EA P L The PVD applied to this case is dv= pi a1=L P证 da ld dx EA,-元1 EA dindar,- pa The second term on the left hand side is zero because we have asked that i1=0 at the support. Note we have not asked for any condition on in at 1=L where the load is applied Ea li,dx The only way this expression can be satisfied for any admissible virtual dis- placement field in is if: P=EA EA 0 which represent the equilibrium conditions at the boundary and inside the bar, respectively P=AE E d 3� � � L P E, A x1 The PVD applied to this case is: du¯1 � σ11 dV = Pu¯1� V dx1 x1=L � L � du1 du¯1 � A E dx1 = Pu¯1� 0 dx1 dx1 x1=L � L� d �du1 � d2u1 � � � EA ¯ ¯ u1� dx1 u1 − dx2 u1 dx1 = P ¯ 0 dx1 1 x1=L � du1 � � du1 � � L d2u1 � � EAdx1 u1 − EAdx1 u1 − EA dx2 ¯ ¯ u¯1dx1 = Pu¯1� x1=L x1=0 0 1 x1=L The second term on the left hand side is zero because we have asked that u¯1 = 0 at the support. Note we have not asked for any condition on u¯1 at x1 = L where the load is applied. � du1 � � � � � � L d2u1 EAdx1 � x1=L − P u1� = EA dx2 ¯ u¯1dx1 x1=L 0 1 The only way this expression can be satisfied for any admissible virtual dis￾placement field u¯1 is if: du1 � P = EA � dx1 x1=L and d2u1 EA = 0 dx2 1 which represent the equilibrium conditions at the boundary and inside the bar, respectively: � du1 �� � � � P = A E � = Aσ11� dx1 x1=L x1=L and d � du1 � d EA = σ11 = 0 dx1 dx1 dx1 3