Beam Reactions - Moments of Inertia - Engineering Assignment Help

Assignment Task

Problem 1 – Beam Reactions
Data: q, l, P= ql, M = ql2
l1 = l3 = l, l2 = l4 = ? l.
M1 = ql2 , M2 =?? ql2
P1 = ql , P2 =?? ql
Free Body Diagram of separated elements (separation at C) - loads and reactions (symbols)

Loads and reactions of separated elements (values)
181306-635000
Loads and reactions of the beam (values) with connected elements
1638304602400
Degree of static indetermination, finding reactions: SN =
Checking of the balance of forces (equations of equilibrium of the beam with connected elements):

Problem 2 – Internal Forces
Known: a, q, P=qa, M =qa2
25844517316450-6351402080Free Body Diagram - loads and reactions (symbols)
Free Body Diagram - loads and reactions (symbols)
Loads and reactions (values), cross-sections for internal force analysis
99060174625N??? qa
Diagrams of internal forces
qa2

Diagrams of internal forces

Degree of static indetermination, finding reactions:
Schemes for analysis of internal forces:
Functions of internal forces:
At least 3 result verifications: boundary conditions, T?? or M? jumps at sections of force or moment application, max M? (x) at sections where T? =0, etc.:

Problem 3 - moments of inertia
Known: a
Mark on the figure: centroid, centroidal axes Cx1C, Cx2C and principal axes Cx1p, Cx2p through centroid
Cross-sectional area:
A =
Static moment with respect to Ox1:
Sx1 =
Static moment with respect to Ox2:
Sx2 =
Coordinates of centroid:
x1C =
x2C =
C (x1C, x2C) =
Moments of inertia with respect to Ox1, Ox2:
Jx1 =
Jx2 =
Jx1x2 = Moments of inertia with respect to Cx1C, Cx2C:
Jx1C =
Jx2C =
Jx1Cx2C =
Mohr’s circle:

r =
?0 =
JC =
J1 =
J2 =

Problem 4 – Stress Analysis
Cross-section Shear stress [MPa] Normal stress [MPa]
g ?r/sc ? = ?g + ?r/sc
635144780(mark the position of centroid)
00(mark the position of centroid)
107823093980T?
M?
N?
T?
M?
N?
-64706657785000 55245889000 273054699000 565157556500 Internal forces bending tension/compression total normal stress
1. Determination of the rolled cross-section depth
m = 6 , n = 7 , ? =0.86 , h = 100 ?? = 86
Chosen cross-section depth (as the nearest to calculated h): ………………..
Dimensions and characteristics of the profile (the sketch of your cross-section draw in the table above)
h s b t x2C A Jx3C W3
[mm] [cm2] [cm4] [cm3]
3100331172085b

2. The cross-section load capacity 3. Internal forces
Symbol stalikdMmax? N?? T?? M??
[MPa] [MPa] [kN] [kN] [kNcm]
Allowable stress (tension/compression) kd [MPa], bending moment M?? =0,3 (-1)n+m W3 kd [kNcm], normal force N??=0,4 (-1)nA kd [kN] , shear force: T?? =0,5(-1)m N?? [kN] .
Attach calculations of the cross-section load capacity in the case of pure bending (Mmax-?), as well as for complex loading, N???????T???????M?????,calculations of normal stress ?max, ?min, ?r/c ,??max????r/c, ??max????r/c and shear stress  ?C , ?K , ?K’ , across the given cross-section. Calculations should be done with the use of units [kN] and [cm], result obtained in [kN/m2] should be converted into [MPa].

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