Chapter 2
Specifications, Loads, and Methods of Design
Problem 2-1
1. 1.41.4(40) D == 56 psf
2. 1.21.6(0.5or0.3or0.5) r DLLSR ++
a. 1.2(40) ++1.6(0)0.5(20) = 58 psf
b. 1.2(40) ++1.6(0)0.3(30) = 57 psf
c. 1.2(40) ++1.6(0)0.5(14) = 55 psf
3. 1.2 D ++ (1.6 L r or1.0 S or1.6 R )( L *or0.5W )
a. 1.2(40) + 1.6(20) + 0 =
b. 1.2(40) + 1.0(30) + 0 =
c. 1.2(40) + 1.6(14) + 0 =
4. 1.2 D + 1.0(WW + Tr ) ++ L *(0.5 L or0.3S or0.5 R )
a. 1.2(40) + 1.0(0) + 0 + 0.5(20) = 58 psf
b. 1.2(40) + 1.0(0) + 0 + 0.3(30) = 57 psf
c. 1.2(40) + 1.0(0) + 0 + 0.5(14) = 55 psf
5. 0.9 D + 1.0(WW or T ) 0.9(40) + 1.0(0) = 36 psf (No Uplift)
6. 1.2 DE++ L * + 0.15 S 1.2(40) + 0 + 0 + 0.15(30) = 52.5 psf
7. 0.9 DE + 1.0
0.9(40) + 1.0(0) = 36 psf (No Uplift)
Governingfactoredload = 80psf(Case3a)
Problem 2-2
1. 1.4 D = 1.4(15,000) = 21,000 lb
2. 1.21.6(0.5or0.3or0.5) r DLLSR ++ 1.2(15,000)1.6(0)0.3(18,000) ++ = 23,400 lb
3. 1.2 D ++ (1.6 L r or1.0 S or1.6 R )( L *or0.5W ) 1.2(15,000) ++1.0(18,000)0.5(36,000) = 54,000 lb
4. 1.2 D + 1.0(WW or Tr ) ++ L *(0.5 L or0.3S + 0.5 R )
1.2(15,000) + 1.0(36,000) + 0 + 0.3(18,000) =
5. 0.9 D + 1.0(WW or T ) (Uplift)
0.9(15,000) +−1.0(36,000) =
6. 1.2 DE++ L * + 0.15 S
1.2(15,000) + 40,000 + 0 + 0.15(18,000) =
7. 0.9 DE + (Uplift)
0.9(15,000) −= 40,000
Maximumfactoredloads: +60,700lb Case6 26,500lb Case7
Problem 2-3
1. 1.4 D = 1.4(40) =
2. 1.2 DL++1.6(0.5 L r or0.3S or0.5 R ) 1.2(40) ++1.6(56)0.5(16) =
3. 1.2 DL++ (1.6or1.0 S or1.6 R )(0.5 L or0.5W ) r a. 1.2(40) + 1.6(16) + 0.5(56) =
lb
k b. 1.2(40) + 1.6(16) + 0.5(64) =
4. 1.2 D + 1.0(WWor) ++ 0.5 L (0.5 L or0.3S or0.5 R ) Tr
k
1.2(40) ++1.0(64)0.5(56) + 0.5(16) = 148.0 k
5. 0.9 D + 1.0 (Uplift) W
0.9(40) + 1.0( 40) = −4 k
6. 1.2 DE++ 0.5 L + 0.15 S 1.2(40) ++750.5(56) + 0.15(0) = 151 k ←
7. 0.9 DE + 0.9(40) + ( 50) = −14 k ←
Maximumfactoredloads: +151k downward (Case6) 14k upward (Case7)
Problem 2-4
1. 1.4 D = 1.4(30) = 42 psf
2. 1.2 DL++1.6(0.5 L r or0.3S or0.5 R )
1.2(30) ++1.6(0)0.5(16) =
3. 1.2 D ++ (1.6 L r or1.0 S or1.6 R )( L *or0.5W )
1.2(30) + 1.6(16) + 0.5(40) =
4. 1.2 D + 1.0(WWor) ++ L *(0.5 L r or0.3S + 0.5 R ) T
1.2(30) + 1.0(40) + 0 + 0.5(16) =
5. 0.9 D + 1.0(WW or T ) 0.9(30) + 1.0( 40) =
6. Does not control (By inspection)
7. Does not control (By inspection)
Maximumfactoredloads: +84psf Case4 13psf Case5
Problem 2-5
DL
= 70psf = 80psf
Beamspacing = 7 ′-6′′ or7.5ft
D (plf) = 70(7.5) = 525plf
L (plf) = 80(7.5) = 600plf
1. 1.4 D = 1.4(525) = 735 plf
2. 1.2 DL++1.6(0.5 L r or0.3S + 0.5 R )
1.2(525) + 1.6(600) + 0 = 1590 plf ←
3. 1.2 DL + (1.6 r or1.0 S or1.6 R )(0.5 L or0.5) W + 1.2(525) + 0 + 0.5(600) = 930 plf
4. 1.2 D + 1.0(WW or Tr ) ++ 0.5 L (0.5 L or0.3S or0.5 R )
1.2(525) + 0 + 0.5(600) + 0 = 930 plf
5. 0.9 D + 1.0(W or T ) (NoUplift) W
0.9(525) + 1.0(0) = 472.5 plf
6. 1.2 DE++ 0.5 L + 0.15 S 1.2(525) + 0 + 0.5(600) + 0 = 930 plf
7. 0.9 DE + (No Uplift)
0.9(525) + 0 = 472.5 plf
Maximumfactoredload = 1590plf Case2
Problem 2-6
DS = 25psf, = 36psf, L r = 20psf, W = 36psf ↑−()or18psf ↓+()
Beamspacing = 5′-0′′ or5ft
(plf) = 25(5) = 125plf (plf) = 36(5) = 180plf r (plf) = 20(5) = 100plf DS L W (plf) =−= 36(5) 180plf or 18(5) =+90plf
1. 1.4 D = 1.4(125) = 175 plf
2. 1.2 DL++1.6(0.5 L r or0.3S + 0.5 R )
a. 1.2(125) ++1.6(0)0.5(100) = 200 plf
b. 1.2(125) ++1.6(0)0.3(180) = 204 plf
3. 1.2 D ++ (1.6 L r or1.0 S or1.6 R )( L *or0.5W )
a. 1.2(125) + 1.6(100) + 0.5(90) =
b. 1.2(125) + 1.0(180) + 0.5(90) =
4. 1.2 D + 1.0(WW or Tr ) ++ L *(0.5 L or0.3S or0.5 R )
plf
a. 1.2(125) + 1.0(90) + 0 + 0.5(100) = 290 plf b. 1.2(125) + 1.0(90) + 0 + 0.3(180) =
5. 0.9 D + 1.0(WW or T )
plf
0.9(125) +−1.0(180) = −67.5 plf ←
6. 1.2 DE++ L * + 0.15 S
1.2(125) + 0 + 0 + 0.15(180) = 177 plf
7. Does not control – By inspection No earthquake force, E forces
Maximumfactoredloads:
375plf Case3b Downward 67.5plf Case5 Upward
Problem 2-7
1. D = 40 = _ 40 psf
2. DL+= 40 + 0 = 40 psf
3. DL + ( r or0.7 S or R )
a. 40 + 20 =
b. 40 + 0.7(30) =
c. 40 += 14
4. D ++ 0.75 L 0.75( L r or0.7 SRor)
psf
a. 40 + 0.75(0) + 0.75(20) = 55 psf
b. 40 + 0.75(0) + 0.75(0.7)(30) =
c. 40 + 0.75(0) + 0.75(14) = 50.5 psf
5. D + 0.6(WW or T ) 40 + 0.6(0) = 40 psf
6. D ++ 0.75 L 0.75(0.6(WW or Tr )) + 0.75( L or0.7 S or R )
a. 40 + 0.75(0) + 0.75(0.6)(0) + 0.75(20) =
b. 40 + 0.75(0) + 0.75(0.6)(0) + 0.75(0.7)(30) =
c. 40 + 0.75(0) + 0.75(0.6)(0) + 0.75(14) =
7. 0.6 D + 0.6(WW or T )(NoUplift)
psf
0.6(40) += 0.6(0) 24 psf
8. 1.0 DE + 0.7
1.0(40) += 0.7(0) 40 psf
9. 1.0 D + 0.525 E ++ 0.75 LS 0.1
1.0(40) + 0.525(0) + 0.75(0) += 0.1(30) 43 psf
10. 0.6 D + 0.7 (NoUplift) E
0.6(40) + 0.7(0) = 24 psf
Governingload = 61psf Case3b
Problem 2-8
1. D = 15,000 = 15,000 lb
2. DL+= 15,000 + 0 = 15,000 lb
3. DL + ( r or0.7 S or R ) 15,000 + 0.7(18,000) = 27,600 lb
Problem 2-11
D = 70psf L = 80psf
Beamspacing = 7 ′-6′′ or7.5ft
D (plf) = 70(7.5) = 525plf
L (plf) = 80(7.5) = 600plf
1. D = 525 = 525 plf
2. DL+= 525 + 600 = 1125 plf ←
3. DL + ( r or0.7 S or R )
525 + 0 = _ 525 plf
4. D ++ 0.75 L 0.75( L r or0.7 SRor)
525 + 0.75(600) + 0.75(0) = 975 plf
5. D + 0.6(WW or T )
525 + 0.6(0) = 525 plf
6. D ++ 0.75 L 0.75(0.6(WW or Tr )) + 0.75( L or0.7 S or R )
525 + 0.75(600) + 0.75(0.6)(0) + 0.75(0) = 975 plf
7. 0.6 D + 0.6(W or T ) (Uplift) W
0.6(525) + 0.6(0) = 315 plf (No Uplift)
8. 1.0 DE + 0.7
1.0(525) + 0.7(0) = 525 plf
9. 1.0 D + 0.525 E ++ 0.75 LS 0.1
1.0(525) + 0.525(0) + 0.75(600) + 0.1(0) = 975 plf
10. 0.6 D + 0.7 (Uplift) E
0.6(525) += 0.7(0) 315 plf (No Uplift)
Governingload: 1125plf Case2
Problem 2-12
DS = 25psf; = 36psf; L r = 20psf, W = 36psf ↑−() 18psf ↓+()
Beamspacing = 5′-0′′
D (plf) = 25(5) = 125plf
L r (plf) = 20(5) = 100plf
S (plf) = 36(5) = 180plf
W (plf) = 36(5) =−180plf = 18(5) =+90plf
1. D = 125 = 125 plf
2. DL+= 125 + 0 = 125 plf
3. DL + ( r or0.7 S or R )
a. 125 + 100 = 225 plf
b. 125 + 0.7(180) = 251 plf
4. D ++ 0.75 L 0.75( L r or0.7 SRor)
a. 125 + 0.75(0) + 0.75(100) =
b. 125 + 0.75(0) + 0.75(0.7)(180) =
plf
plf
5. D + 0.6(WW or T ) 125 + 0.6(90) = 179 plf
6. D + 0.75 L 0.75(0.6(WW or Tr )) + 0.75( L or0.7 S or R ) +
a. 125 + 0.75(0) + 0.75(0.6)(90) + 0.75(100) =
b. 125 + 0.75(0) + 0.75(0.6)(90) + 0.75(0.7)(180) =
7. 0.6 D + 0.6(WW or T )
0.6(125) + 0.6( 180) = −33 plf ←
Cases 8, 9, 10 Do not control – By inspection No earthquake force, E forces
Governingloads: 260plf Case6b 33plf Case7