(1)Consider the truss loaded and supportedAs shownForce (KN)T/CMember3 m3m--4 m--4 m---4 m-(1) AB(2) AC(3) BD(4) BC(5) CD(6) DE(7) CE(8) EF(9) GE(10)1.75 KN2kN1.5 IN2IN0.75 IN6m-6 m.6mG(11)GF(12)H,

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Answered: (1)Consider the truss loaded and…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Principle of superposition The beam is indeterminate to the first degree. Force in member BD is chosen as the redundant. This member is therefore sectioned to eliminate its capacity to sustain a force. 60 kN/m 3 m 1.8 m 60 45 actual structure (a) 160 kN/m B FaD redundant рrimary structure ARD FRp applied Fap D
PROBLEM 1 P1, P2 = 30 kips %3D Determine the internal force in the ff. members: [ Select] 1. BD [ Select ] 2. CD [ Select] v 3. CE P1 P2 В D F H. 7.5 ft E 4 panels at 10 ft = 40 ft- >
x Determin the design Compressive 12-5 Strength of the Cross- sectiona Column shown with unbraced length kala= 35m, Kyly=2m A-118:75 I= 10872 Iy= 3799 J-247 8a e =4.13 c ? J= 247-8 %3D Tu = 9.57 C 5.746 E=2-1x
De term ine the kin d and am ount of stress in e ach mem ber of the truss shown by the method of jo in ts in figure P 5-2 (page 76).- 1000lb 1000lb 1000lb 1000Ib 1000 Ib U. h = 20 ft 500lb 500 Ib L. L. RAV 6 pane ls at 10' - 6 ft RBy Note: 1. F o llo w strictly the lecture notes g iv en. Draw the FBD of e ach jo in t 2. Sum m a rize and tabu la te your answers. 3. Compu te the bar forces up to member U,L3 on ly because the truss lo a d in g is sym metric aland the bar forces of right sid e members are id en tic al w ith th e left s ide members. But you in c lu de th e rig ht sid e mem bers in have to the tabu la tion. 4. Show the com puta tio n ho w the leng th s were obta in ed. 5. Look for your own data in the table provided. 6. Copy the problem , data of problem and fig ure of the proble m TABULAT ION FOR SUMMAR Y OF COMPUTATION OF BAR FORCES METHOD OF JOINTS (PROBLEM P5-2 PAGE 76) Mem ber Identification Length Bar Forc e Туре of ft Ib force LOL1 10 3750 Tensio n L1L 2 10 Tension…
954. The 2014-T6 aluminum rod AC is reinforced with the firmly bonded A992 steel tube BC. When no laad is applied to the assembly, the gap between end C and the rigid support is 0.5 mm. Determine the support reactions when the axial force of 400 kN is applied. 400 mm 400 kN A992 steel So mm S00 mm 25 mi2014-T6 aluminum alloy Section a-a
Higher Order L.D.E (D3+6D2+5D-12)y=0
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A steel pipe (1) is attached to an aluminum pipe (2) at flange B. The pipes are attached to rigid supports at A and C, respectively 18 m 14 m Member (a) has a cross-sectional area A,-3,600 mm2, an elastic modul as a cross-sectional area A 3,600 mm2, an elastic modulus E-200 GPa, and an allowable has a cross-sectional area A2 2,o00 mm2, an elastie modulus E-70 GPa, and an allowable normal stress of 120 MPa. Determine the maximum load Pthat can be applied to fange B without exceeding either allowable stress
TTTT Draw AFD, SFD and BMD for the following frame, and determine the following • Vertical Reaction at A • Vertical Reaction at D • Horizontal Reaction at G • Maximum Axial Force • Maximum Shear Force • Maximum Bending Moment 40kN/m F H 30kN/m 6m 4m Зт 4m C 12m I 40kNm 6m 7m 6m 75KN B y D 3m TTTT >x A
|%A• In. ? Asst. Prof. Muhammed Abbas المعلم منذ ۱۲ ساعات مضت. H.W.2: Find the 40 k force in each G member of the 5 k truss. 16 ft V17 F 10 k 10 k 16 ft 3 B - 8 ft- 8 ft- 4 ft 4 ft 4 ft 4 ft 4.
DO NO OWAVE www.oukpak.com F06 unic 30KN 150 KN 20KN 3m =30" SOLVE FOR ACL SUPPORT 2m HINGE REACTIONS LN DICATE THE DIRECTICNS e UNITC 20 A B im
-55./Draw the influence line for the force in member CJ. then compute the maximum live force (tension or compression) that can be developed in this member due to a uniform live load of 3 kN/m which is transmitted to the *6-60. E DE of th force (te- member truss along the bottom cord. transmi

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(1)Consider the truss loaded and supported As shown Member Force (KN) T/C 3 m3 m -4m--4 m---4 m- (1) AB (2) AC (3) BD (4) BC (5) CD (6) DE (7) CE (8) EF (9) GE (10) 1.75 KN 2kN 15 IN 2AN IAN 0.75 IN 6m -6 m -6 m G (11)GF (12)H,