>149</p> <p>150</p> <p>151</p> <p>152</p> <p>153</p> <p>154</p> <p>EXAMPLE 615 KN 30 KN 1 3 b 4m 6m 8m 8m d a 4m c 2 KN/m 6 KN/M</p> <p>10 KN</p> <p>155</p> <p>15 KN 10KN b 4m</p> <p>15 KN 26 KN 10 m 8m</p> <p>16 KN c 4m d</p> <p>2.67 m 24 KN 5.33 m</p> <p>Rdx a Ray 4m</p> <p>Rdy</p> <p>156</p> <p>30 x 1/2</p> <p>30 x 3 /2</p> <p>157</p> <p>Taking the moment at point d a clock wise positive Ray x 18 + 10 x 8 - 15 x 18 - 15 x 14 - 16 x 4 - 24 x 5.33 -26 x 8 = 0 Ray = 44.4 KN Summation of Y axis forces equal to zero up positive Ray + Rdy - 15 -15 -16 = 0 44.4 + Rdy - 46 = 0 Rdy = 1.6 KN Summation of X axis equal to zero right is positive 10 - 26 -24 Rdx = 0 Rdx = 40 KN</p> <p>Solution</p> <p>158</p> <p>Example 7</p> <p>12 KN</p> <p>4 KN/m</p> <p>2 1 a 5m 5m</p> <p>6 KN/m</p> <p>b 5m</p> <p>c</p> <p>159</p> <p>Free body diagram10.73 KN 4 x 5=20 KN (2x15)/2=15 KN</p> <p>4 KN Rax a</p> <p>5.36 KN 5m 7.5 m 10 m Ray Rby b c</p> <p>4 KN</p> <p>160</p> <p>Solution Summation of forces in the X- axis equal to zero right is positive Rax 5.36 = 0 Summation of moment at a is equal to zero clock wise is positive 10.73 x 5 + 60 x 7.5 + 15 x 10 Rby x 10 = 0 Rby = 65.36 KN Summation of forces in the Y axis is equal to zero up is positive Ray + Rby 10.73 60 15 = 0 Ray = 20.4 KN161</p> <p>162</p> <p>163</p> <p>164</p> <p>165</p> <p>166</p> <p>Example 1</p> <p>167</p> <p>168</p> <p>169</p> <p>170</p> <p>171</p> <p>172</p> <p>173</p> <p>174</p> <p>Question31-What are the reactions at the supports A and C? 2-Draw the shear force diagram? 3-Draw the bending force diagram ?</p> <p>175</p> <p>Solution</p> <p>176</p> <p>Shear force diagram</p> <p>177</p> <p>Bending Moment Diagram</p> <p>178</p> <p>Question 41- What are...</p>

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