SS-V: 5030 Reactions at the Ends of Axially Loaded Plastic Bar

Test No. VNL04 Find reactions at the fixed ends and maximum displacement of a bar axially loaded beyond plasticity.

Definition



Figure 1.

Bar dimensions are 10 x 10 x 200 mm. Distance between loaded point and left end A=50 mm. Strain-stress curve of the bar material is defined by the power law:

σ=Kεn

Where,
K
Strength coefficient
n
Must be in the range [0,1]
n =0
Material is perfectly plastic.
n =1
Material is elastic.
The material properties are:
Properties
Value
K
530 MPa
n
0.26
Poisson's Ratio
0


Figure 2. Corresponded strain-stress curve

The study was performed for the following load F values: 30000 N, 47000 N, 55000 N, and 60000 N. These loads cover the full range of elastic-plastic response of the bar.

Reference Solution

One-dimensional analytical reference solution is described here.

The length of the bar does not change under the load.

A∫0ε1dx 

or,

∫ 0 A N / ( K * A ) n d x   −   ∫ 0 L − A ( F − N ) / ( K * A ) n d x   =   0 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape WaaybCaeqal8aabaWdbiaaicdaa8aabaWdbiaadgeaa0WdaeaapeGa ey4kIipaaOWaaOqaa8aabaWdbiaad6eacaGGVaWaaeWaa8aabaWdbi aadUeacaGGQaGaamyqaaGaayjkaiaawMcaaaWcpaqaa8qacaWGUbaa aOGaamizaiaadIhacaGGGcGaeyOeI0IaaiiOamaawahabeWcpaqaa8 qacaaIWaaapaqaa8qacaWGmbGaeyOeI0Iaamyqaaqdpaqaa8qacqGH RiI8aaGcdaGcbaWdaeaapeWaaeWaa8aabaWdbiaadAeacqGHsislca WGobaacaGLOaGaayzkaaGaai4lamaabmaapaqaa8qacaWGlbGaaiOk aiaadgeaaiaawIcacaGLPaaaaSWdaeaapeGaamOBaaaakiaadsgaca WG4bGaaiiOaiabg2da9iaacckacaaIWaaaaa@5C73@

Where,
ε 1 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaeqyTduMaaGymaaaa@386E@
Tensile strain at the left span of the bar.
ε 2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaeqyTduMaaGymaaaa@386E@
Compressive strain at the right span of the bar,
N MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqefqvATv2CG4uz3bIuV1wyUbqedmvETj2BSbqefm0B1jxALjhi ov2DaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaacaGacmGadaWaaiqacaabaiaafaaake aacaWGlbaaaa@3987@
Reaction force at left end of the bar.
R = F − N MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqefqvATv2CG4uz3bIuV1wyUbqedmvETj2BSbqefm0B1jxALjhi ov2DaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaacaGacmGadaWaaiqacaabaiaafaaake aacaWGsbGaeyypa0JaamOraiabgkHiTiaad6eaaaa@3D1F@
Reaction force at the right end of the bar.
A MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqefqvATv2CG4uz3bIuV1wyUbqedmvETj2BSbqefm0B1jxALjhi ov2DaebbnrfifHhDYfgasaacH8qrps0lbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaacaGacmGadaWaaiqacaabaiaafaaake aacaWGlbaaaa@3987@
Bar cross-section area

From this equation you can find the reaction at the left end of the bar.

N = F/( 1+ ( a/b ) n ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaamOtaiaacckacqGH9aqpcaGGGcGaamOraiaac+cadaqadaWdaeaa peGaaGymaiabgUcaRmaabmaapaqaa8qacaWGHbGaai4laiaadkgaai aawIcacaGLPaaapaWaaWbaaSqabeaapeGaamOBaaaaaOGaayjkaiaa wMcaaaaa@4461@

and R   =   F − N MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaabaaaaaaaaape GaamOuaiaacckacqGH9aqpcaGGGcGaamOraiabgkHiTiaad6eaaaa@3CBC@ at the right end.

Results

Bar was modeled as a 3D solid with immovable ends. Axial force F could not be applied precisely at the solid bar axis, so four line spots were created at the bar sides and total load F was uniformly distributed over the spots (Figure 3).
Figure 3.

The following table summarizes the reaction force results.
Force F [N] SOL Reference, Reaction [N] SimSolid, Reaction [N] % Difference
30000 17128 17522 2.3%
47000 26834 26987 0.5%
55000 31401 31660 0.8%
60000 34256 34490 0.6%