7/30/2019 clasificare stalpi
1/1000
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2/1000
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3/1000
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4/1000
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5/1000
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7/1000
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81/1000
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82/1000
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83/1000
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84/1000
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86/1000
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87/1000
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88/1000
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89/1000
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90/1000
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91/1000
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92/1000
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98/1000
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101/1000
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102/1000
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103/1000
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106/1000
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107/1000
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108/1000
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109/1000
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110/1000
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111/1000
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112/1000
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113/1000
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114/1000
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115/1000
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116/1000
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117/1000
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118/1000
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119/1000
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120/1000
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121/1000
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122/1000
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123/1000
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124/1000
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126/1000
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127/1000
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128/1000
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129/1000
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130/1000
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131/1000
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135/1000
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136/1000
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137/1000
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138/1000
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139/1000
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140/1000
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141/1000
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142/1000
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143/1000
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144/1000
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145/1000
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146/1000
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148/1000
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149/1000
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151/1000
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152/1000
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153/1000
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154/1000
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155/1000
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156/1000
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157/1000
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158/1000
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159/1000
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160/1000
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161/1000
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162/1000
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163/1000
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164/1000
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165/1000
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166/1000
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167/1000
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168/1000
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169/1000
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170/1000
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171/1000
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172/1000
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173/1000
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174/1000
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175/1000
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176/1000
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177/1000
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178/1000
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181/1000
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182/1000
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183/1000
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184/1000
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185/1000
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186/1000
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187/1000
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188/1000
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189/1000
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202/1000
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203/1000
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204/1000
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205/1000
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206/1000
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207/1000
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208/1000
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211/1000
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212/1000
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213/1000
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214/1000
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215/1000
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216/1000
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217/1000
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226/1000
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228/1000
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229/1000
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231/1000
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234/1000
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239/1000
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240/1000
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241/1000
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242/1000
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243/1000
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244/1000
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245/1000
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246/1000
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248/1000
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257/1000
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261/1000
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262/1000
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263/1000
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264/1000
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268/1000
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269/1000
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270/1000
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271/1000
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272/1000
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273/1000
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274/1000
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275/1000
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276/1000
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277/1000
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278/1000
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279/1000
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280/1000
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281/1000
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282/1000
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283/1000
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284/1000
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285/1000
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286/1000
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287/1000
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288/1000
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289/1000
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290/1000
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291/1000
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292/1000
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293/1000
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294/1000
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295/1000
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296/1000
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297/1000
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298/1000
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299/1000
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301/1000
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302/1000
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303/1000
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304/1000
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306/1000
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307/1000
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308/1000
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309/1000
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310/1000
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311/1000
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312/1000
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313/1000
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314/1000
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315/1000
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316/1000
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317/1000
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318/1000
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319/1000
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321/1000
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323/1000
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327/1000
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328/1000
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329/1000
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330/1000
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331/1000
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332/1000
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333/1000
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334/1000
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335/1000
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336/1000
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338/1000
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341/1000
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342/1000
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343/1000
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344/1000
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345/1000
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346/1000
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347/1000
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348/1000
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351/1000
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352/1000
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353/1000
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355/1000
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356/1000
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357/1000
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358/1000
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359/1000
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360/1000
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361/1000
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362/1000
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363/1000
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364/1000
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365/1000
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366/1000
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367/1000
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368/1000
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369/1000
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370/1000
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371/1000
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372/1000
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373/1000
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374/1000
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375/1000
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376/1000
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377/1000
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378/1000
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379/1000
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380/1000
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381/1000
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382/1000
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383/1000
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384/1000
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385/1000
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386/1000
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387/1000
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388/1000
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389/1000
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390/1000
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391/1000
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392/1000
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393/1000
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394/1000
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395/1000
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396/1000
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397/1000
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398/1000
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399/1000
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400/1000
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401/1000
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402/1000
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403/1000
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404/1000
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405/1000
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406/1000
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407/1000
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408/1000
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409/1000
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410/1000
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411/1000
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412/1000
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413/1000
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414/1000
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415/1000
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416/1000
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417/1000
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418/1000
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419/1000
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420/1000
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421/1000
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422/1000
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423/1000
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425/1000
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426/1000
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427/1000
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428/1000
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429/1000
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430/1000
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431/1000
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432/1000
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433/1000
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434/1000
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435/1000
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436/1000
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437/1000
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438/1000
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439/1000
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440/1000
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441/1000
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442/1000
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443/1000
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444/1000
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445/1000
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446/1000
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447/1000
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448/1000
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449/1000
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450/1000
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451/1000
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452/1000
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453/1000
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454/1000
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455/1000
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CHAPTER 22
DESIGN AND ANALYSIS FLOWCHARTS
FLOWCHARTS
To help the student and design engineer to prepare their own programs, flowcharts are
given for most chapters.
1. Flowcharts 22.1, 22.2, and 22.3 explain the analysis of single, double, and T-sections(Chapter 3).
2. Flowcharts 22.4, 22.5, and 22.6 explain the design of single, double, and T-sections(Chapter 4).
3. Flowchart 22.7 explains the calculation of development length (Chapter 7).
4. Flowchart 22.8 explains shear design (Chapter 8).
5. Flowchart 22.9 explains the analysis of rectangular columns at balanced condition(Chapter 11).
6. Flowchart 22.10 explains the analysis of rectangular columns (Chapter 11).
7. Flowchart 22.11 explains the design of rectangular or square footings (Chapter 13).
8. Flowchart 22.12 explains the design for combined shear and torsion (Chapter 15).
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No
YesNo
Given: b, d, As, fc, fyRequired: Mn
Let = As / bd
> min(1)
Increase max(2)
Yes
Reduce
a = Asfy/ (0.85fcb)
c = a /1
= 0.9
Mn = Asfy(d-a/2)
End
Flowchart 22.1 Analysis of single reinforced rectangular section.
(1) min = 3 'cf / fy 200 / fy
(2) b = (0.851)(fc / fy)87
87y
f
+
max=0.003 /
0.008
y s
b
f E
+
t = 0.003(dt c)/c
t 0.005
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No
No
No
No
No
Given: b, d, d, As, As, fc, fyRequired: Mn
Let = As / bd, = As / bd
- ' K(1)
fs < fy
min -maxYes
Yes
Change
a = ( ' ' )0.85 '
s y s s
c
A f A ff b
Flowchart 22.2 Analysis of double reinforced rectangular section.
(1) K = (0.851)(fc / fy)(d / d)87
87yf
(2) Refer to flowchart 22.1 to calculate min and max.
fs = fy
Solve for c:A1c
2 + A2c + A3 = 0
A1 = 0.851fcbA2 = As(87 0.85fc) - AsfyA3 = -87Asd
Calculate fs < fy;
s = 0.003(c-d)/c
fs = Ess =( ')
87c d
c
Mn = [Asfy Asfs)(d-a/2) + Asfs(d-d)] End
min Yes
Increase
t = 0.003(dt -c)/c
t 0.005Yes
t 0.002
= 0.9
a=(AsAs)fy/(0.85fcb)
Yes
= 0.65 (othermembers)
= 0.65+(t 0.002)(250/3)(other members)
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No
No
No
No
Given: b, bw, d, t, As, fc, fyRequired: Mn
As A
s max(1)
Reduce As As As min(2)
Yes
Yes
Increase As
Let a = Asfy/(0.85fcbe)
Mn = [(As Asf)fy(d-a/2) + Asffy(d-t/2)
End
Flowchart 22.3 Analysis of T- and L-sections.
(1) As max = 0.6375(fc/fy)[t(b-bw) + 0.375bw1d]
(2) dbf
dbf
fA
w
y
w
y
c
s
=
2003'
min
Rectangular analysis
a = a
Mn = Asfy (d - a/2)
End
a t
T-section analysis
Yes
Asf= 0.85fct(b-bw) / fy
a = (As Asf)fy/(0.85fcbw)
For T-section, be is the smallest value of: Span / 4 or Center to
center of adjacent slabs or (bw + 16t); t = slab thickness.For L-section, be is the smallest value of: Span / 12 or half clear
distance to next web + web width or (bw + 6t); t = slab thickness
t =0.003(dt-c)/c
t 0.005
Yes
Enlarge
section
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YesNo
NoYes
Given: Mu, fc, fyRequired: As
b and d are given
Calculate
Ru = Mu/(bd2)
minmax(1)
Assume .minmaxor let b / 2
Assume b (12-20) in.
Or d / b 2
As = bd
h = d+2.5 in. (one row of bars)
h = d+3.5 in. (two rows of bars)
Round h to the next higher inch.
Flowchart 22.4 Design of single reinforced rectangular section.
(1) Refer to flowchart 22.1 to calculate min and max
= 0.85 ' 1 1 (2 / 0.85 ' )c u cy
fR f
f
Change section
Calculate Ru:
Ru = fy (1-fy/1.7fc)
d = /u uM R b
As = bdChoose bars
End
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No
No
No
Given: Mu, b, d, d, fc, fyRequired: As, As
Calculate min and max
(1)
Mu M
u1
Comp. steel is not
required.
fs fyYes
Yes
Mu2 = Mu - Mu1
As2 = Mu2/fy(d-d)
f's < fy
As= As2fy/ fs
End
As1 = maxbd
Check if comp. steel yields.
Let a = As1fy / 0.85fcb
c = a /1Calculate fs = 87(c - d) / c
Go to flowchart 22.4
End
Calculate Ru maxRu max = maxfy(1-maxfy/1.7fc)
Calculate Mn = Mu1 as singly reinforcedMu1 = Ru maxbd
2
Total As = As1 + As2As As max
f's = fy
As= As2
Flowchart 22.5 Design of rectangular sections with compression steel.
(1) Refer to flowchart 22.1 to calculate min and max
t =0.003(dt-c)/c
t 0.005
Changesection
Yes
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No
No
No
Given: Mu, b, bw, d, t, fc, fyRequired: As
d is given
Assume a = t
Mu MuftYes
Yes
Change section
Flowchart 22.6 Design of T- or L-sections.
Rectangular sectiondesign, a t
As = bd
AsminAsAsmax
Ort = (dt - c)/c 0.005
T-section design,a > t
Yes
d = Mu/(0.85fcbt) + t/2
As = Mu/ [ fy (d - t/2)]
If steel is high,assume
a < t or a t / 2
Calculate Muft (total flange).
a = tCft = 0.85fcbt
Muft = Cft(d t/2)
Cf= 0.85fct(b bw)
Asf= Cf/ fy
Muf= Cf(d t/2)Muw (web) = Mu - Muf
Ru = Mu / bd
As = Mu / [fy(d a/2)]
d = Mu / (0.85fcba) + a/2
Ruw = Muw / bwd
= (0.85fc/fy)[1- 1 2 / 0.85 'u cR f ]
w =(0.85fc/fy)[1- 1 2 / 0.85 'uw cR f ]
Asw = wbwdAs = Asf+ Asw
As is OK
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No
Given: bw, d, fc, fy, bars
Required: Development length
Compression bars
Isc db
s dbor s > 2db(1)
Yes
ld = ld1 12 in.
(for no 7 and large bars)
ld1 / db ='20/
cyetff
(for no-6 and smaller bars)
ld1 / db ='25/
cetff
Flowchart 22.7 Calculation of development length
(1)c = clear covers = clear spacing
Tension bars
ldc = (0.02dbfy) / 'cf
ldc 0.0003dbfy
ld = ldc*(Rs or Rs1, if applicable)
ld 8 in.
Rs = As (required) / As (provided)
Rs1 = 0.75 for spiral columns
'cf 100 psi
ld = 1.5 ld1 12 in.
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No
No
No
Yes No
Given: bw, d, fc, fy, VuRequired: shear reinforcement
Vc = 2 'cf bwd = 0.75
VuVc/2
No shearreinforcement
is required
Vu > VcYes
Yes
Choose minimumshear reinforcement.
Av 50bwS / fy
Av =0.75 'cf (bwS/fy)
S d/2 24 in.Minimum stirrups
no.3 at Smax
Vs = (Vu - Vc) /
Av = VsS / fyd
OrS = Avfyd/Vs
End
Flowchart 22.8 Shear Design.
Vs > 4Vc Yes
Increase
section
Vs > 2Vc
S d/4 12 in. S d/2 24 in.
S Avfy/ 50bw
End
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cb = 87d / (87 + fy)ab = 1cb
fs = [87(cb d) / cb]
fs fy (Ksi)
Mb = Cc(d ab/2 d) + Cs(d d d) + Td
End
Flowchart 22.9 Balanced load, moment, and eccentricity for rectangularcolumn sections
(Use Ksi for fc and fy; d = distance from the plastic centroid to As).
Cs = As (fs 0.85fc)Cc = 0.85fcab
T = Asfy
eb = Mb / Pb
Pb = Cc + Cs - T
Given: b, d, d, As, As, fc, fyRequired: Pb, Mb, eb
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No
Given: b, d, d, As, eRequired: Pn, Mn
e > eb YesNo
Tension failure
fs = fya < ab
0.65 0.9
Cc = 0.85fcab
Cs = As(fs 0.85fc) 0
Flowchart 22.10 Analysis of rectangular columns(Use Ksi for fc and fy)
(1) s = 0.003(c-d/c); y = fy / Es
Solve for a:
Aa2 +
Ba + C = 0A = 0.425fcb
B = 2A(e d)C = As(fs 0.85fc)(e d + d) - Asfye
e = e + d, c = a / 1
fs = [87(c d)/c] fy
T = Asfy
End
Compression failure
fs < fya > ab = 0.65
Solve for a:
Aa3 +
Ba2
+ Ca + D = 0A = 0.85fcb/2
B = 0.85fcb(e d)C = As(fy 0.85fc)(e d + d) + 87Ase
D = -87Ase1d
e = e + d, c = a /1Solve b trial or calculator
fs = [87(c d)/c] fyfs = [87(d c)/c] fy
T = Asfs
Pn = Cc +Cs - T
Mn = Pn e
Assume fs = fy
sy(1)
fs = fy
Yes
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941/1000
No
Given: PD, PL, H, qa, fc, fyColumn size c, and bars
Assume footing depth h (in.)
Let Wc = 150 pcf, Ws = 100 pcf
Pu = 1.2 PD + 1.6 PLqu = Pu / A
Let average d = h 4.5 in.
Pn > N1Yes
Check two-way shear: d2
is the largest value of:
d2 = Vu2/(4 'cf b0)
d2 = Vu2/[(sd/b0) + 2] 'cf b0
d2 = Vu2/[(4/ + 2] 'cf b0
s = 40, 30, 20 for interior, edge and corner columnsb0 = 4(c + d) for square columnsb0 = 2(c1+d) + 2(c2 + d) for rectangular columns
= 0.75
Check one-way shear:
d1 = Vu1/(2 'cf b)
Dowel bars
Min. Asd = 0.005 A1Use 4 bars
Pex = Pu N1Asd = Pex/fy 0.005A1
d = the larger of d1 and d2
h = d + 4.5 h
If h < h, increase h and repeat.
End
qe
= qa
h(150)/12 (H - h)(100)/12 ; psf
Footing area A = (PD + PL)/qe
Side = A for square footing
Or L x B for rectangular footingChoose L / B < 2
Long direction MuL = 0.5qu (L/2 c/2)
Short direction MuS = 0.5qu (B/2 c/2)2
Calculate AsL and AsSChoose bars and check ld
Choose dowel bars and check ld in compression.
lowchart 22.11 Desi n of rectan ular or s uare footin s .
N1 = (0.85fc)A1 (column)
N2 = N1( 2 1/A A ) 2N1
= 0.65
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No
No
NoYes
Given: b, h, Vu, TuRequired: Closed stirrups and A1
Let x0 = b, y0 = h, = 0.75
Let x1 = (b 3.5 in.), y1 = (h 3.5 in.)
Tu > QYes
Yes
Spacing of stirrups = area of bar/Atot
Max. S = Ph/8 12 in.
Flowchart 22.12 Design for combined shear and torsion.
Let Q = ( 'cf )A2
cp/ Pcp
Acp = x0y0, Pcp = 2(x0 + y0)
A0 = 0.85 x1y1 = 0.85A0h
Ph = 2(x1 + y1)
Tu is neglected.
Check for Vu
(flowchart 22.7)Use Tu Tu = 4Q
for compability Tu
Increase
section
Total area of closed stirrups
Avt = 2At + AvAvt (50bwS/fyt)
(Av from shear)
At/S = Tn/(2A0fytcot)
End
Tu > 4Q
2 2 2
0( / ) ( /1.7 ) [( / ) (8 ' )]
u w u h h c w cV b d T P A V b d f + +
A1 = (At/S) Ph (fyt/fy) cot2
A1 min = (5 'cf Acp)/fy (At/S) Ph (fyt/fy)
ytwlfbA /25
min
Min. bar diameter =0.042S no. 3
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1
CONTENTS
Solutions Manual Chapter
Chapter
2 Properties of Reinforced Concrete3 Flexural Analysis of Reinforced Concrete Beams4 Flexural Design of Reinforced Concrete Beams6 Deflection and Control of Cracking7 Development Length of Reinforcing Bars8 Shear and Diagonal Tension9 One Way Slabs10 Axially Loaded Columns11 Members in Compression and Bending12 Slender Columns13 Footings14 Retaining Walls15 Design for Torsion16 Continuous Beams and Frames17 Design of Two-Way Slabs18 Stairs19 Introduction to Prestressed Concrete20 Seismic Design of Reinforced Concrete Structures21 Beams curved in Plan
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2
CHAPTER 2
PROPERTIES OF REINFORCED CONCRETE
2.1- 2.8 refer to the relative section in text
2.9 Calculate the modulus of elasticity; Ec (see the table below)
2.10 Calculate the modular ratio; n and the modulus of rupture; fr:
Density fc Ec n fr
160 pcf 5000 psi. 4,723,000 psi. 6.14 530.3 psi.
145 pcf 4000 psi. 3,644,000 psi. 7.96 474.3 psi.
125 pcf 2500 psi. 2,306,000 psi. 12.58 375.0 psi.
2400 kg/m3
35 MPa 29,910 Mpa 6.69 3.668 MPa
2300 kg/m3
30 MPa 25,980 Mpa 7.70 3.396 MPa
2100 kg/m3
25 MPa 20,690 Mpa 9.67 3.10 MPa
2.11a.) See figure below.
b.) Secant modulus (at fc/2 = 1910 psi.)
Ec = 1910 / 6.1010-4
= 3130 ksi.
Approximate Initial Modulus = 2.6(ksi.) / 5.4510-4
= 4771 ksi.
(Possible range 4600 5200)c.) Ec (ACI formula) = 57000 ' 57000 3820 3523 ksi.fc = =
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3
CHAPTER 3
FLEXURAL ANALYSIS OF REINFORCED CONCRETE BEAMS
Problem 3.1
DIM. No.a No.b No.c No.d No.e No.f No.g No.h
b (in.) 14 18 12 12 16 14 10 20
d (in.) 22.5 28.5 23.5 18.5 24.5 26.5 17.5 31.5
As (in.2)
4#10
5.08
6#10
7.62
4#9
4.00
4#8
3.16
5#10
6.35
5#9
5.00
3#9
3.00
4#9
4.00
a(in.) 6.4 7.47 5.88 4.65 7.00 6.3 5.29 3.53
Mn(k.ft) 441.2 849.1 370.1 230.0 600.0 525.3 200.5 535.2
0.0161 0.0149 0.0142 0.0142 0.0162 0.0135 0.0171 0.0063
max
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4
Problem 3.3
DIM. No.a No.b No.c No.d .No.e No.f No.g No.h
b ( in.) 54 48 72 32 44 50 40 42
bw ( in.) 14 14 16 16 12 14 16 12
t ( in. ) 3 4 4 3 4 3 3 3
d ( in. ) 17.5 16.5 18.5 15.5 20.5 16.5 16.5 17.5
As (in.2)
4#10
5.08
4#9
4.00
8#10
10.16
6#9
6.00
8#9
8.00
7#9
7.00
5#10
6.35
6#9
6.00
a ( in. ) 2.213 1.961 3.32 4.41 4.27 3.29 3.74 3.36
Rect. or T R R R T T T T T
Mn (k.ft) 374.7 279.4 769.9
Asmin< As < Asmax Yes Yes Yes
Asf (in.2) 2.04 5.44 4.59 3.06 3.83
a ( in. ) 4.94 5.02 4.05 4.84 4.25
Mn (k.ft) 293.4 660.1 466.8 415 425.9
Asmin< As < Asmax No Yes Yes Yes Yes
Problem 3.4
DIM. No.a No.b No.c No.d No.e No.f No.g No.h
fc 3 ksi 4 ksi 4 ksi 5 ksi 30 Mpa 20 Mpa 30 Mpa 25 Mpa
fy 40 ksi 60 ksi 75 ksi 60 ksi 400 Mpa 300 Mpa 500 Mpa 300 Mpa
b 12 in. 12 in. 12 in. 12 in. 300 mm 300 mm 300 mm 300 mm
d 20 in. 20 in. 20 in. 20 in. 500 mm 500 mm 500 mm 500 mm
As
(in.2)
(mm
2
)
4#8
3.14
4#7
2.41
4#9
4.00
4#9
4.00
3*30 mm
2121mm
2
3*25 mm
1473
4*25 mm
1963
4*20 mm
12571 0.85 0.85 0.85 0.8 0.85 0.85 0.85 0.85
0.037 0.0285 0.0207 0.0335 0.0325 0.0321 0.0236 0.04
max 0.020 0.0181 0.0145 0.0212 0.0203 0.01805 0.0162 0.0225
Rumax0.614
ksi
0.821
ksi
0.822
ksi
0.973
ksi
6.14
Mpa
4.11
Mpa
6.13
Mpa
5.11
Mpa
0.013 0.0100 0.0167 0.0167 0.01414 0.00982 0.01309 0.0084
Ru0.422
7 ksi
0.4942
ksi
0.9182
ksi
0.7941
ksi
4.52
Mpa
2.42
Mpa
5.13
Mpa
2.13
Mpa
a/d ratio 0.205 0.1772 0.3676 0.2353 0.2218 0.1732 0.2567 0.118
(a/d) max 0.318 0.3194 0.3199 0.2993 0.3184 0.3194 0.3176 0.3176
min
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947/1000
5
Problem 3.5:
3.5 a:
As = 1256.6mm2
; = 0.00914 ; Ru = 3.6 Mpa ; Mn = 272.3 KN.m
b = 0.02024 ; max = 0.6375b = 0.0129 Ok
3.5 b:
As = 1472 mm2
; = 0.0107 ; Ru = 3.51 Mpa ; Mn = 265.4 KN.mb = 0.02024 ; max = 0.6375b = 0.0129
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6
Problem 3.10
3.10 a:
= 0.011157 ; = 0.00556 ; - = 0.005597 < max = 0.0203
- < k = 0.0139, Compression steel doesnt yield.
c = 3.07 in. ; a = 1c = 6.61 in. ; fs= 16.15 ksi fy
C1 = 79.85 k ; C2 = 16.32 k ; As1 = 1.996 in.2
; 1 = 0.009242 k = 0.0139, Compression steel yields.
- 60 ; Compression steel yields, fs = 60 ksiC1 = 451.66 k ; C2 = 79.2 k ; T = 530.86 k ; As1b = 7.53 in.
2
As max = 0.63375As1b + As= 6.09 in.2
(2) = 0.02307 ; = 0.005 ; - = 0.01807 > k, Compression steel yields.
- = max ; Ok ; a = 7.01 in. ; Mn = 512.8 k.ft ; Wl = 8.516 k/ft
Problem 3.12:
= 0.01805 ; = 0.0031 ; - = 0.015 < max = 0.01806 - < k = 0.0194 , Compression steel doesnt yield.
c = 6.44 in. ; a = 5.47 in. ; fs= 53.23 ksi < fy
C1 = 186 k ; C2 = 30.4 k ; As1 = 3.1 in.2; 1 = 0.0155 t ; T section ; Asmax = 4.99 in.2
> As Ok
(2) Asf= 2.55 in.2
; a = 5.08 in. ; Mn = 339.6 k.ft
Problem 3.15:a = 3.69 in. > t, T-section, Asf= 2.55 in.
2, Asw = 2.16 in.2, As(max) = 5.80 in.
2,
a = 5.08 in., Mu = 466.78 k-ft.
Problem 3.16:
Same analysis and answer as 3.14
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7
CHAPTER 4
DESIGN OF REINFORCED CONCRETE BEAMS
Problem 4.1:
DIM. No.a No.b No.c No.d No.e No.f No.g No.h
Mu (k.ft) 272.7 969.2 816 657 559.4 254.5 451.4 832
b (in) 12 18 16 16 14 10 14 18
d (in) 21.5 32 29.52 26.5 24.5 21.5 21.75 28
Ru psi 589.9 631 780 700 799 660.7 818 708
(%
) 1.168 1.206 1.7 1.5 1.75 1.115 1.8 1.27
AS (in2) 3.013 6.947 8.029 6.36 6.0 2.39 5.48 6.42
Bars
rows
5#7
One
7#9
two
7#10
two
5#10
one
6#9
two
6#6
one
6#9
two
8#9
two
h (in) 24 36 32 29 28 24 25 32
DIM. No.i No.j No.k No.l
Mu(k.ft) 345 510 720 605
b (in) 15 12 12 16
d (in) 18.5 24.9 29.67 23.6
(%) 1.77 1.81 1.8 1.8
AS (in)2
4.91 5.4 6.4 6.8
Barsrows
5#9one
6#9two
5#10two
7#9two
h (in.) 21 29 33 28
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8
Problem 4.2:
Assume c/dt = 0.375 for fy = 60Ksi,(all problem)dt = (d + 1)in. for 2 rows of bars; d = 2.5 in.
DIM.Prob.
No.a
Prob.
No.b
Prob.
No.c
Prob.
No.d
Prob.
No.e
Prob.
No.f
Prob.
No.g
Prob.
No.h
Prob.
No.i
Prob.
No.j
Prob.
No.k
Prob.
No.l
Mu(k.ft) 554 790 448 520 765 855 555 300 400 280 290 400
b (in) 14 16 12 12 16 18 16 12 16 12 14 14
d (in) 20.5 24.5 18.5 20.5 20.5 22 18.5 16.5 16.5 16.5 14.5 17.5
c (in) 8.39 9.156 6.91 7.66 7.66 8.22 3.258 6.166 6.166 6.187 5.41 6.54
a (in) 7.13 7.78 5.87 6.511 6.512 6.988 2.769 5.24 5.241 5.26 4.606 5.56
Mu1(k.ft) 402 656.3 280.6 344.6 459.5 595.3 374.2 223.2 297.7 223.2 201.1 293.0
Mu2(k.ft) 152 133.7 167.4 175.4 305.5 259.7 180.8 76.76 102.3 56.76 88.86 107.02
As1(in2) 5.44 7.37 3.996 4.428 5.904 7.128 5.328 3.564 4.75 3.576 3.654 4.41
As2(in2) 1.68 1.13 2.324 2.165 3.77 2.959 2.511 1.22 1.62 1.35 1.645 1.585
As (in2) 7.12 8.50 6.38 6.66 9.674 10.09 7.839 4.784 6.376 4.925 5.299 5.995
fs (ksi) 60 60 55.52 58.6 58.6 60 20.24 51.73 51.73 51.84 46.8 53.74
As (in2) 1.68 1.35 2.51 2.22 3.862 2.96 7.44 1.413 1.88 1.0425 2.12 1.77
Tension
Barsrows
6#10
Two
2#8
Two
2#10
Two
2#10
Two
9#6
Two
10#5
Two
6#10
Two
5#5
Two
10#4
Two
2#7
Two
7#5
Two
4#6
Two
Comp. bars 1#9 2#7 2#10 2#9 4#9 3#9 3#8 2#9 2#8 2#6 2#9 2#8
h (in) 24 28 22 24 24 26 22 20 20 20 18 21
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Problems 4.3:
DIM. No.a No.b No.c No.d No.e No.f No.g No.h
Mu (ft) 394 800 250 327 577 559 388 380
b (in) 48 60 44 50 54 48 44 46
bw (in) 14 16 15 14 16 14 12 14
t (in) 3 4 3 3 4 4 3 3
d (in) 18.5 19.5 15 13 18.5 17.5 16 15
Mn(k.ft) 468.2 803.3 340.8 329.9 681.6 569.2 366 356.3
R or_T R R R R R R T T
Ru (psi) 288 421 303 464 375 456
(%) 0.0063 0.769 0.646 0.558 0.76 0.804AS (in
2) 5.577 9.006 4.264 3.631 7.59 6.754
Bars
rows
10#7
Two
4#14
one
4#9
one
12#5
two
6#10
two
3#14
one
6#9
two
5#10
two
c/ dt 0.375 Yes Yes Yes Yes Yes Yes Yes Yes
Cf(k) 244.8 244.8
ASf(in2) 4.08 4.08
Muf(ft) 266.2 247.9
Muw (ft) 121.8 132.1
Ru (web) (psi) 476 503
w(%) 1.0 1.07
ASw (in2) 1.92 2.25
AS (in2) 6.0 6.33
AS (max) (in2) 7.85 11.7 6.75 7.06 10.50 9.10 6.68 6.93
h (in.) 21 23 18 18 22 21 20 19
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Problems 4.3: (continued)
DIM. No.i No .j No.k No.l No.m No.n No.o No.p
Mu (ft) 537 515 361 405 378 440 567 507
b (in) 60 54 44 50 44 36 48 46
bw (in) 16 16 15 14 16 16 12 14
t (in) 3 3 3 3 3 4 3 3
d (in) 16.5 17.5 15 15.5 13.5 18 22.5 18
Mn(k.ft) 516.4 495.7 340.8 401.6
R or_T T T T T R R R T
Ru (psi) 379 452 280
Barsrows
8#9two
6#10two
6#9two
7#9two
6#9two
5#10one
6#9two
7#9two
c/ dt 0.375 Yes Yes Yes Yes Yes Yes Yes Yes
Cf(k) 336.6 290.7 221.9 275.4
ASf(in2) 5.61 4.85 3.7 4.59
Muf(ft) 378.7 348.8 224.6 289.2
Muw (ft) 158.3 166.2 136.44 115.8
Ru (web) (psi) 436.12 407 485 413
w(%) 0.9 0.84 1.02 0.85
ASw (in2) 2.38 2.35 2.3 1.84
AS (in2) 7.99 7.2 6.0 6.43
AS (max) (in2) 9.19 8.64 6.75 7.53 6.49 7.30 8.25 3.58
h (in.) 20 21 19 19 20 21 26 22
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Problem 4.4:
(a) = max , d = 18.4 in. ; AS = 3.31 in.2
; Use 4#9 bars ( two rows. AS = 4.0 in.2
);h = 22 in. ; a = 5.84 in.; c = 6.87 in.; dt = 19.5 in. => c / dt =0.352 < 0.375 ; OK
(b) =0.016, d = 19.4 in. ; AS = 3.1 in.2
; Use 4#8 bars ( two rows. AS = 3.16 in.2
);h = 23 in. ; a = 5.47 in.; c = 6.43 in.; dt = 20.5 in. => c / dt =0.314 < 0.375 ; OK
(c) = 0.012, d = 21.9 in. ; AS = 2.63 in.2
; Use 3#9 bars ( AS = 3 in.2
);h = 25 in. ; a = 4.64 in.; c = 5.46 in.; dt = 22.5 in. => c / dt =0.243 < 0.375 ; OK
Problem 4.5:
Ru = 465 psi ; =0.0146 ; As = 3.5 in2
; Use 3# 10 bars (As = 3.79 in2
)
Problem 4.6:
= 0.0094 ; Ru = 465 psi ; As = 2.27 in2
; Use 2#10 bars ( AS = 2.53 in2)
Problem 4.7:
Assume a total depth = L / 15 = 2012 / 15 =16 in.
Mu = 132 k.ft(a) Using max = 0.01806, Ru(max) = 820 psi ; d = 12.7 in. ; AS = 2.75 in
2; Use 3#9 ;
h =16 in ; min < = max , t = 0.005
(b) Using s = 0.014, Ru = 662 psi ; d = 14.2 in. ; AS = 2.38 in2
; Use 4#7 bars (As = 2.41 in2
).
h= 17 in. ; min < < max , t > 0.005 OK
Problem 4.8:
Case 1:
(1) D.L. on ABC, L.L. on AB for maximum positive B.M. ; U = 4.84 k/ft.
(2) Positive ultimate B.M. = 305.15 k.ft; Let =0.015, Ru =700 psi, d = 20.88 in.(3) AS = 3.76 in
2; use 4#9 bars in one row.; bmin= 11.625 in.< 12 in. ; h = 24 in.
Case 2:D.L. on ABC, L.L. on BC only for maximum negative B.M.
d = 21.5 in. ; Mu = 154.88 k.ft ; Ru = 335 psi
= 0.00674; min < < max , t > 0.005 ; AS = 1.74 in2
; Use 2#9 bars or 3#7
Problem 4.9:
Mu = 317 k.ft ; Ru = 622 psi, min < < max , t > 0.005 ; d = 20.27 in.AS= 3.97 in
2; Use 4#9 bars ( AS = 4.0 in
2)
in one row. ; h = 23 in.
Problem 4.10:
Mu = 151.6 k.ft ; Ru = 476 psi, min < < max , t > 0.005 ; d = 17.8 in.
AS= 3.2 in2 ; Use 3#10 bars ( AS = 3.79 in 2 )in one row. ; h = 21 in.
Problem 4.11:
Mn = Rumaxbd2
= 222 k.ft < Mu ; Compression steel is needed.Mu1 = 222 k.ft ; Mu2 = 68 k.ft ; AS1= 3.10 in.
2; AS2 = 0.92 in.
2; AS = 4.02 i n .
2; Use 4 # 9
a = 6.08 in. ; c = 7.15 in.; dt = 20.5 in.; t =0.0056 > 0.005; OKA S' = 0.98 in.
2; AS(max)=5.27 in
2> As = 4.02 in
2; h = 23 in. OK.
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Problem 4.12:
a)d = 22.5 in. and AS = 3.05 in2;Use 3#9 bars in one row (AS = 3.0 in
2); h =26 in
Mu = 269.6 k.ft.; a = 7.06 in.; c = 8.3 in.; dt = 23.5 in.; t =0.0055>0.005;
b) Mu1 =195 k.ft, M u 2 = 65 k.ftFor Mu1 = 195 k.ft ; d = 19.5 in. and AS1 = 2.64 in
2;Let h = 23 in. and d = 2.5 in.
For M u 2 = 65 k.ft. ; As= AS2 = 0.85 in.
2
; AS = 3.49 in .
2
Use 5 # 8c) Mu = 260 k.ft
1) For two rows of steel h = 26 in, d =22.5 in and d t = 23.5 in.2) Section behaves as a T-section.
ASf= 1.28 in.2
; M u2 =120.5 k.ft ; Mu1 = 139.5 k.ft ; AS1 = 1.53 in.2
AS(max)=4.33 in2
> Total As = 2.81 in2
; OK ; Choose 3#9 bars, ( AS = 3.0 in2
)
Problem 4.13:
d = 19.8 in.; As = 6.8 in.2
; Use 7#9 bars (two rows. AS = 7.0 in.2
)if d = 22 in. is used; a=3.53 in. < 4 in. ; h = 26in.
AS(min) = 0.86 in.2
< AS < AS(max) = 7.78 in.2
OK
Problem 4.14:Ultimate load W = 7.9 k/ft
MB( maximum negative) = -395 K.ft
MDmax = 270.8 k.ft
b) Mn = 344.6 k.ft < Mu Compression steel is needed.
AS = 5.04 in .2
Use 7 # 8 ( two rows)A S' = 0.64 in.
2; AS1 = 3.12 in.
2;Use 4#8 bars in one row, ( A S = 3.16 in
2)
Problem 4.15:
a) W = 8.56 k/ft ; HA = HD = 34.24 k ; RA = RD = 154.08 k
MB = MC = - 616.32 k.ft. ; MBC (midspan) = 770.4 k.ftDesign beam BC for positive moment,AS = 8.86 in
2Use 9#9 bars (two rows, As = 9.0 in
2)
ASmax = 26.65 in.2
; AS(min)= mi nb w d = 1 . 0 9 in2
AS(min) < AS < AS(max) OK
Design beam BC for negative moment,
Mu =616.32 k.ft
Compression steel is needed.
AS = 7.93 in .2
Use 8 # 9 in two rows, ( AS = 8 in2
)A S'= 1.99 in.
2Use 2 # 9 in one row, ( A S = 2 in
2)
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CHAPTER 5
ALTERNATE DESIGN METHODS
Extra Problems
Problem E5.1
Tie AB kipsFu 75.58= 231.1 inAts = Provide 5 No. 5 Bar
255.1 inAts =
Tie CD kipsFu 0.15= 233.0 inAts = Provide 1 No. 4 Bar, 2 legs
240.0 inAts =
Tie BD & DF kipsFu 21.68= 252.1 inAts = Provide 5 No. 5 Bar
255.1 inAts =
Calculate strut widths
75.0=s psifce 2868382575.0 ==
Strut AC, inwkipsPu 26.1,83.86 ==
Strut BC, inwkipsPu 18.1,93.80 ==
Strut CE, inwkipsPu 90.1,67.130 ==
Strut DE, inwkipsPu 26.1,16 ==
sin
2 0.20
24 4.5 0.0032 0.003
Problem E5.2: Deep Beam Design Example: Bridge Bent Cap Using Strut and Tie Method
Load Combination: U = 1.8 [DL + (LL+I)]
PGirder= 1.8 x 580 kips = 1044 kip
WDL = 1.8 x 8.5 kips/lf = 15.3 kip/lfWLL = 1.8 x 7.6 kips/lf = 13.6 kip/ft
Strut And Tie Model:
P1 = 1195 kipLclr / Ds = 1.21 < 4 Bent Cap is considered a deep beam
Angle between strut and tie = 46d > 26d Good
Check Maximum Shear Strength of Beam X-Section: (ACI 318-08 11.7.3)Vu = 1195 kip
Assume d = 0.9 * h = 75.6 in
Vn = 2886 kip > Vu Good
Force Resultants:From Truss Analysis, internal forces are presented below. Since the bent cap is symmetrical, the
left side is designed and reinforcement will be applied symmetrically.
Member 1 and 2 (Tie) T1 = T2= 1020 kip TensionMember 3 (Strut) C3 = 1570 kip Compression
Member 4 (Strut) C4 = 1195 kip Compression
Member 5 (Strut) C5 = 910 kip CompressionMember 6, 7 and 8 (Tie) T6 = T7 =T8 = 0 kip
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Support Reaction D1x = -1016 kip, D1y = 1195 kip
Support Reaction D2x = 0 kip, D2y = 1195 kipSupport Reaction D3x = 627 kip, D3y = 655 kip
Calculate Effective Strength, fce:
Member 3 and Member 5 Bottle-shape struts fce = 3.18 ksi
Member 4 Uniform x-section strut fce = 4.25 ksiNodal Zone A C-C-T fce = 3.4 ksi
Nodal Zone B C-T-T fce = 2.55 ksi
Nodal Zone C C-C-T fce = 3.4 ksiNodal Zone D1 C-C-T fce = 3.4 ksi
Nodal Zone D3 C-C-T fce = 3.4 ksi
Dimensions of nodal zones:Nodal zone A Horizontal width wc = 1570 kip / 0.75 * 3.4ksi *72in = 8.55 in
Tie 1 width w1 = 8.55 * (1020kip / 1570kip) = 5.55 in
Strut 3 width w3 = 10.2 inNodal zone B Horizontal width wc = 1195 kip / 0.75 * 2.55 ksi *72in = 8.68 in
Nodal zone C Horizontal width wc = 910 kip / 0.75 * 3.4ksi *72in = 4.96 inTie 2 width w2 = 6.5 * (1020kip / 910kip) = 5.6 in
Strut 5 width w5 = 7.45 inNodal zone D1 and D3 Assume same widths as Nodal zone A and C
Nodal zone D2 Assume same widths as Nodal zone B
Check Capacity of Struts:Capacity of Strut 3 fns = *fce * Acs = 1751 kips > 1570 Good
Capacity of Strut 4 fns = *fce * Acs = 1195 kips = 1195 Good
Capacity of Strut 5 fns = *fce * Acs = 1280 kips > 910 GoodDesign of vertical and horizontal reinforcement:
Vertical Bars Use #6 @ 12 4 legged stirrups, As = 4 * 0.44 = 1.76 in^2Sin 50d = 0.76, As / (b * s) * sin = 0.0015
Horizontal Bars Use #8 @ 9 2 sides, As = 2 * 0.79 = 1.58 in^2
Sin 40d = 0.64, As / (b * s) * sin = 0.00156Design of horizontal tie 1 and 2:
Fu = *As*fy As = 1020 kip / (0.75 * 60 ksi) = 22.7 in^2
22.7 in^2 / (1 in^2) #9 Bars = 23
Use #9 Tot 24, Vertically Bundled in 2sAnchorage Length Ldh = 17 in
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CHAPTER 6
DEFLECTION AND CONTROL OF CRACKING
Problem 6.1(a)-6.1(f):
Description Prob. (a) Prob. (b) Prob. (c) Prob. (d) Prob. (e) Prob. (f)
b (in.) 14 20 12 18 16 14
d (in.) 17.5 27.5 19.5 20.5 22.5 20.5
h (in.) 20 30 23 24 26 24
As (in2.) 5 7.59 4.71 7.59 9.37 8
As (in
2
.) 0 0 0 2 2.53 2
Wd (k/ft.) 2.2 7 3 6 5 3.8
Wl (k/ft.) 1.8 3.6 1.5 2 3.2 2.8
Pd (k.) 0 0 0 0 12 8
Pl (k.) 0 0 0 0 10 6
Ig (in4.) 9333.3 45000 12167 20736 23434.7 16128
x (in.) 7.543 10.238 8.363 0 0 0
Icr (in4.) 5968.5 25247.2 7013.2 0 0 0
x (in.) 0 0 0 8.475 10.017 9.377
Icr (in4.) 0 0 0 12932.3 18042 12427.9
Mcr (k-ft.) 36.89 118.59 41.82 68.31 71.26 53.13
Ma (k-ft.) 200 530 225 400 520 400
Ie (in4.) 5989.6 25468.5 7046.3 12971.1 18055.8 12436.6
I (in.) 0.667 0.416 0.638 0.616 0.551 0.62
2 2 2 1.57 1.48 1.48
I (in.) 0.397 0.289 0.447 0.478 0.352 0.384
a (in.) 0.794 0.577 0.893 0.751 0.521 0.568
T = I +a (in.) 1.46 0.99 1.53 1.37 1.07 1.19
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Problem 6.2(a)-6.2(d):
Discription Prob. (a) Prob. (b) Prob. (c) Prob. (d)
b (in.) 15 18 12 14
d (in.) 20.5 22.5 19.5 20.5
h (in.) 24 26 23 24
As (in2.) 8 7.59 6.28 8
As (in2.) 2 0 1.57 2
Wd (k/ft.) 3.5 2 2.4 3
Wl (k/ft.) 2 1.5 1.6 1.1
Pd (k.) 0 7.4 0 5.5
Pl (k.) 0 5 0 4
Ig (in4.) 17280 26364 12167 16128
x (in.) 9.175 9.401 8.822 9.377
Icr (in4.) 12693.9 1540.3 8914 12427.9
x (in.) 0 0 0 0
Icr (in4.) 0 0 0 0
Mcr (k-ft.) 56.92 80.16 41.82 53.13
Ma (k-ft.) 396 381 288 393
Ie (in4.) 12707.5 15505.9 8923.9 12437.1
I (in.) 0.538 0.457 0.558 0.577
1.51 2 1.5 1.48
I (in.) 0.342 0.29 0.33 0.42
a (in.) 0.516 0.58 0.5 0.63
T = I +a (in.) 1.05 1.042 1.06 1.203
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Problem 6.3:1.) d = 23.27 in. ; As = 5.0 in.
2use 5 #9 ; say h = 27 in.
2.) Ig = 19683 in.4
; Ma = 3916 k.in. ; Mcr= 691.1 k.in.4 411268 in. ; 11217 in.cr e g I I I= = <
in.12.1=
Problem 6.4:1.) Mu = 444 k.ft.
d = 20.82 in. ; Total As = 5.57 in.2 Use 6 #9, As = 6.0 in.
2
Compression steel, fs = 59 ksi.
h = 24.32 in., say 25 in. ; d = 21.5 in. ; For comp. steel, use 2 #7, As = 1.2 in.2
2.) Ig = 15625 in.4
; Ma = 3916 k.in. ; Mcr= 592.93 k.in.4 410750 in. ; 10767 .cr e g I I in I= = <
in.17.1=
Additional long-term deflection = 1.29 in.Total deflection = 2.46 in.
Problem 6.5:
a) S = 10.31 in less than 12 in.; provided spacing = 3.5 in., OK.W=0.0121 in.
b) W = 0.011 in.
c) W = 0.0103 in.
d) W = 0.011 in.
Problem 6.6:
As (min.) = 0.126 in
2
/ft./side.; Max. spacing = 6.75 in., say 6.5 in.As/side = (0.12640.5) / (212) = 0.212 in
2; Choose 3 #3 bars each side.
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CHAPTER 7
DEVELOPMENT LENGTH OF REINFORCING BARS
Problems 7.1(a) - 7.1(j)
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)
fc(ksi.) 3 4 5 3 4 5 5 3 4 4
fy (ksi.) 60 60 60 40 60 60 60 40 60 60
Bar No. 5 6 7 8 9 10 11 9 8 6
db (in.) 0.625 0.75 0.875 1.0 1.128 1.27 1.41 1.128 1.0 0.75
Clear
Cover (in.)2.0 2.0 2.0 2.5 1.5 2.0 3.0 2.0 2.0 1.5
Clear
Spacing(in.)2.25 2.5 2.13 2.3 1.5 2.5 3.0 1.5 1.75 1.65
Cond. met Y Y Y Y N Y Y N N Y
t 1.0 1.0 1.0 1.3 1.0 1.0 1.0 1.0 1.0 1.3
e 1.0 1.0 1.5 1.0 1.0 1.0 1.0 1.5 1.0 1.5
1.0 0.75
Recommended