Question

task 3

TASK 3 - Beams & Theory of Bending Aircraft beam structures can be represented using free-body diagrams in order to carry out certain calculations of the loads exerted on them. a) Define and Explain Neutral As, Section Modulus and Bending Stress Support you explanation with diagrams and provide the derivation of equations where applicable. b) The following diagram represents a beam, and shows the various loads that act upon in when it is placed on its ends. 16.0 KN 20 kN/m --1.0m -1.0 m 2 .0 m 4.0 m For the given beam: a. Calculate the support reactions at B and E b. Construct the shear force diagram c Construct the bending moment diagram d. Given that the section modulus of the spar is 3250 cm, determine the allowable bending stress, while ignoring the weight of the spar. The below diagram represents the distributed load acting on the wing of a small airplane. Calculate the shear force and bending moment at the inboard end of the wing. 1600 N

Answer 1

Beam 0 0 a) Neutral anis - When a latercel lead is appled on the bean. The beam bends and due to bending bending stresses are induced on bean I Bending stresses are basically I tensile or compressure stress ou can see in the NL sigure due to load P. there will be compression, Gossection of Bean or upper fibres of bean while there is compression on lower fibres of beam. There will be only one section on between them where there is neather compression nor tensia. That fibre vor correponds to neutral anes of beam. It passes through ca of crossection!

6) is first of all we will calculate reaction of the beam. Let RA cand Ro be the reaction fornes cecting on beam in upward direction. EF=0 (summation of vertical force to zero since beam es in equilibrin) 6+ 2x2 =RAT RB 10. RATRE Taking moment about A Rox 4= 6x1+ 2x2x3 Ro= 6 + 12 = 4.50 KN 4 So Ra = 10-4.50 = 5.50 kN. Now we will draw shear forme and bending moment diagram for ( this

6 KN 2KN/m SHAB – 5.5kN NAB A Suco - (5.5-6) KN Stitclean 14.5 SADR= +5.5-6-2(x-2) KN 5.5kNE O.NET S.FD I (Shea fore chagran) e chergran 4.5KN Iron Benden moment ego B.M= 5.52 B.MCD-5.5n-6(x-1) 5.5kNet – 5KNM B.MDB= 5.5 -6(n-1) - 2x(2-2² 2 B.M ( = 5.561 -5.5KN. B.M. - 5.5x2 - 6x1 = 5 KN

d. Greien z. I. 3250 cm? - 3250x100 m W nan Bending moment is nasuminat pt. c so we will calculate bending stress at this point, We will use this formular M-5 I gran Mx non a co non = 5.5x10²x 3250 x 10-6 B = 16.92307.692 Parcere 15.-1.692 MPa manimum bending stress endured in the becam,

w c) Wing is acting like a beam on which lateral fous are acting (cantilever becam) 16000 900N) At-2.6 - 216 -- - We have to calculate shear force and bendery moment at pt A of bean 700 N/m nou We broke trapezoid into U.D. and no vrangular load of haght 700 N so sha forn at pt A SA = 1x 700 x 2.6 + 900 X (2.6 +26) + 4x900x! Sn = 6040 KN

Bendery moment at pt AA : Considuers only magnitude B.Motx 700 x 2.6 x 2.6 + 900 x 5.2% 2.6 + 4x900x1x (3.6 +246 + ) (B.Mn = 15431.66 kN-m) 2 çig of all loads acting -2.6 (2.6+216 + 5) obou oblained value care shear foru & Bendery moment acting at inboard end of wing