Given: L1 = 8m S1 = 2.5 m L2 = 8 m S2 = 3 m L3 = 8 m L4 = 8 m Total dead load = 4.6 kPa Live load = 4.8 kPa The interior beam KLMNO is to be analyzed for the maximum forces at ultimate condition, U= 1.2D+1.6L Calculate the maximum reaction (kN) at N. 332 345 354 362 (continuation) Calculate the minimum reaction (kN) at N. 139 235 125 92 (continuation) Which of the following most nearly gives the maximum positive moment (kN-m) for span KL. 180 210 230 170

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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Given: L1 = 8m
S1 = 2.5 m
L2= 8 m
S2 3 m
L3 = 8 m
L4 = 8 m
Total dead load = 4.6 kPa
Live load = 4.8 kPa
The interior beam KLMNO is to be analyzed for the maximum forces at ultimate condition, U=
1.2D+ 1.6L
Calculate the maximum reaction (kN) at N.
332
345
354
362
(continuation)
Calculate the minimum reaction (kN) at N.
139
235
125
92
(continuation)
Which of the following most nearly gives the maximum positive moment (kN-m) for span KL.
180
210
230
170
Transcribed Image Text:Given: L1 = 8m S1 = 2.5 m L2= 8 m S2 3 m L3 = 8 m L4 = 8 m Total dead load = 4.6 kPa Live load = 4.8 kPa The interior beam KLMNO is to be analyzed for the maximum forces at ultimate condition, U= 1.2D+ 1.6L Calculate the maximum reaction (kN) at N. 332 345 354 362 (continuation) Calculate the minimum reaction (kN) at N. 139 235 125 92 (continuation) Which of the following most nearly gives the maximum positive moment (kN-m) for span KL. 180 210 230 170
SITUATION.
Refer to Fig. RC-001 for the floor framing plan and SE-103 for the load diagrams.
K
Fig. RC-001
BEAM DIAGRAMS AND DEFLECTIONS
For various static loading conditions
For meang el synbe, s pa 2.293
37. CONTINUOUS BEAM-FOUR EQUAL SPANS-THIRD SPAN UNLOADED
wl
C
RA-0.380 u Ra1.223 wl Rc0.357 uel Ro 0.598 wl
18
1.
RE 0.442 wl
0.603 wt
0.620 w
-0.1205 uf
0.558 ul
0.040 u
0.380 ul
0.42 ul
SHEAR
0.397 wl
-0.0179 wl
-0.058 w
+0.072 w
+0.0611
+0.0977 ul
MOMENT
0.380
0.603
0.442
A Ma. ( Ifrem )- /
38. CONTINUOUS BEAM-FOUR EQUAL SPANS-LOAD FIRST AND THIRD SPANS
el
E
A.
B.
1.
RA0.446 ul Re0.572 wl Rc 0.464 ul Ro =0.572 wl Re-0.054 wl
0.482 sol
0.446 wl
0.018 e
0.054 ul
0.054 al
0.554 wl
0.518 wl
SHEAR
-0.0536 ul
-0.0357 ul
-0.0536 ul
+0.0996 ul
+0.0805 wl
MOMENT
0.446
0.518 /
AMa. 47 I trom A)-0.007 /
39. CONTINUOUS BEAM-FOUR EQUAL SPANS ALL SPANS LOADED
wl
1.
RA 0.393 wt Re1.143 al Rc- 0.928 wl Rp-1.143 ied RE 0.393 wl
0.464 ul
0.607 l
0.464 sol
-0.0714 o
0.393 ul
0.536 ul
0.536 ul
0.393 ud
0.607 sol
SHEAR
-0.1071 w
-0.1071
+0.0772 w
+0.0772 w
t0.0364 u t0.0364 w
MOMENT
0.393
0.536
0.3932
0.536I
A Max. ( frem A and -.0S/E
FIG SE-103
Transcribed Image Text:SITUATION. Refer to Fig. RC-001 for the floor framing plan and SE-103 for the load diagrams. K Fig. RC-001 BEAM DIAGRAMS AND DEFLECTIONS For various static loading conditions For meang el synbe, s pa 2.293 37. CONTINUOUS BEAM-FOUR EQUAL SPANS-THIRD SPAN UNLOADED wl C RA-0.380 u Ra1.223 wl Rc0.357 uel Ro 0.598 wl 18 1. RE 0.442 wl 0.603 wt 0.620 w -0.1205 uf 0.558 ul 0.040 u 0.380 ul 0.42 ul SHEAR 0.397 wl -0.0179 wl -0.058 w +0.072 w +0.0611 +0.0977 ul MOMENT 0.380 0.603 0.442 A Ma. ( Ifrem )- / 38. CONTINUOUS BEAM-FOUR EQUAL SPANS-LOAD FIRST AND THIRD SPANS el E A. B. 1. RA0.446 ul Re0.572 wl Rc 0.464 ul Ro =0.572 wl Re-0.054 wl 0.482 sol 0.446 wl 0.018 e 0.054 ul 0.054 al 0.554 wl 0.518 wl SHEAR -0.0536 ul -0.0357 ul -0.0536 ul +0.0996 ul +0.0805 wl MOMENT 0.446 0.518 / AMa. 47 I trom A)-0.007 / 39. CONTINUOUS BEAM-FOUR EQUAL SPANS ALL SPANS LOADED wl 1. RA 0.393 wt Re1.143 al Rc- 0.928 wl Rp-1.143 ied RE 0.393 wl 0.464 ul 0.607 l 0.464 sol -0.0714 o 0.393 ul 0.536 ul 0.536 ul 0.393 ud 0.607 sol SHEAR -0.1071 w -0.1071 +0.0772 w +0.0772 w t0.0364 u t0.0364 w MOMENT 0.393 0.536 0.3932 0.536I A Max. ( frem A and -.0S/E FIG SE-103
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