Copper

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

45 formulas

Reference page

~60 min

All Mechanics of Materials lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

a SPECIFIC VALUES MAY VARY FOR A PARTICULAR MATERIAL DUE TO ALLOY OR MINERAL COMPOSITION, MECHANICAL WORKING OF THE SPECIMEN, OR HEAT TREATMENT. FOR A MORE EXACT VALUE REFERENCE
BOOKS FOR THE MATERIAL SHOULD BE CONSULTED.
THE YIELD AND ULTIMATE STRENGTHS FOR DUCTILE MATERIALS CAN BE ASSUMED EQUAL FOR BOTH TENSION AND COMPRESSION.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Copper

Handbook notes for this section

Core formulas for this FE topic

22θmax⁡=vmax=v=(3L2−4x2)Mmaxatcenterh=42 2 \theta \max = vmax = v= (3L_{2} - 4x_{2}) Mmax ^at cente rh = 422θmax=vmax=v=(3L2​−4x2​)Mmaxatcenterh=4

θmax vmax 0 ≤ x ≤ L/2Needs technical review

θ1=v=v=(L−b−x)\theta_{1} = v = v= (L - b - x )θ1​=v=v=(L−b−x)

6EILx=a6EIL(L−b−a)6EIL6EIL x=a 6EIL (L - b - a ) 6EIL6EILx=a6EIL(L−b−a)6EIL

xPab(L+a)Mmaxatpointofloadi=Lx Pab(L+a) Mmax _at point of load i = LxPab(L+a)Mmaxa​tpointofloadi=L

θ2 = 0≤x≤aNeeds technical review

−M0Lvmax=−M0x- M_{0} L vmax = -M_{0} x−M0​Lvmax=−M0​x

L θ1 =Needs technical review

v=(x2−3Lx+2L2)v= (x 2 - 3 L x + 2L_{2})v=(x2−3Lx+2L2​)

θ1θ2Mmaxatx=0h=M0\theta_{1} \theta_{2} Mmax ^at x = 0 h = M_{0}θ1​θ2​Mmaxatx=0h=M0​

θ2 =Needs technical review

vmax=−wx3wL2vmax = - wx 3 wL_{2}vmax=−wx3wL2​

θ max =Needs technical review

24EIv=(x−2Lx2+L3)Mmaxatcenterh=824EI v= ( x - 2 L x 2 + L 3) Mmax ^at center h = 824EIv=(x−2Lx2+L3)Mmaxatcenterh=8

θ1=vx=L/2=768EIv=16x−24Lx2+9L3)\theta_{1} = v x = L/2 = 768EI v= 16x - 24Lx_{2} + 9L_{3})θ1​=vx=L/2=768EIv=16x−24Lx2​+9L3​)

θ20≤x≤L/2\theta_{2} 0 \le x \le L/2θ2​0≤x≤L/2

θ2=7wL\theta_{2} = 7wLθ2​=7wL

−wLMmaxcatx=8lm=128wL2-wL Mmax c at x = 8 l m = 128 wL_{2}−wLMmaxcatx=8lm=128wL2​

xvmax=−0.006563wLx vmax = - 0.006563 wLxvmax=−0.006563wL

v=384(8x3−24Lx2v = 384 (8x_{3} - 24Lx_{2}v=384(8x3​−24Lx2​

22atx=0.4598LL/2≤x<L2 2 at x = 0.4598L L/2 \le x < L22atx=0.4598LL/2≤x<L

θ1 =Needs technical review

−7w0L3vmax=−0.00652w0L-7w_{0} L_{3} vmax = - 0.00652 w_{0} L−7w0​L3​vmax=−0.00652w0​L

v=4224Mmax⁡datx=n=0v= 4 2 2 4 M \max d at x = n= 0v=4224Mmaxdatx=n=0

θ2=w0L\theta_{2} = w_{0} Lθ2​=w0​L

atx=0.5193Lat x = 0.5193Latx=0.5193L

a b vmax v=Needs technical review

−Pa2−Pa2(3L−a)6EIMmaxatx=0h=Pa-Pa_{2} -Pa_{2} (3L - a) 6EI Mmax ^at x = 0 h = Pa−Pa2​−Pa2​(3L−a)6EIMmaxatx=0h=Pa

x θmax = vmax =Needs technical review

2EI6EI−Px2(−x+3a),forx≤a2EI 6EI -Px_{2} (-x + 3a), for x \le a2EI6EI−Px2​(−x+3a),forx≤a

θmax =Needs technical review

− wL 3 vmax =Needs technical review

xv=(x−4Lx+6L2)Mmaxatx=0h=2x v= (x - 4L x + 6 L_{2}) Mmax ^at x = 0 h = 2xv=(x−4Lx+6L2​)Mmaxatx=0h=2

vmaxM0Lvmax=M0Lvmax M_{0} L vmax = M_{0} LvmaxM0​Lvmax=M0​L

x θmax = Mmax ^at all x h = M 0Needs technical review

M0 EI 2EI v=Needs technical review

v=24EIx2−2Lx+32L2(v = 24EI x 2 - 2L x + 32 L_{2} (v=24EIx2−2Lx+32L2​(

−−0≤x≤L/2- - 0 \le x \le L/2−−0≤x≤L/2

θmax = wL vmax = 7wLNeeds technical review

vmax48EI384EIMmaxatx=0h=8vmax 48EI 384EI Mmax ^at x = 0 h = 8vmax48EI384EIMmaxatx=0h=8

LLv=(4x−L/2)L L v= (4x - L/2)LLv=(4x−L/2)

2 2 θmax L/2 ≤ x ≤ LNeeds technical review

θmax = vmax = Mmax ^at x = 0 h = 06Needs technical review

x24EI30EIv=120EIL(10L3−10L2x+5Lx2−x3)x 24EI 30EI v = 120EIL (10L_{3} - 10 L 2 x + 5 Lx_{2} - x 3)x24EI30EIv=120EIL(10L3​−10L2x+5Lx2​−x3)

ν max ν Mmax (at x = 0) =Needs technical review