钢结构设计英文课程 作业及答案 Assignment 1_Solution
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x = 0 ⇒ Maximum positive bendng moment and the corresponding axial force
M B =1.25 MD +1.4 Mw =1.25(0) + 1.4(+20) = +28 KNm
N B = α D N D + α w N w = 0. + 1.4(−2.5) =+ −3.5kN compression
Combination 3: Combination 3 is non-critical.
1
The shown structural system is subjected to two types of loading, dead load (D) and wind load (W). Wind load W can act either from left to right or from right to left. a) List all the factored load combinations that need to be considered in designing the structural system according to the National Building Code of Canada requirements. b) For each of the two loading cases (Dead, Wind Left to Right) determine the NFD, QFD, and BMD c) For Member A, determine the maximum factored tensile force and the corresponding bending moments d) For Member A, determine the maximum factored internal compressive force and corresponding bending moment e) For Member B, determine the maximum factored bending moment and corresponding axial forces.
Assignment 1_Solution
D= 2 kN/m
W Member A
Member B W= 5 kN
4m
W
4m
Solution: a) Combination 1: dead load alone= 1.4D Combination 2: dead load + wind load left to right= 1.25D+1.4W Combination 3: dead load + wind load left to right= 0.9D+1.4W
F
= 0 , YF = 5kN
20 kNm
2.5kN
2.5kN
+
5kN
2.5kN
+
5kN
+
5kN QFD 5kN
+
20 kNm
+
BMD
5kN 5kN 5kN NFD
c) Refer to NFD and BMD for member A in (b) Combination 1: N A = α D N D = 1.4(−4) = −5.6 KN MA = 0 Combination 2: N A = 1.25 N D + 1.4 N w = 1.25(−4) + 1.4(+5) = +2.0 KN M A = 1.25M D + 1.4 M w = 1.25(0) + 1.4(+20) = +28 KNm Combination 3: N A = 0.90 N D + 1.40 N w = 0.9(−4) + 1.40(+5) = +3.4 KN Æ Maximum Tensile Force M A = 0.90M D + 1.40M w = 1.25(0) + 1.4(+20) = +28KNm Maximum factored internal compressive force and corresponding bending moment appears for combination 1: N A = -5.6 KN (compression) , MA = 0 d) From (c) maximum factored internal tensile force and corresponding bending moment appears for combination 3: N A =3.4KN (tension), M A = +28.KNm
3
e) Refer to NFD and BMD for member B in b Combination 1: M B = α D M D =1.4(4) = 5.6 KNm NB = 0
Combination 2:
Maximum positive bending moment occurs within the member but at the left end.
(ii) Wind loads (left to right) Alone W ∑ X = 0 ⇒ X E − 2.5 − 2.5 = 0 , X E = −5kN
∑ M = 0 ⇒ ( −2.5)(4) + (−2.5)(4) + 4Y ∑Y = 0 ⇒ Y + 5 = 0 , Y = −5kN
E E E
b) (i) Dead load Alone ∑ X = 0 ⇒ XE = 0
1 Given the symmetry, YE = YF = (2)(4) = 4kN (Compression) 2
D=2 kN/m 4 kN 4 kN
4m
XE
E YE 4m
F YF
4 kN
NFD
4 kN
QFD
BMD
2
M B =1.25 MD +1.4 Mw =1.25(0) + 1.4(+20) = +28 KNm
N B = α D N D + α w N w = 0. + 1.4(−2.5) =+ −3.5kN compression
Combination 3: Combination 3 is non-critical.
1
The shown structural system is subjected to two types of loading, dead load (D) and wind load (W). Wind load W can act either from left to right or from right to left. a) List all the factored load combinations that need to be considered in designing the structural system according to the National Building Code of Canada requirements. b) For each of the two loading cases (Dead, Wind Left to Right) determine the NFD, QFD, and BMD c) For Member A, determine the maximum factored tensile force and the corresponding bending moments d) For Member A, determine the maximum factored internal compressive force and corresponding bending moment e) For Member B, determine the maximum factored bending moment and corresponding axial forces.
Assignment 1_Solution
D= 2 kN/m
W Member A
Member B W= 5 kN
4m
W
4m
Solution: a) Combination 1: dead load alone= 1.4D Combination 2: dead load + wind load left to right= 1.25D+1.4W Combination 3: dead load + wind load left to right= 0.9D+1.4W
F
= 0 , YF = 5kN
20 kNm
2.5kN
2.5kN
+
5kN
2.5kN
+
5kN
+
5kN QFD 5kN
+
20 kNm
+
BMD
5kN 5kN 5kN NFD
c) Refer to NFD and BMD for member A in (b) Combination 1: N A = α D N D = 1.4(−4) = −5.6 KN MA = 0 Combination 2: N A = 1.25 N D + 1.4 N w = 1.25(−4) + 1.4(+5) = +2.0 KN M A = 1.25M D + 1.4 M w = 1.25(0) + 1.4(+20) = +28 KNm Combination 3: N A = 0.90 N D + 1.40 N w = 0.9(−4) + 1.40(+5) = +3.4 KN Æ Maximum Tensile Force M A = 0.90M D + 1.40M w = 1.25(0) + 1.4(+20) = +28KNm Maximum factored internal compressive force and corresponding bending moment appears for combination 1: N A = -5.6 KN (compression) , MA = 0 d) From (c) maximum factored internal tensile force and corresponding bending moment appears for combination 3: N A =3.4KN (tension), M A = +28.KNm
3
e) Refer to NFD and BMD for member B in b Combination 1: M B = α D M D =1.4(4) = 5.6 KNm NB = 0
Combination 2:
Maximum positive bending moment occurs within the member but at the left end.
(ii) Wind loads (left to right) Alone W ∑ X = 0 ⇒ X E − 2.5 − 2.5 = 0 , X E = −5kN
∑ M = 0 ⇒ ( −2.5)(4) + (−2.5)(4) + 4Y ∑Y = 0 ⇒ Y + 5 = 0 , Y = −5kN
E E E
b) (i) Dead load Alone ∑ X = 0 ⇒ XE = 0
1 Given the symmetry, YE = YF = (2)(4) = 4kN (Compression) 2
D=2 kN/m 4 kN 4 kN
4m
XE
E YE 4m
F YF
4 kN
NFD
4 kN
QFD
BMD
2