A Figura 1 representa duas barras de aço soldadas na secção BB.
Figura 1
Barras soldadas
A carga de tração que atua na peça é 5,0 kN.
A seção 1 da peça possui D1 = 20 mm e comprimento L1= 50mm, sendo que a seção 2 possui D2 = 30 mm e L2 = 90 mm.
Despreze o efeito do peso próprio do material.
Considere:
E subscript A Ç O end subscript space equals space 210 space G P A
nu subscript A Ç O end subscript space equals space 0 comma 3
Aplicando a Lei de Hooke é possível descobrir:
A tensão normal (σ1 e σ2);
O alongamento (∆L1 e ∆L2);
A deformação longitudinal (ε1 e ε2);
A deformação transversal (εt1 e εt2).
Assinale a alternativa que apresenta respectivamente estas características.
Escolha uma:
a.
σ1=16,92MPa ; σ2=8,07MPa ; ∆L1=0,0058mm ; ∆L2=0,0080mm ; ε1=0 comma 076.10 to the power of negative 3 end exponent; ε2=0 comma 033.10 to the power of negative 3 end exponent; εt1=negative space 9 comma 9.10 to the power of negative 6 end exponent; εt2=negative space 2 comma 28.10 to the power of negative 5 end exponent.

b.
σ1=16,92MPa ; σ2=8,07MPa ; ∆L1=0,0038mm ; ∆L2=0,0030mm ; ε1=0 comma 033.10 to the power of negative 3 end exponent; ε2=0 comma 076.10 to the power of negative 3 end exponent; εt1=negative space 2 comma 28.10 to the power of negative 5 end exponent; εt2=negative space 9 comma 9.10 to the power of negative 6 end exponent.

c.
σ1=15,92MPa ; σ2=7,07MPa ; ∆L1=0,0038mm ; ∆L2=0,0030mm ; ε1=0 comma 076.10 to the power of negative 3 end exponent; ε2=0 comma 033.10 to the power of negative 3 end exponent; εt1= negative space 2 comma 28.10 to the power of negative 5 end exponent ; εt2=negative space 9 comma 9.10 to the power of negative 6 end exponent.

d.
σ1=15,92MPa ; σ2=7,07MPa ; ∆L1=0,0038mm ; ∆L2=0,0030mm ; ε1=0 comma 033.10 to the power of negative 3 end exponent; ε2=0 comma 076.10 to the power of negative 3 end exponent; εt1=negative space 9 comma 9.10 to the power of negative 6 end exponent; εt2=negative space 2 comma 28.10 to the power of negative 5 end exponent.

e.
σ1=16,92MPa ; σ2=8,07MPa ; ∆L1=0,0058mm ; ∆L2=0,0080mm ; ε1=0 comma 076.10 to the power of negative 3 end exponent; ε2=0 comma 033.10 to the power of negative 3 end exponent; εt1=negative space 2 comma 28.10 to the power of negative 5 end exponent; εt2=negative space 9 comma 9.10 to the power of negative 6 end exponent.